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\input{preamble}
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\begin{document}
\title{Topology}
\maketitle
\phantomsection
\label{section-phantom}
\tableofcontents
\section{Introduction}
\label{section-introduction}
\noindent
Basic topology will be explained in this document.
A reference is \cite{Engelking}.
\section{Basic notions}
\label{section-topology-basic}
\noindent
The following is a list of basic notions in topology. Some of these notions
are discussed in more detail in the text that follows and some
are defined in the list, but others are considered basic and will not
be defined. If you are not familiar with most of the italicized concepts,
then we suggest looking at an introductory text on topology before
continuing.
\begin{enumerate}
\item
\label{item-space}
$X$ is a {\it topological space},
\item
\label{item-point}
$x\in X$ is a {\it point},
\item
\label{item-locally-closed}
$E \subset X$ is a {\it locally closed} subset,
\item
\label{item-closed-point}
$x\in X$ is a {\it closed point},
\item
\label{item-dense}
$E \subset X$ is a {\it dense} subset,
\item
\label{item-continuous}
$f : X_1 \to X_2$ is {\it continuous},
\item
\label{item-upper-semi-continuous}
an extended real function $f : X \to \mathbf{R} \cup \{\infty, -\infty\}$
is {\it upper semi-continuous} if $\{x \in X \mid f(x) < a\}$ is open for
all $a \in \mathbf{R}$,
\item
\label{item-lower-semi-continuous}
an extended real function $f : X \to \mathbf{R} \cup \{\infty, -\infty\}$
is {\it lower semi-continuous} if $\{x \in X \mid f(x) > a\}$ is open for
all $a \in \mathbf{R}$,
\item a continuous map of spaces $f : X \to Y$ is
{\it open} if $f(U)$ is open in $Y$ for $U \subset X$ open,
\item a continuous map of spaces $f : X \to Y$ is
{\it closed} if $f(Z)$ is closed in $Y$ for $Z \subset X$ closed,
\item
\label{item-neighbourhood}
a {\it neighbourhood of $x \in X$} is any subset
$E \subset X$ which contains an open subset that
contains $x$,
\item
\label{item-induced-topology}
the {\it induced topology} on a subset $E \subset X$,
\item
\label{item-covering}
$\mathcal{U} : U = \bigcup_{i \in I} U_i$ is an
{\it open covering of} $U$ (note: we allow any $U_i$ to be empty
and we even allow, in case $U$ is empty, the empty set for $I$),
\item
\label{item-refinement}
the open covering $\mathcal{V}$ is a {\it refinement}
of the open covering $\mathcal{U}$ (if
$\mathcal{V} : V = \bigcup_{j \in J} V_j$ and
$\mathcal{U} : U = \bigcup_{i \in I} U_i$
this means each $V_j$ is completely contained in one of the $U_i$),
\item
\label{item-fundamental-system}
{\it $\{ E_i \}_{i \in I}$ is a fundamental system of neighbourhoods
of $x$ in $X$},
\item
\label{item-Hausdorff}
a topological space $X$ is called {\it Hausdorff} or {\it separated}
if and only if for every distinct pair of points $x, y \in X$ there exist
disjoint opens $U, V \subset X$ such that $x \in U$, $y \in V$,
\item the {\it product} of two topological spaces,
\label{item-product}
\item
\label{item-fibre-product}
the {\it fibre product $X \times_Y Z$} of a pair of continuous maps
$f : X \to Y$ and $g : Z \to Y$,
\item
\label{item-discrete-indiscrete}
the {\it discrete topology} and the {\it indiscrete topology} on a set,
\item etc.
\end{enumerate}
\section{Hausdorff spaces}
\label{section-Hausdorff}
\noindent
The category of topological spaces has finite products.
\begin{lemma}
\label{lemma-Hausdorff}
Let $X$ be a topological space. The following are equivalent:
\begin{enumerate}
\item $X$ is Hausdorff,
\item the diagonal $\Delta(X) \subset X \times X$ is closed.
\end{enumerate}
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{lemma}
\label{lemma-graph-closed}
\begin{slogan}
Graphs of maps to Hausdorff spaces are closed.
\end{slogan}
Let $f : X \to Y$ be a continuous map of topological spaces.
If $Y$ is Hausdorff, then the graph of $f$ is closed in $X \times Y$.
\end{lemma}
\begin{proof}
The graph is the inverse image of the diagonal under the map
$X \times Y \to Y \times Y$. Thus the lemma follows from
Lemma \ref{lemma-Hausdorff}.
\end{proof}
\begin{lemma}
\label{lemma-section-closed}
Let $f : X \to Y$ be a continuous map of topological spaces.
Let $s : Y \to X$ be a continuous map such that $f \circ s = \text{id}_Y$.
If $X$ is Hausdorff, then $s(Y)$ is closed.
\end{lemma}
\begin{proof}
This follows from Lemma \ref{lemma-Hausdorff} as
$s(Y) = \{x \in X \mid x = s(f(x))\}$.
\end{proof}
\begin{lemma}
\label{lemma-fibre-product-closed}
Let $X \to Z$ and $Y \to Z$ be continuous maps of topological spaces.
If $Z$ is Hausdorff, then $X \times_Z Y$ is closed in $X \times Y$.
\end{lemma}
\begin{proof}
This follows from Lemma \ref{lemma-Hausdorff} as
$X \times_Z Y$ is the inverse image of $\Delta(Z)$
under $X \times Y \to Z \times Z$.
\end{proof}
\section{Separated maps}
\label{section-separated}
\noindent
Just the definition and some simple lemmas.
\begin{definition}
\label{definition-separated}
A continuous map $f : X \to Y$ of topological spaces is called
{\it separated} if and only if the diagonal $\Delta : X \to X \times_Y X$
is a closed map.
\end{definition}
\begin{lemma}
\label{lemma-separated}
Let $f : X \to Y$ be continuous map of topological spaces.
The following are equivalent:
\begin{enumerate}
\item $f$ is separated,
\item $\Delta(X) \subset X \times_Y X$ is a closed subset,
\item given distinct points $x, x' \in X$ mapping to the same point of
$Y$, there exist disjoint open neighbourhoods of $x$ and $x'$.
\end{enumerate}
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{lemma}
\label{lemma-from-hausdorff}
Let $f : X \to Y$ be continuous map of topological spaces.
If $X$ is Hausdorff, then $f$ is separated.
\end{lemma}
\begin{proof}
Clear from Lemma \ref{lemma-separated}.
\end{proof}
\begin{lemma}
\label{lemma-base-change-separated}
Let $f : X \to Y$ and $Z \to Y$ be continuous maps of topological spaces.
If $f$ is separated, then $f' : Z \times_Y X \to Z$ is separated.
\end{lemma}
\begin{proof}
Follows from characterization (3) of Lemma \ref{lemma-separated}.
\end{proof}
\section{Bases}
\label{section-bases}
\noindent
Basic material on bases for topological spaces.
\begin{definition}
\label{definition-base}
Let $X$ be a topological space. A collection of subsets $\mathcal{B}$ of $X$
is called a {\it base for the topology on $X$} or a {\it basis for the
topology on $X$} if the following conditions hold:
\begin{enumerate}
\item Every element $B \in \mathcal{B}$ is open in $X$.
\item For every open $U \subset X$ and every $x \in U$,
there exists an element $B \in \mathcal{B}$ such that
$x \in B \subset U$.
\end{enumerate}
\end{definition}
\noindent
The following lemma is sometimes used to define a topology.
\begin{lemma}
\label{lemma-make-base}
Let $X$ be a set and let $\mathcal{B}$ be a collection of subsets.
Assume that $X = \bigcup_{B \in \mathcal{B}} B$ and that given
$x \in B_1 \cap B_2$ with $B_1, B_2 \in \mathcal{B}$ there is a
$B_3 \in \mathcal{B}$ with $x \in B_3 \subset B_1 \cap B_2$.
Then there is a unique topology on $X$ such that $\mathcal{B}$
is a basis for this topology.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{lemma}
\label{lemma-refine-covering-basis}
Let $X$ be a topological space.
Let $\mathcal{B}$ be a basis for the topology on $X$.
Let $\mathcal{U} : U = \bigcup_i U_i$ be an open covering of
$U \subset X$. There exists an open covering $U = \bigcup V_j$
which is a refinement of $\mathcal{U}$ such that each
$V_j$ is an element of the basis $\mathcal{B}$.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{definition}
\label{definition-subbase}
Let $X$ be a topological space. A collection of subsets $\mathcal{B}$ of $X$
is called a {\it subbase for the topology on $X$} or a {\it subbasis for the
topology on $X$} if the finite intersections of
elements of $\mathcal{B}$ form a basis for the topology on $X$.
\end{definition}
\noindent
In particular every element of $\mathcal{B}$ is open.
\begin{lemma}
\label{lemma-subbase}
Let $X$ be a set. Given any collection $\mathcal{B}$ of subsets of $X$
there is a unique topology on $X$ such that $\mathcal{B}$ is a subbase
for this topology.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{lemma}
\label{lemma-create-map-from-subcollection}
Let $X$ be a topological space. Let $\mathcal{B}$ be a collection
of opens of $X$. Assume $X = \bigcup_{U \in \mathcal{B}} U$ and
for $U, V \in \mathcal{B}$ we have
$U \cap V = \bigcup_{W \in \mathcal{B}, W \subset U \cap V} W$,
Then there is a continuous map $f : X \to Y$ of topological spaces
such that
\begin{enumerate}
\item for $U \in \mathcal{B}$ the image $f(U)$ is open,
\item for $U \in \mathcal{B}$ we have $f^{-1}(f(U)) = U$, and
\item the opens $f(U)$, $U \in \mathcal{B}$
form a basis for the topology on $Y$.
\end{enumerate}
\end{lemma}
\begin{proof}
Define an equivalence relation $\sim$ on points of $X$
by the rule
$$
x \sim y \Leftrightarrow
(\forall U \in \mathcal{B} : x \in U \Leftrightarrow y \in U)
$$
Let $Y$ be the set of equivalence classes and $f : X \to Y$
the natural map. Part (2) holds by construction.
The assumptions on $\mathcal{B}$ exactly
mirror the assumptions in Lemma \ref{lemma-make-base}
on the set of subsets $f(U)$, $U \in \mathcal{B}$.
Hence there is a unique topology on $Y$ such that (3) holds.
Then (1) is clear as well.
\end{proof}
\section{Submersive maps}
\label{section-submersive}
\noindent
If $X$ is a topological space and $E \subset X$ is a subset, then
we usually endow $E$ with the {\it induced topology}.
\begin{lemma}
\label{lemma-induced}
Let $X$ be a topological space. Let $Y$ be a set and let
$f : Y \to X$ be an injective map of sets. The induced
topology on $Y$ is the topology characterized by
each of the following statements:
\begin{enumerate}
\item it is the weakest topology on $Y$ such that $f$ is continuous,
\item the open subsets of $Y$ are $f^{-1}(U)$ for $U \subset X$ open,
\item the closed subsets of $Y$ are the sets $f^{-1}(Z)$ for $Z \subset X$
closed.
\end{enumerate}
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\noindent
Dually, if $X$ is a topological space and $X \to Y$ is a surjection of
sets, then $Y$ can be endowed with the {\it quotient topology}.
\begin{lemma}
\label{lemma-quotient}
Let $X$ be a topological space. Let $Y$ be a set and let $f : X \to Y$
be a surjective map of sets. The quotient topology on $Y$ is the
topology characterized by each of the following statements:
\begin{enumerate}
\item it is the strongest topology on $Y$ such that $f$ is continuous,
\item a subset $V$ of $Y$ is open if and only if $f^{-1}(V)$ is open,
\item a subset $Z$ of $Y$ is closed if and only if $f^{-1}(Z)$ is closed.
\end{enumerate}
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\noindent
Let $f : X \to Y$ be a continuous map of topological spaces.
In this case we obtain a factorization $X \to f(X) \to Y$
of maps of sets. We can endow $f(X)$ with the
quotient topology coming from the surjection $X \to f(X)$
or with the induced topology coming from the injection $f(X) \to Y$.
The map
$$
(f(X), \text{quotient topology})
\longrightarrow
(f(X), \text{induced topology})
$$
is continuous.
\begin{definition}
\label{definition-submersive}
Let $f : X \to Y$ be a continuous map of topological spaces.
\begin{enumerate}
\item We say $f$ is a {\it strict map of topological spaces}
if the induced topology and the quotient topology on $f(X)$ agree
(see discussion above).
\item We say $f$ is {\it submersive}\footnote{This is very different from
the notion of a submersion between differential manifolds! It is probably
a good idea to use ``strict and surjective'' in stead of ``submersive''.}
if $f$ is surjective and strict.
\end{enumerate}
\end{definition}
\noindent
Thus a continuous map $f : X \to Y$ is submersive if $f$
is a surjection and for any $T \subset Y$ we have
$T$ is open or closed if and only if $f^{-1}(T)$ is so.
In other words, $Y$ has the
quotient topology relative to the surjection $X \to Y$.
\begin{lemma}
\label{lemma-open-morphism-quotient-topology}
Let $f : X \to Y$ be surjective, open, continuous map of topological spaces.
Let $T \subset Y$ be a subset. Then
\begin{enumerate}
\item $f^{-1}(\overline{T}) = \overline{f^{-1}(T)}$,
\item $T \subset Y$ is closed if and only $f^{-1}(T)$ is closed,
\item $T \subset Y$ is open if and only $f^{-1}(T)$ is open, and
\item $T \subset Y$ is locally closed if and only $f^{-1}(T)$ is locally closed.
\end{enumerate}
In particular we see that $f$ is submersive.
\end{lemma}
\begin{proof}
It is clear that $\overline{f^{-1}(T)} \subset f^{-1}(\overline{T})$.
If $x \in X$, and $x \not \in \overline{f^{-1}(T)}$, then there
exists an open neighbourhood $x \in U \subset X$ with
$U \cap f^{-1}(T) = \emptyset$. Since $f$ is open we see that
$f(U)$ is an open neighbourhood of $f(x)$ not meeting $T$.
Hence $x \not \in f^{-1}(\overline{T})$. This proves (1).
Part (2) is an easy consequence of (1).
Part (3) is obvious from the fact that $f$ is open and surjective.
For (4), if $f^{-1}(T)$ is locally closed, then
$f^{-1}(T) \subset \overline{f^{-1}(T)} = f^{-1}(\overline{T})$
is open, and hence by (3) applied to the map
$f^{-1}(\overline{T}) \to \overline{T}$ we see that
$T$ is open in $\overline{T}$, i.e., $T$ is locally closed.
\end{proof}
\begin{lemma}
\label{lemma-closed-morphism-quotient-topology}
Let $f : X \to Y$ be surjective, closed, continuous map of topological spaces.
Let $T \subset Y$ be a subset. Then
\begin{enumerate}
\item $\overline{T} = f(\overline{f^{-1}(T)})$,
\item $T \subset Y$ is closed if and only $f^{-1}(T)$ is closed,
\item $T \subset Y$ is open if and only $f^{-1}(T)$ is open, and
\item $T \subset Y$ is locally closed if and only $f^{-1}(T)$ is locally closed.
\end{enumerate}
In particular we see that $f$ is submersive.
\end{lemma}
\begin{proof}
It is clear that $\overline{f^{-1}(T)} \subset f^{-1}(\overline{T})$.
Then $T \subset f(\overline{f^{-1}(T)}) \subset \overline{T}$
is a closed subset, hence we get (1). Part (2) is obvious from
the fact that $f$ is closed and surjective.
Part (3) follows from (2) applied to the complement of $T$.
For (4), if $f^{-1}(T)$ is locally closed, then
$f^{-1}(T) \subset \overline{f^{-1}(T)}$ is open.
Since the map $\overline{f^{-1}(T)} \to \overline{T}$ is surjective
by (1) we can apply part (3) to the map $\overline{f^{-1}(T)} \to \overline{T}$
induced by $f$ to conclude that $T$ is open in
$\overline{T}$, i.e., $T$ is locally closed.
\end{proof}
\section{Connected components}
\label{section-connected-components}
\begin{definition}
\label{definition-connected-components}
Let $X$ be a topological space.
\begin{enumerate}
\item We say $X$ is {\it connected} if $X$ is not empty and whenever
$X = T_1 \amalg T_2$ with $T_i \subset X$ open and closed, then either
$T_1 = \emptyset$ or $T_2 = \emptyset$.
\item We say $T \subset X$ is a {\it connected component} of $X$ if
$T$ is a maximal connected subset of $X$.
\end{enumerate}
\end{definition}
\noindent
The empty space is not connected.
\begin{lemma}
\label{lemma-image-connected-space}
Let $f : X \to Y$ be a continuous map of topological spaces.
If $E \subset X$ is a connected subset, then $f(E) \subset Y$
is connected as well.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{lemma}
\label{lemma-connected-components}
Let $X$ be a topological space.
\begin{enumerate}
\item If $T \subset X$ is connected, then so is its closure.
\item Any connected component of $X$ is closed (but not necessarily open).
\item Every connected subset of $X$ is contained in a connected
component of $X$.
\item Every point of $X$ is contained in a connected component, in other
words, $X$ is the union of its connected components.
\end{enumerate}
\end{lemma}
\begin{proof}
Let $\overline{T}$ be the closure of the connected subset $T$.
Suppose $\overline{T} = T_1 \amalg T_2$ with $T_i \subset \overline{T}$
open and closed. Then $T = (T\cap T_1) \amalg (T \cap T_2)$. Hence
$T$ equals one of the two, say $T = T_1 \cap T$. Thus clearly
$\overline{T} \subset T_1$ as desired.
\medskip\noindent
Pick a point $x\in X$. Consider the set $A$ of connected subsets
$x \in T_\alpha \subset X$. Note that $A$ is nonempty since
$\{x\} \in A$. There is a partial ordering on $A$ coming from
inclusion: $\alpha \leq \alpha' \Leftrightarrow T_\alpha \subset T_{\alpha'}$.
Choose a maximal totally ordered subset $A' \subset A$, and let
$T = \bigcup_{\alpha \in A'} T_\alpha$. We claim that $T$ is
connected. Namely, suppose that $T = T_1 \amalg T_2$ is a disjoint
union of two open and closed subsets of $T$.
For each $\alpha \in A'$ we have either $T_\alpha \subset T_1$
or $T_\alpha \subset T_2$, by connectedness of $T_\alpha$.
Suppose that for some $\alpha_0 \in A'$ we have
$T_{\alpha_0} \not\subset T_1$ (say, if not we're done anyway).
Then, since $A'$ is totally ordered we see immediately that
$T_\alpha \subset T_2$ for all $\alpha \in A'$. Hence $T = T_2$.
\medskip\noindent
To get an example where connected components are not open, just take
an infinite product $\prod_{n \in \mathbf{N}} \{0, 1\}$
with the product topology. Its connected components are singletons,
which are not open.
\end{proof}
\begin{lemma}
\label{lemma-connected-fibres-quotient-topology-connected-components}
Let $f : X \to Y$ be a continuous map of topological spaces.
Assume that
\begin{enumerate}
\item all fibres of $f$ are connected, and
\item a set $T \subset Y$ is closed if and only if $f^{-1}(T)$ is closed.
\end{enumerate}
Then $f$ induces a bijection between the sets of connected
components of $X$ and $Y$.
\end{lemma}
\begin{proof}
Let $T \subset Y$ be a connected component.
Note that $T$ is closed, see Lemma \ref{lemma-connected-components}.
The lemma follows if we show that $f^{-1}(T)$ is connected
because any connected subset of $X$ maps into a connected component
of $Y$ by Lemma \ref{lemma-image-connected-space}.
Suppose that $f^{-1}(T) = Z_1 \amalg Z_2$
with $Z_1$, $Z_2$ closed. For any $t \in T$ we see that
$f^{-1}(\{t\}) = Z_1 \cap f^{-1}(\{t\}) \amalg Z_2 \cap f^{-1}(\{t\})$.
By (1) we see $f^{-1}(\{t\})$ is connected we conclude that
either $f^{-1}(\{t\}) \subset Z_1$ or $f^{-1}(\{t\}) \subset Z_2$.
In other words $T = T_1 \amalg T_2$ with $f^{-1}(T_i) = Z_i$.
By (2) we conclude that $T_i$ is closed in $Y$.
Hence either $T_1 = \emptyset$ or $T_2 = \emptyset$ as desired.
\end{proof}
\begin{lemma}
\label{lemma-connected-fibres-connected-components}
Let $f : X \to Y$ be a continuous map of topological spaces.
Assume that
(a) $f$ is open,
(b) all fibres of $f$ are connected.
Then $f$ induces a bijection between the sets of connected
components of $X$ and $Y$.
\end{lemma}
\begin{proof}
This is a special case of
Lemma \ref{lemma-connected-fibres-quotient-topology-connected-components}.
\end{proof}
\begin{lemma}
\label{lemma-finite-fibre-connected-components}
Let $f : X \to Y$ be a continuous map of nonempty topological spaces. Assume
that
(a) $Y$ is connected,
(b) $f$ is open and closed, and
(c) there is a point $y\in Y$ such that the fiber $f^{-1}(y)$ is a finite set.
Then $X$ has at most $|f^{-1}(y)|$ connected components. Hence any connected
component $T$ of $X$ is open and closed, and $p(T)$ is a nonempty open and
closed subset of $Y$, which is therefore equal to $Y$.
\end{lemma}
\begin{proof}
If the topological space $X$ has at least $N$ connected components for some
$N \in \mathbf{N}$, we find by induction a decomposition
$X = X_1 \amalg \ldots \amalg X_N$ of $X$ as a disjoint union of $N$ nonempty
open and closed subsets $X_1, \ldots , X_N$ of $X$. As $f$ is open and closed,
each $f(X_i)$ is a nonempty open and closed subset of $Y$ and is hence equal to
$Y$. In particular the intersection $X_i \cap f^{-1}(y)$ is nonempty for each
$1 \leq i \leq N$. Hence $f^{-1}(y)$ has at least $N$ elements.
\end{proof}
\begin{definition}
\label{definition-totally-disconnected}
A topological space is {\it totally disconnected} if the connected components
are all singletons.
\end{definition}
\noindent
A discrete space is totally disconnected.
A totally disconnected space need not be discrete, for example
$\mathbf{Q} \subset \mathbf{R}$ is totally disconnected but not discrete.
\begin{lemma}
\label{lemma-space-connected-components}
Let $X$ be a topological space. Let $\pi_0(X)$ be the set of connected
components of $X$. Let $X \to \pi_0(X)$ be the map which sends
$x \in X$ to the connected component of $X$ passing through $x$.
Endow $\pi_0(X)$ with the quotient topology. Then $\pi_0(X)$ is a
totally disconnected space and any continuous map $X \to Y$
from $X$ to a totally disconnected space $Y$ factors through $\pi_0(X)$.
\end{lemma}
\begin{proof}
By Lemma
\ref{lemma-connected-fibres-quotient-topology-connected-components}
the connected components of $\pi_0(X)$ are the singletons.
We omit the proof of the second statement.
\end{proof}
\begin{definition}
\label{definition-locally-connected}
A topological space $X$ is called {\it locally connected} if
every point $x \in X$ has a fundamental system of connected neighbourhoods.
\end{definition}
\begin{lemma}
\label{lemma-locally-connected}
Let $X$ be a topological space. If $X$ is locally connected, then
\begin{enumerate}
\item any open subset of $X$ is locally connected, and
\item the connected components of $X$ are open.
\end{enumerate}
So also the connected components of open subsets of $X$ are open.
In particular, every point has a fundamental system of open connected
neighbourhoods.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\section{Irreducible components}
\label{section-irreducible-components}
\begin{definition}
\label{definition-irreducible-components}
Let $X$ be a topological space.
\begin{enumerate}
\item We say $X$ is {\it irreducible}, if $X$ is not empty, and whenever
$X = Z_1 \cup Z_2$ with $Z_i$ closed, we have $X = Z_1$ or $X = Z_2$.
\item We say $Z \subset X$ is an {\it irreducible component} of $X$
if $Z$ is a maximal irreducible subset of $X$.
\end{enumerate}
\end{definition}
\noindent
An irreducible space is obviously connected.
\begin{lemma}
\label{lemma-image-irreducible-space}
Let $f : X \to Y$ be a continuous map of topological spaces.
If $E \subset X$ is an irreducible subset, then $f(E) \subset Y$
is irreducible as well.
\end{lemma}
\begin{proof}
Suppose $f(E)$ is the union of $Z_1 \cap f(E)$ and $Z_2 \cap f(E)$, for two
distinct closed subsets $Z_1$ and $Z_2$ of $Y$; this is equal to the
intersection $(Z_1 \cup Z_2) \cap f(E)$, so $f(E)$ is then contained in the
union $Z_1 \cup Z_2$. For the irreducibility of $f(E)$ it suffices to show
that it is contained in either $Z_1$ or $Z_2$. The relation
$f(E) \subset Z_1 \cup Z_2$ shows that
$f^{-1}(f(E)) \subset f^{-1}(Z_1 \cup Z_2)$; as the right-hand side is
clearly equal to $f^{-1}(Z_1) \cup f^{-1}(Z_2)$ and since
$E \subset f^{-1}(f(E))$, it follows that
$E \subset f^{-1}(Z_1) \cup f^{-1}(Z_2)$, from which one concludes by the
irreducibility of $E$ that $E \subset f^{-1}(Z_1)$ or
$E \subset f^{-1}(Z_2)$. Hence one sees that either
$f(E) \subset f(f^{-1}(Z_1)) \subset Z_1$ or $f(E) \subset Z_2$.
\end{proof}
\begin{lemma}
\label{lemma-irreducible}
Let $X$ be a topological space.
\begin{enumerate}
\item If $T \subset X$ is irreducible so is its closure in $X$.
\item Any irreducible component of $X$ is closed.
\item Any irreducible subset of $X$ is contained in an
irreducible component of $X$.
\item Every point of $X$ is contained in some irreducible component
of $X$, in other words, $X$ is the union of its irreducible components.
\end{enumerate}
\end{lemma}
\begin{proof}
Let $\overline{T}$ be the closure of the irreducible subset $T$.
If $\overline{T} = Z_1 \cup Z_2$ with $Z_i \subset \overline{T}$
closed, then $T = (T\cap Z_1) \cup (T \cap Z_2)$ and hence
$T$ equals one of the two, say $T = Z_1 \cap T$. Thus clearly
$\overline{T} \subset Z_1$. This proves (1). Part (2) follows
immediately from (1) and the definition of irreducible components.
\medskip\noindent
Let $T \subset X$ be irreducible. Consider the set $A$ of irreducible subsets
$T \subset T_\alpha \subset X$. Note that $A$ is nonempty since
$T \in A$. There is a partial ordering on $A$ coming from
inclusion: $\alpha \leq \alpha' \Leftrightarrow T_\alpha \subset T_{\alpha'}$.
Choose a maximal totally ordered subset $A' \subset A$, and let
$T' = \bigcup_{\alpha \in A'} T_\alpha$. We claim that $T'$ is
irreducible. Namely, suppose that $T' = Z_1 \cup Z_2$ is a union
of two closed subsets of $T'$. For each $\alpha \in A'$ we have
either $T_\alpha \subset Z_1$ or $T_\alpha \subset Z_2$, by irreducibility
of $T_\alpha$. Suppose that for some $\alpha_0 \in A'$ we have
$T_{\alpha_0} \not\subset Z_1$ (say, if not we're done anyway).
Then, since $A'$ is totally ordered we see immediately that
$T_\alpha \subset Z_2$ for all $\alpha \in A'$. Hence $T' = Z_2$.
This proves (3). Part (4) is an immediate consequence of (3)
as a singleton space is irreducible.
\end{proof}
\noindent
A singleton is irreducible. Thus if $x \in X$ is a point
then the closure $\overline{\{x\}}$ is an irreducible closed
subset of $X$.
\begin{definition}
\label{definition-generic-point}
Let $X$ be a topological space.
\begin{enumerate}
\item Let $Z \subset X$ be an irreducible closed subset.
A {\it generic point} of $Z$ is a point $\xi \in Z$ such
that $Z = \overline{\{\xi\}}$.
\item The space $X$ is called {\it Kolmogorov}, if for every $x, x' \in X$,
$x \not = x'$ there exists a closed subset of $X$ which contains
exactly one of the two points.
\item The space $X$ is called {\it quasi-sober} if every
irreducible closed subset has a generic point.
\item The space $X$ is called {\it sober} if every
irreducible closed subset has a unique generic point.
\end{enumerate}
\end{definition}
\noindent
A topological space $X$ is Kolmogorov, quasi-sober, resp.\ sober if and
only if the map $x\mapsto\overline{\{x\}}$ from $X$ to the set of
irreducible closed subsets of $X$ is injective, surjective, resp.\ bijective.
Hence we see that a topological space is sober if and only if it is
quasi-sober and Kolmogorov.
\begin{lemma}
\label{lemma-sober-subspace}
Let $X$ be a topological space and let $Y\subset X$.
\begin{enumerate}
\item If $X$ is Kolmogorov then so is $Y$.
\item Suppose $Y$ is locally closed in $X$. If $X$ is quasi-sober then
so is $Y$.
\item Suppose $Y$ is locally closed in $X$. If $X$ is sober then so is $Y$.
\end{enumerate}
\end{lemma}
\begin{proof}
Proof of (1). Suppose $X$ is Kolmogorov. Let $x,y\in X$ with $x\neq y$. Then
$\overline{\overline{\{x\}}\cap Y}=\overline{\{x\}}\neq\overline{\{y\}}=
\overline{\overline{\{y\}}\cap Y}$. Hence
$\overline{\{x\}}\cap Y\neq\overline{\{y\}}\cap Y$. This shows that $Y$ is
Kolmogorov.
\medskip\noindent
Proof of (2). Suppose $X$ is quasi-sober. It suffices to consider the
cases $Y$ is closed and $Y$ is open. First, suppose $Y$ is closed. Let
$Z$ be an irreducible closed subset of $Y$. Then $Z$ is an irreducible closed
subset of $X$. Hence there exists $x\in Y$ with $\overline{\{x\}}=Y$. It
follows $\overline{\{x\}}\cap Y=Y$. This shows $Y$ is quasi-sober. Second,
suppose $Y$ is open. Let $Z$ be an irreducible closed subset of $Y$. Then
$\overline{Z}$ is an irreducible closed subset of $X$. Hence there
exists $x\in Z$ with $\overline{\{x\}}=\overline{Z}$. If
$x\notin Y$ we get the contradiction
$Z=Z\cap Y\subset\overline{Z}\cap Y=\overline{\{x\}}\cap Y=\emptyset$.
Therefore $x\in Y$. It follows $Z=\overline{Z}\cap Y=\overline{\{x\}}\cap Y$.
This shows $Y$ is quasi-sober.
\medskip\noindent
Proof of (3). Immediately from (1) and (2).
\end{proof}
\begin{lemma}
\label{lemma-sober-local}
Let $X$ be a topological space and let $(X_i)_{i\in I}$ be a covering of $X$.
\begin{enumerate}
\item Suppose $X_i$ is locally closed in $X$ for every $i\in I$. Then, $X$ is
Kolmogorov if and only if $X_i$ is Kolmogorov for every $i\in I$.
\item Suppose $X_i$ is open in $X$ for every $i\in I$. Then, $X$ is
quasi-sober if and only if $X_i$ is quasi-sober for every $i\in I$.
\item Suppose $X_i$ is open in $X$ for every $i\in I$. Then, $X$ is sober if
and only if $X_i$ is sober for every $i\in I$.
\end{enumerate}
\end{lemma}
\begin{proof}
Proof of (1). If $X$ is Kolmogorov then so is $X_i$ for every $i\in I$ by
Lemma \ref{lemma-sober-subspace}. Suppose $X_i$ is Kolmogorov for every
$i\in I$. Let $x,y\in X$ with $\overline{\{x\}}=\overline{\{y\}}$. There exists
$i\in I$ with $x\in X_i$. There exists an open subset $U\subset X$ such that
$X_i$ is a closed subset of $U$. If $y\notin U$ we get the contradiction
$x\in\overline{\{x\}}\cap U=\overline{\{y\}}\cap U=\emptyset$. Hence $y\in U$.
It follows $y\in\overline{\{y\}}\cap U=\overline{\{x\}}\cap U\subset X_i$.
This shows $y\in X_i$. It follows
$\overline{\{x\}}\cap X_i=\overline{\{y\}}\cap X_i$. Since $X_i$ is
Kolmogorov we get $x=y$. This shows $X$ is Kolmogorov.
\medskip\noindent
Proof of (2). If $X$ is quasi-sober then so is $X_i$ for every $i\in I$ by
Lemma \ref{lemma-sober-subspace}. Suppose $X_i$ is quasi-sober for every
$i\in I$. Let $Y$ be an irreducible closed subset of $X$. As $Y\neq\emptyset$
there exists $i\in I$ with $X_i\cap Y\neq\emptyset$. As $X_i$ is open in $X$ it
follows $X_i\cap Y$ is non-empty and open in $Y$, hence irreducible
and dense in $Y$. Thus $X_i\cap Y$ is an irreducible closed subset of $X_i$. As
$X_i$ is quasi-sober there exists $x\in X_i\cap Y$ with
$X_i\cap Y=\overline{\{x\}}\cap X_i\subset\overline{\{x\}}$. Since
$X_i\cap Y$ is dense in $Y$ and $Y$ is closed in $X$ it follows
$Y=\overline{X_i\cap Y}\cap Y\subset\overline{X_i\cap Y}\subset
\overline{\{x\}}\subset Y$. Therefore
$Y=\overline{\{x\}}$. This shows $X$ is quasi-sober.
\medskip\noindent
Proof of (3). Immediately from (1) and (2).
\end{proof}
\begin{example}
\label{example-quasi-sober-not-kolmogorov}
Let $X$ be an indiscrete space of cardinality at least $2$. Then $X$ is
quasi-sober but not Kolmogorov. Moreover, the family of its singletons is a
covering of $X$ by discrete and hence Kolmogorov spaces.
\end{example}
\begin{example}
\label{example-kolmogorov-not-quasi-sober}
Let $Y$ be an infinite set, furnished with the topology whose closed sets are
$Y$ and the finite subsets of $Y$. Then $Y$ is Kolmogorov but not quasi-sober.
However, the family of its singletons (which are its irreducible components) is
a covering by discrete and hence sober spaces.
\end{example}
\begin{example}
\label{example-not-kolmogorov-not-quasi-sober}
Let $X$ and $Y$ be as in Example \ref{example-quasi-sober-not-kolmogorov} and
Example \ref{example-kolmogorov-not-quasi-sober}. Then, $X\amalg Y$ is neither
Kolmogorov nor quasi-sober.
\end{example}
\begin{example}
\label{example-sober-subspace}
Let $Z$ be an infinite set and let $z\in Z$. We furnish $Z$ with the topology
whose closed sets are $Z$ and the finite subsets of $Z\setminus\{z\}$. Then $Z$
is sober but its subspace $Z\setminus\{z\}$ is not quasi-sober.
\end{example}
\begin{example}
\label{example-Hausdorff}
Recall that a topological space $X$ is Hausdorff iff for every
distinct pair of points $x, y \in X$ there exist disjoint
opens $U, V \subset X$ such that $x \in U$, $y \in V$.
In this case $X$ is irreducible if and only if $X$ is
a singleton. Similarly, any subset of $X$ is irreducible
if and only if it is a singleton. Hence a Hausdorff space is
sober.
\end{example}
\begin{lemma}
\label{lemma-irreducible-on-top}
Let $f : X \to Y$ be a continuous map of topological spaces.
Assume that
(a) $Y$ is irreducible,
(b) $f$ is open, and
(c) there exists a dense collection of points $y \in Y$ such
that $f^{-1}(y)$ is irreducible.
Then $X$ is irreducible.
\end{lemma}
\begin{proof}
Suppose $X = Z_1 \cup Z_2$ with $Z_i$ closed.
Consider the open sets $U_1 = Z_1 \setminus Z_2 = X \setminus Z_2$ and
$U_2 = Z_2 \setminus Z_1 = X \setminus Z_1$. To get a contradiction
assume that $U_1$ and $U_2$ are both nonempty. By (b) we see that $f(U_i)$
is open. By (a) we have $Y$ irreducible and hence
$f(U_1) \cap f(U_2) \not = \emptyset$. By (c) there is a point $y$ which
corresponds to a point of this intersection such that the fibre
$X_y = f^{-1}(y)$ is irreducible. Then $X_y \cap U_1$ and
$X_y \cap U_2$ are nonempty disjoint open subsets of $X_y$ which is
a contradiction.
\end{proof}
\begin{lemma}
\label{lemma-irreducible-fibres-irreducible-components}
Let $f : X \to Y$ be a continuous map of topological spaces.
Assume that (a) $f$ is open, and
(b) for every $y \in Y$ the fibre $f^{-1}(y)$ is irreducible.
Then $f$ induces a bijection between irreducible components.
\end{lemma}
\begin{proof}
We point out that assumption (b) implies that $f$ is surjective (see
Definition \ref{definition-irreducible-components}).
Let $T \subset Y$ be an irreducible component.
Note that $T$ is closed, see Lemma \ref{lemma-irreducible}.
The lemma follows if we show that $f^{-1}(T)$ is irreducible
because any irreducible subset of $X$ maps into an irreducible component
of $Y$ by Lemma \ref{lemma-image-irreducible-space}.
Note that $f^{-1}(T) \to T$ satisfies the assumptions
of Lemma \ref{lemma-irreducible-on-top}. Hence we win.
\end{proof}
\noindent
The construction of the following lemma is sometimes called
the ``soberification''.
\begin{lemma}
\label{lemma-make-sober}
Let $X$ be a topological space. There is a canonical continuous map
$$
c : X \longrightarrow X'
$$
from $X$ to a sober topological space $X'$ which is universal
among continuous maps from $X$ to sober topological spaces.
Moreover, the assignment $U' \mapsto c^{-1}(U')$ is a bijection
between opens of $X'$ and $X$ which commutes with finite intersections
and arbitrary unions.
The image $c(X)$ is a Kolmogorov topological space and the
map $c : X \to c(X)$ is universal for maps of $X$ into Kolmogorov spaces.
\end{lemma}
\begin{proof}
Let $X'$ be the set of irreducible closed subsets of $X$ and let
$$
c : X \to X', \quad x \mapsto \overline{\{x\}}.
$$
For $U \subset X$ open, let $U' \subset X'$ denote the set
of irreducible closed subsets of $X$ which meet $U$.
Then $c^{-1}(U') = U$. In particular, if $U_1 \not = U_2$ are open in
$X$, then $U'_1 \not = U_2'$. Hence $c$ induces
a bijection between the subsets of $X'$ of the form $U'$ and the
opens of $X$.
\medskip\noindent
Let $U_1, U_2$ be open in $X$. Suppose that $Z \in U'_1$ and
$Z \in U'_2$. Then $Z \cap U_1$ and $Z \cap U_2$ are nonempty
open subsets of the irreducible space $Z$ and hence $Z \cap U_1 \cap U_2$
is nonempty. Thus $(U_1 \cap U_2)' = U'_1 \cap U'_2$.
The rule $U \mapsto U'$ is also compatible with arbitrary unions
(details omitted). Thus it is clear that the collection of
$U'$ form a topology on $X'$ and that we have a bijection as
stated in the lemma.
\medskip\noindent
Next we show that $X'$ is sober. Let $T \subset X'$ be an irreducible
closed subset. Let $U \subset X$ be the open such that $X' \setminus T = U'$.
Then $Z = X \setminus U$ is irreducible because of the properties
of the bijection of the lemma. We claim that $Z \in T$ is the unique generic
point. Namely, any open of the form $V' \subset X'$
which does not contain $Z$ must come from an open $V \subset X$
which misses $Z$, i.e., is contained in $U$.
\medskip\noindent
Finally, we check the universal property. Let $f : X \to Y$ be a continuous
map to a sober topological space. Then we let $f' : X' \to Y$ be the map
which sends the irreducible closed $Z \subset X$ to the unique generic
point of $\overline{f(Z)}$. It follows immediately that
$f' \circ c = f$ as maps of sets, and the properties of $c$ imply that
$f'$ is continuous. We omit the verification that the continuous
map $f'$ is unique. We also omit the proof of the statements on
Kolmogorov spaces.
\end{proof}
\section{Noetherian topological spaces}
\label{section-noetherian}
\begin{definition}
\label{definition-noetherian}
A topological space is called {\it Noetherian}
if the descending chain condition holds for
closed subsets of $X$. A topological space is called
{\it locally Noetherian} if every point has a neighbourhood
which is Noetherian.
\end{definition}
\begin{lemma}
\label{lemma-Noetherian}
Let $X$ be a Noetherian topological space.
\begin{enumerate}
\item Any subset of $X$ with the induced topology is Noetherian.
\item The space $X$ has finitely many irreducible components.
\item Each irreducible component of $X$ contains a nonempty open of $X$.
\end{enumerate}
\end{lemma}
\begin{proof}
Let $T \subset X$ be a subset of $X$.
Let $T_1 \supset T_2 \supset \ldots$
be a descending chain of closed subsets of $T$.
Write $T_i = T \cap Z_i$ with $Z_i \subset X$ closed.
Consider the descending chain of closed subsets
$Z_1 \supset Z_1\cap Z_2 \supset Z_1 \cap Z_2 \cap Z_3 \ldots$
This stabilizes by assumption and hence the original sequence
of $T_i$ stabilizes. Thus $T$ is Noetherian.
\medskip\noindent
Let $A$ be the set of closed subsets of $X$ which do not
have finitely many irreducible components. Assume that
$A$ is not empty to arrive at a contradiction.
The set $A$ is partially ordered by inclusion: $\alpha \leq \alpha'
\Leftrightarrow Z_{\alpha} \subset Z_{\alpha'}$.
By the descending chain condition we may find a
smallest element of $A$, say $Z$. As $Z$ is not a finite
union of irreducible components, it is not irreducible.
Hence we can write $Z = Z' \cup Z''$ and both are strictly smaller
closed subsets. By construction $Z' = \bigcup Z'_i$ and
$Z'' = \bigcup Z''_j$ are finite unions of their irreducible
components. Hence $Z = \bigcup Z'_i \cup \bigcup Z''_j$ is
a finite union of irreducible closed subsets.
After removing redundant members of this expression,
this will be the decomposition of $Z$ into its irreducible
components, a contradiction.
\medskip\noindent
Let $Z \subset X$ be an irreducible component of $X$.
Let $Z_1, \ldots, Z_n$ be the other irreducible components
of $X$. Consider $U = Z \setminus (Z_1\cup\ldots\cup Z_n)$.
This is not empty since otherwise the irreducible space
$Z$ would be contained in one of the other $Z_i$.
Because $X = Z \cup Z_1 \cup \ldots Z_n$ (see Lemma \ref{lemma-irreducible}),
also $U = X \setminus (Z_1\cup\ldots\cup Z_n)$
and hence open in $X$. Thus $Z$ contains a nonempty
open of $X$.
\end{proof}
\begin{lemma}
\label{lemma-image-Noetherian}
Let $f : X \to Y$ be a continuous map of topological spaces.
\begin{enumerate}
\item If $X$ is Noetherian, then $f(X)$ is Noetherian.
\item If $X$ is locally Noetherian and $f$ open, then $f(X)$ is
locally Noetherian.
\end{enumerate}
\end{lemma}
\begin{proof}
In case (1), suppose that $Z_1 \supset Z_2 \supset Z_3 \supset \ldots$
is a descending chain of closed subsets of $f(X)$ (as usual with the induced
topology as a subset of $Y$). Then
$f^{-1}(Z_1) \supset f^{-1}(Z_2) \supset f^{-1}(Z_3) \supset \ldots$ is
a descending chain of closed subsets of $X$. Hence this chain stabilizes.
Since $f(f^{-1}(Z_i)) = Z_i$ we conclude that
$Z_1 \supset Z_2 \supset Z_3 \supset \ldots$
stabilizes also. In case (2), let $y \in f(X)$. Choose $x \in X$ with
$f(x) = y$. By assumption there exists a neighbourhood $E \subset X$ of
$x$ which is Noetherian. Then $f(E) \subset f(X)$ is a neighbourhood
which is Noetherian by part (1).
\end{proof}
\begin{lemma}
\label{lemma-finite-union-Noetherian}
Let $X$ be a topological space.
Let $X_i \subset X$, $i = 1, \ldots, n$ be a finite collection of subsets.
If each $X_i$ is Noetherian (with the induced topology), then
$\bigcup_{i = 1, \ldots, n} X_i$ is Noetherian (with the induced topology).
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{example}
\label{example-locally-Noetherian-no-closed-point}
Any nonempty, Kolmogorov Noetherian topological space has a closed point
(combine Lemmas \ref{lemma-quasi-compact-closed-point} and
\ref{lemma-Noetherian-quasi-compact}).
Let $X = \{1, 2, 3, \ldots \}$. Define a topology on $X$
with opens $\emptyset$, $\{1, 2, \ldots, n\}$, $n \geq 1$
and $X$. Thus $X$ is a locally Noetherian topological space,
without any closed points. This space cannot be the underlying
topological space of a locally Noetherian scheme, see
Properties, Lemma \ref{properties-lemma-locally-Noetherian-closed-point}.
\end{example}
\begin{lemma}
\label{lemma-locally-Noetherian-locally-connected}
Let $X$ be a locally Noetherian topological space.
Then $X$ is locally connected.
\end{lemma}
\begin{proof}
Let $x \in X$. Let $E$ be a neighbourhood of $x$.
We have to find a connected neighbourhood of $x$ contained
in $E$. By assumption there exists a neighbourhood $E'$ of $x$
which is Noetherian. Then $E \cap E'$ is Noetherian, see
Lemma \ref{lemma-Noetherian}.
Let $E \cap E' = Y_1 \cup \ldots \cup Y_n$ be the decomposition
into irreducible components, see
Lemma \ref{lemma-Noetherian}.
Let $E'' = \bigcup_{x \in Y_i} Y_i$. This is a connected
subset of $E \cap E'$ containing $x$. It contains the open
$E \cap E' \setminus (\bigcup_{x \not \in Y_i} Y_i)$ of $E \cap E'$
and hence it is a neighbourhood of $x$ in $X$. This proves the lemma.
\end{proof}
\section{Krull dimension}
\label{section-krull-dimension}
\begin{definition}
\label{definition-Krull}
Let $X$ be a topological space.
\begin{enumerate}
\item A {\it chain of irreducible closed subsets} of $X$
is a sequence $Z_0 \subset Z_1 \subset \ldots \subset Z_n \subset X$
with $Z_i$ closed irreducible and $Z_i \not = Z_{i + 1}$ for
$i = 0, \ldots, n - 1$.
\item The {\it length} of a chain
$Z_0 \subset Z_1 \subset \ldots \subset Z_n \subset X$
of irreducible closed subsets of $X$ is the
integer $n$.
\item The {\it dimension} or more precisely the {\it Krull dimension}
$\dim(X)$ of $X$ is the element of
$\{-\infty, 0, 1, 2, 3, \ldots, \infty\}$ defined by the formula:
$$
\dim(X) =
\sup \{\text{lengths of chains of irreducible closed subsets}\}
$$
Thus $\dim(X) = -\infty$ if and only if $X$ is the empty space.
\item Let $x \in X$.
The {\it Krull dimension of $X$ at $x$} is defined as
$$
\dim_x(X) = \min \{\dim(U), x\in U\subset X\text{ open}\}
$$
the minimum of $\dim(U)$ where $U$ runs over the open
neighbourhoods of $x$ in $X$.
\end{enumerate}
\end{definition}
\noindent
Note that if $U' \subset U \subset X$ are open then
$\dim(U') \leq \dim(U)$. Hence if $\dim_x(X) = d$ then $x$
has a fundamental system of open neighbourhoods $U$ with
$\dim(U) = \dim_x(X)$.
\begin{lemma}
\label{lemma-dimension-supremum-local-dimensions}
Let $X$ be a topological space. Then $\dim(X) = \sup \dim_x(X)$
where the supremum runs over the points $x$ of $X$.
\end{lemma}
\begin{proof}
It is clear that $\dim(X) \geq \dim_x(X)$ for all $x \in X$ (see
discussion following Definition \ref{definition-Krull}).
Thus an inequality in one direction. For the converse, let $n \geq 0$
and suppose that $\dim(X) \geq n$. Then we can find a chain of irreducible
closed subsets $Z_0 \subset Z_1 \subset \ldots \subset Z_n \subset X$.
Pick $x \in Z_0$. Then we see that every open neighbourhood $U$ of $x$
has a chain of irreducible closed subsets
$Z_0 \cap U \subset Z_1 \cap U \subset \ldots Z_n \cap U \subset U$.
In this way we see that $\dim_x(X) \geq n$ which proves the other
inequality.
\end{proof}
\begin{example}
\label{example-Krull-Rn}
The Krull dimension of the usual Euclidean space
$\mathbf{R}^n$ is $0$.
\end{example}
\begin{example}
\label{example-krull-2set}
Let $X = \{s, \eta\}$ with open sets given
by $\{\emptyset, \{\eta\}, \{s, \eta\}\}$.
In this case a maximal chain of irreducible
closed subsets is $\{s\} \subset \{s, \eta\}$.
Hence $\dim(X) = 1$. It is easy to generalize
this example to get a $(n + 1)$-element topological
space of Krull dimension $n$.
\end{example}
\begin{definition}
\label{definition-equidimensional}
Let $X$ be a topological space.
We say that $X$ is {\it equidimensional} if every irreducible
component of $X$ has the same dimension.
\end{definition}
\section{Codimension and catenary spaces}
\label{section-catenary-spaces}
\noindent
We only define the codimension of irreducible closed subsets.
\begin{definition}
\label{definition-codimension}
Let $X$ be a topological space.
Let $Y \subset X$ be an irreducible closed subset.
The {\it codimension} of $Y$ in $X$ is the supremum of
the lengths $e$ of chains
$$
Y = Y_0 \subset Y_1 \subset \ldots \subset Y_e \subset X
$$
of irreducible closed subsets in $X$ starting with $Y$.
We will denote this $\text{codim}(Y, X)$.
\end{definition}
\noindent
The codimension is an element of $\{0, 1, 2, \ldots\} \cup \{\infty\}$.
If $\text{codim}(Y, X) < \infty$, then every chain can be extended to
a maximal chain (but these do not all have to have the same length).
\begin{lemma}
\label{lemma-codimension-at-generic-point}
Let $X$ be a topological space.
Let $Y \subset X$ be an irreducible closed subset.
Let $U \subset X$ be an open subset such that $Y \cap U$ is nonempty.
Then
$$
\text{codim}(Y, X) = \text{codim}(Y \cap U, U)
$$
\end{lemma}
\begin{proof}
The rule $T \mapsto \overline{T}$ defines a bijective
inclusion preserving map between the closed irreducible subsets
of $U$ and the closed irreducible subsets of $X$ which meet $U$.
Using this the lemma easily follows. Details omitted.
\end{proof}
\begin{example}
\label{example-Noetherian-infinite-codimension}
Let $X = [0, 1]$ be the unit interval with the following
topology: The sets $[0, 1]$, $(1 - 1/n, 1]$ for $n \in \mathbf{N}$, and
$\emptyset$ are open. So the closed sets are
$\emptyset$, $\{0\}$, $[0, 1 - 1/n]$ for $n > 1$ and $[0, 1]$.
This is clearly a Noetherian topological space.
But the irreducible closed subset $Y = \{0\}$ has infinite
codimension $\text{codim}(Y, X) = \infty$.
To see this we just remark that all the closed sets
$[0, 1 - 1/n]$ are irreducible.
\end{example}
\begin{definition}
\label{definition-catenary}
Let $X$ be a topological space. We say $X$ is {\it catenary} if
for every pair of irreducible closed subsets $T \subset T'$
we have $\text{codim}(T, T') < \infty$ and every maximal chain
of irreducible closed subsets
$$
T = T_0 \subset T_1 \subset \ldots \subset T_e = T'
$$
has the same length (equal to the codimension).
\end{definition}
\begin{lemma}
\label{lemma-catenary}
Let $X$ be a topological space.
The following are equivalent:
\begin{enumerate}
\item $X$ is catenary,
\item $X$ has an open covering by catenary spaces.
\end{enumerate}
Moreover, in this case any locally closed subspace of $X$ is catenary.
\end{lemma}
\begin{proof}
Suppose that $X$ is catenary and that $U \subset X$ is an open
subset. The rule $T \mapsto \overline{T}$ defines a bijective
inclusion preserving map between the closed irreducible subsets
of $U$ and the closed irreducible subsets of $X$ which meet $U$.
Using this the lemma easily follows. Details omitted.
\end{proof}
\begin{lemma}
\label{lemma-catenary-in-codimension}
Let $X$ be a topological space. The following are equivalent:
\begin{enumerate}
\item $X$ is catenary, and
\item for every pair of irreducible closed subsets $Y \subset Y'$ we have
$\text{codim}(Y, Y') < \infty$ and for every triple
$Y \subset Y' \subset Y''$ of irreducible closed subsets we have
$$
\text{codim}(Y, Y'') = \text{codim}(Y, Y') + \text{codim}(Y', Y'').
$$
\end{enumerate}
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\section{Quasi-compact spaces and maps}
\label{section-quasi-compact}
\noindent
The phrase ``compact'' will be reserved
for Hausdorff topological spaces. And many spaces occurring
in algebraic geometry are not Hausdorff.
\begin{definition}
\label{definition-quasi-compact}
Quasi-compactness.
\begin{enumerate}
\item We say that a topological space $X$ is {\it quasi-compact}
if every open covering of $X$ has a finite refinement.
\item We say that a continuous map $f : X \to Y$ is {\it quasi-compact}
if the inverse image $f^{-1}(V)$ of every quasi-compact open $V \subset Y$
is quasi-compact.
\item We say a subset $Z \subset X$ is {\it retrocompact}
if the inclusion map $Z \to X$ is quasi-compact.
\end{enumerate}
\end{definition}
\noindent
In many texts on topology a space is called {\it compact} if it
is quasi-compact and Hausdorff; and in other texts the Hausdorff
condition is omitted. To avoid confusion in algebraic geometry
we use the term quasi-compact. Note that the notion of quasi-compactness
of a map is very different from the notion of a ``proper map''
in topology, since there one requires the inverse image of any
(quasi-)compact subset of the target to be (quasi-)compact,
whereas in the definition above we only consider quasi-compact
{\it open} sets.
\begin{lemma}
\label{lemma-composition-quasi-compact}
A composition of quasi-compact maps is quasi-compact.
\end{lemma}
\begin{proof}
This is immediate from the definition.
\end{proof}
\begin{lemma}
\label{lemma-closed-in-quasi-compact}
A closed subset of a quasi-compact topological space is quasi-compact.
\end{lemma}
\begin{proof}
Let $E \subset X$ be a closed subset of the quasi-compact space $X$.
Let $E = \bigcup V_j$ be an open covering. Choose $U_j \subset X$
open such that $V_j = E \cap U_j$. Then $X = (X \setminus E) \cup \bigcup U_j$
is an open covering of $X$. Hence
$X = (X \setminus E) \cup U_{j_1} \cup \ldots \cup U_{j_n}$ for some
$n$ and indices $j_i$. Thus $E = V_{j_1} \cup \ldots \cup V_{j_n}$
as desired.
\end{proof}
\begin{lemma}
\label{lemma-quasi-compact-in-Hausdorff}
Let $X$ be a Hausdorff topological space.
\begin{enumerate}
\item If $E \subset X$ is quasi-compact, then it is closed.
\item If $E_1, E_2 \subset X$ are disjoint quasi-compact subsets
then there exists opens $E_i \subset U_i$ with $U_1 \cap U_2 = \emptyset$.
\end{enumerate}
\end{lemma}
\begin{proof}
Proof of (1). Let $x \in X$, $x \not \in E$.
For every $e \in E$ we can find disjoint opens $V_e$ and $U_e$
with $e \in V_e$ and $x \in U_e$. Since $E \subset \bigcup V_e$
we can find finitely many $e_1, \ldots, e_n$ such that
$E \subset V_{e_1} \cup \ldots \cup V_{e_n}$. Then
$U = U_{e_1} \cap \ldots \cap U_{e_n}$ is an open neighbourhood
of $x$ which avoids $V_{e_1} \cup \ldots \cup V_{e_n}$. In particular
it avoids $E$. Thus $E$ is closed.
\medskip\noindent
Proof of (2). In the proof of (1) we have seen that given $x \in E_1$
we can find an open neighbourhood $x \in U_x$ and an open
$E_2 \subset V_x$ such that $U_x \cap V_x = \emptyset$. Because
$E_1$ is quasi-compact we can find a finite number $x_i \in E_1$
such that $E_1 \subset U = U_{x_1} \cup \ldots \cup U_{x_n}$.
We take $V = V_{x_1} \cap \ldots \cap V_{x_n}$ to finish the proof.
\end{proof}
\begin{lemma}
\label{lemma-closed-in-compact}
Let $X$ be a quasi-compact Hausdorff space. Let $E \subset X$.
The following are equivalent: (a) $E$ is closed in $X$, (b)
$E$ is quasi-compact.
\end{lemma}
\begin{proof}
The implication (a) $\Rightarrow$ (b) is
Lemma \ref{lemma-closed-in-quasi-compact}.
The implication (b) $\Rightarrow$ (a) is
Lemma \ref{lemma-quasi-compact-in-Hausdorff}.
\end{proof}
\noindent
The following is really a reformulation of the
quasi-compact property.
\begin{lemma}
\label{lemma-intersection-closed-in-quasi-compact}
Let $X$ be a quasi-compact topological space.
If $\{Z_\alpha\}_{\alpha \in A}$ is a collection of closed subsets
such that the intersection of each finite subcollection
is nonempty, then $\bigcap_{\alpha \in A} Z_\alpha$ is nonempty.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{lemma}
\label{lemma-image-quasi-compact}
Let $f : X \to Y$ be a continuous map of topological spaces.
\begin{enumerate}
\item If $X$ is quasi-compact, then $f(X)$ is quasi-compact.
\item If $f$ is quasi-compact, then $f(X)$ is retrocompact.
\end{enumerate}
\end{lemma}
\begin{proof}
If $f(X) = \bigcup V_i$ is an open covering, then $X = \bigcup f^{-1}(V_i)$
is an open covering. Hence if $X$ is quasi-compact then
$X = f^{-1}(V_{i_1}) \cup \ldots \cup f^{-1}(V_{i_n})$ for some
$i_1, \ldots, i_n \in I$ and hence
$f(X) = V_{i_1} \cup \ldots \cup V_{i_n}$. This proves (1).
Assume $f$ is quasi-compact, and let $V \subset Y$ be quasi-compact open.
Then $f^{-1}(V)$ is quasi-compact, hence by (1) we see that
$f(f^{-1}(V)) = f(X) \cap V$ is quasi-compact. Hence $f(X)$
is retrocompact.
\end{proof}
\begin{lemma}
\label{lemma-quasi-compact-closed-point}
Let $X$ be a topological space. Assume that
\begin{enumerate}
\item $X$ is nonempty,
\item $X$ is quasi-compact, and
\item $X$ is Kolmogorov.
\end{enumerate}
Then $X$ has a closed point.
\end{lemma}
\begin{proof}
Consider the set
$$
\mathcal{T} =
\{Z \subset X \mid Z = \overline{\{x\}} \text{ for some }x \in X\}
$$
of all closures of singletons in $X$. It is nonempty since $X$ is
nonempty. Make $\mathcal{T}$ into a
partially ordered set using the relation of inclusion.
Suppose $Z_\alpha$, $\alpha \in A$ is a totally ordered subset of $\mathcal{T}$.
By Lemma \ref{lemma-intersection-closed-in-quasi-compact} we see
that $\bigcap_{\alpha \in A} Z_\alpha \not = \emptyset$. Hence there exists
some $x \in \bigcap_{\alpha \in A} Z_\alpha$ and we see that
$Z = \overline{\{x\}}\in \mathcal{T}$ is a lower bound for
the family. By Zorn's lemma there exists a minimal element
$Z \in \mathcal{T}$. As $X$ is Kolmogorov we conclude that
$Z = \{x\}$ for some $x$ and $x \in X$ is a closed point.
\end{proof}
\begin{lemma}
\label{lemma-closed-points-quasi-compact}
Let $X$ be a quasi-compact Kolmogorov space. Then the set $X_0$ of
closed points of $X$ is quasi-compact.
\end{lemma}
\begin{proof}
Let $X_0 = \bigcup U_{i, 0}$ be an open covering.
Write $U_{i, 0} = X_0 \cap U_i$ for some open $U_i \subset X$.
Consider the complement $Z$ of $\bigcup U_i$. This is a closed subset of
$X$, hence quasi-compact (Lemma \ref{lemma-closed-in-quasi-compact})
and Kolmogorov. By Lemma \ref{lemma-quasi-compact-closed-point}
if $Z$ is nonempty it would have a closed
point which contradicts the fact that $X_0 \subset \bigcup U_i$.
Hence $Z = \emptyset$ and $X = \bigcup U_i$. Since $X$ is quasi-compact
this covering has a finite subcover and we conclude.
\end{proof}
\begin{lemma}
\label{lemma-connected-component-intersection}
Let $X$ be a topological space.
Assume
\begin{enumerate}
\item $X$ is quasi-compact,
\item $X$ has a basis for the topology consisting of quasi-compact opens, and
\item the intersection of two quasi-compact opens is quasi-compact.
\end{enumerate}
For any $x \in X$ the connected component of $X$ containing
$x$ is the intersection of all open and closed subsets
of $X$ containing $x$.
\end{lemma}
\begin{proof}
Let $T$ be the connected component containing $x$.
Let $S = \bigcap_{\alpha \in A} Z_\alpha$ be the intersection of all
open and closed subsets $Z_\alpha$ of $X$ containing $x$.
Note that $S$ is closed in $X$.
Note that any finite intersection of $Z_\alpha$'s is a $Z_\alpha$.
Because $T$ is connected and $x \in T$ we have $T \subset S$.
It suffices to show that $S$ is connected.
If not, then there exists a disjoint union decomposition
$S = B \amalg C$ with $B$ and $C$ open and closed in $S$.
In particular, $B$ and $C$ are closed in $X$, and so quasi-compact by
Lemma \ref{lemma-closed-in-quasi-compact} and assumption (1).
By assumption (2) there exist quasi-compact opens
$U, V \subset X$ with $B = S \cap U$ and $C = S \cap V$ (details omitted).
Then $U \cap V \cap S = \emptyset$.
Hence $\bigcap_\alpha U \cap V \cap Z_\alpha = \emptyset$.
By assumption (3) the intersection $U \cap V$ is quasi-compact.
By Lemma \ref{lemma-intersection-closed-in-quasi-compact}
for some $\alpha' \in A$ we have $U \cap V \cap Z_{\alpha'} = \emptyset$.
Since $X \setminus (U \cup V)$ is disjoint from $S$
and closed in $X$ hence quasi-compact, we can use the same lemma
to see that $Z_{\alpha''} \subset U \cup V$ for some $\alpha'' \in A$.
Then $Z_\alpha = Z_{\alpha'} \cap Z_{\alpha''}$ is contained
in $U \cup V$ and disjoint from $U \cap V$.
Hence $Z_\alpha = U \cap Z_\alpha \amalg V \cap Z_\alpha$
is a decomposition into two open pieces,
hence $U \cap Z_\alpha$ and $V \cap Z_\alpha$ are open and closed in $X$.
Thus, if $x \in B$ say, then we see that $S \subset U \cap Z_\alpha$
and we conclude that $C = \emptyset$.
\end{proof}
\begin{lemma}
\label{lemma-connected-component-intersection-compact-Hausdorff}
Let $X$ be a topological space. Assume $X$ is quasi-compact and Hausdorff.
For any $x \in X$ the connected component of $X$ containing
$x$ is the intersection of all open and closed subsets
of $X$ containing $x$.
\end{lemma}
\begin{proof}
Let $T$ be the connected component containing $x$.
Let $S = \bigcap_{\alpha \in A} Z_\alpha$ be the intersection of all
open and closed subsets $Z_\alpha$ of $X$ containing $x$.
Note that $S$ is closed in $X$.
Note that any finite intersection of $Z_\alpha$'s is a $Z_\alpha$.
Because $T$ is connected and $x \in T$ we have $T \subset S$.
It suffices to show that $S$ is connected.
If not, then there exists a disjoint union decomposition
$S = B \amalg C$ with $B$ and $C$ open and closed in $S$.
In particular, $B$ and $C$ are closed in $X$, and so quasi-compact by
Lemma \ref{lemma-closed-in-quasi-compact}.
By Lemma \ref{lemma-quasi-compact-in-Hausdorff}
there exist disjoint opens $U, V \subset X$ with $B \subset U$ and
$C \subset V$. Then $X \setminus U \cup V$ is closed in $X$
hence quasi-compact (Lemma \ref{lemma-closed-in-quasi-compact}).
It follows that $(X \setminus U \cup V) \cap Z_\alpha = \emptyset$
for some $\alpha$ by Lemma \ref{lemma-intersection-closed-in-quasi-compact}.
In other words, $Z_\alpha \subset U \cup V$. Thus
$Z_\alpha = Z_\alpha \cap V \amalg Z_\alpha \cap U$
is a decomposition into two open pieces,
hence $U \cap Z_\alpha$ and $V \cap Z_\alpha$ are open and closed in $X$.
Thus, if $x \in B$ say, then we see that $S \subset U \cap Z_\alpha$
and we conclude that $C = \emptyset$.
\end{proof}
\begin{lemma}
\label{lemma-closed-union-connected-components}
Let $X$ be a topological space.
Assume
\begin{enumerate}
\item $X$ is quasi-compact,
\item $X$ has a basis for the topology consisting of quasi-compact opens, and
\item the intersection of two quasi-compact opens is quasi-compact.
\end{enumerate}
For a subset $T \subset X$ the following are equivalent:
\begin{enumerate}
\item[(a)] $T$ is an intersection of open and closed subsets of $X$, and
\item[(b)] $T$ is closed in $X$ and is a union of connected components of $X$.
\end{enumerate}
\end{lemma}
\begin{proof}
It is clear that (a) implies (b).
Assume (b). Let $x \in X$, $x \not \in T$. Let $x \in C \subset X$
be the connected component of $X$ containing $x$. By
Lemma \ref{lemma-connected-component-intersection}
we see that $C = \bigcap V_\alpha$ is the intersection of all open and
closed subsets $V_\alpha$ of $X$ which contain $C$.
In particular, any pairwise intersection $V_\alpha \cap V_\beta$
occurs as a $V_\alpha$.
As $T$ is a union of connected components
of $X$ we see that $C \cap T = \emptyset$. Hence
$T \cap \bigcap V_\alpha = \emptyset$. Since $T$ is quasi-compact as a
closed subset of a quasi-compact space (see
Lemma \ref{lemma-closed-in-quasi-compact})
we deduce that $T \cap V_\alpha = \emptyset$ for some $\alpha$, see
Lemma \ref{lemma-intersection-closed-in-quasi-compact}.
For this $\alpha$ we see that $U_\alpha = X \setminus V_\alpha$
is an open and closed subset of $X$ which contains $T$ and not $x$.
The lemma follows.
\end{proof}
\begin{lemma}
\label{lemma-Noetherian-quasi-compact}
Let $X$ be a Noetherian topological space.
\begin{enumerate}
\item The space $X$ is quasi-compact.
\item Any subset of $X$ is retrocompact.
\end{enumerate}
\end{lemma}
\begin{proof}
Suppose $X = \bigcup U_i$ is an open covering of $X$ indexed
by the set $I$ which does not have a refinement by a finite
open covering. Choose $i_1, i_2, \ldots $ elements of $I$ inductively
in the following way: Choose $i_{n + 1}$ such that $U_{i_{n + 1}}$
is not contained in $U_{i_1} \cup \ldots \cup U_{i_n}$. Thus we see that
$X \supset (X \setminus U_{i_1}) \supset
(X \setminus U_{i_1} \cup U_{i_2}) \supset \ldots$ is a strictly
decreasing infinite sequence of closed subsets. This contradicts
the fact that $X$ is Noetherian. This proves the first assertion.
The second assertion is now clear since every subset of $X$ is Noetherian by
Lemma \ref{lemma-Noetherian}.
\end{proof}
\begin{lemma}
\label{lemma-quasi-compact-locally-Noetherian-Noetherian}
A quasi-compact locally Noetherian space is Noetherian.
\end{lemma}
\begin{proof}
The conditions imply immediately that $X$ has a finite covering by
Noetherian subsets, and hence is Noetherian by
Lemma \ref{lemma-finite-union-Noetherian}.
\end{proof}
\begin{lemma}[Alexander subbase theorem]
\label{lemma-subbase-theorem}
Let $X$ be a topological space. Let $\mathcal{B}$ be a subbase for $X$.
If every covering of $X$ by elements of $\mathcal{B}$ has a finite
refinement, then $X$ is quasi-compact.
\end{lemma}
\begin{proof}
Assume there is an open covering of $X$ which does not have a finite
refinement. Using Zorn's lemma we can choose a maximal open covering
$X = \bigcup_{i \in I} U_i$ which does not have a finite refinement
(details omitted).
In other words, if $U \subset X$ is any open which does not occur as
one of the $U_i$, then the covering $X = U \cup \bigcup_{i \in I} U_i$
does have a finite refinement. Let $I' \subset I$ be the set of indices
such that $U_i \in \mathcal{B}$. Then $\bigcup_{i \in I'} U_i \not = X$,
since otherwise we would get a finite refinement covering $X$ by our
assumption on $\mathcal{B}$. Pick $x \in X$,
$x \not \in \bigcup_{i \in I'} U_i$. Pick $i \in I$ with $x \in U_i$.
Pick $V_1, \ldots, V_n \in \mathcal{B}$ such that
$x \in V_1 \cap \ldots \cap V_n \subset U_i$. This is
possible as $\mathcal{B}$ is a subbasis for $X$. Note that
$V_j$ does not occur as a $U_i$. By maximality of the chosen
covering we see that for each $j$ there exist
$i_{j, 1}, \ldots, i_{j, n_j} \in I$ such that
$X = V_j \cup U_{i_{j, 1}} \cup \ldots \cup U_{i_{j, n_j}}$.
Since $V_1 \cap \ldots \cap V_n \subset U_i$ we conclude that
$X = U_i \cup \bigcup U_{i_{j, l}}$ a contradiction.
\end{proof}
\section{Locally quasi-compact spaces}
\label{section-locally-quasi-compact}
\noindent
Recall that a neighbourhood of a point need not be open.
\begin{definition}
\label{definition-locally-quasi-compact}
A topological space $X$ is called
{\it locally quasi-compact}\footnote{This may not be standard notation.
Alternative notions used in the literature are: (1) Every point has some
quasi-compact neighbourhood, and (2) Every point has a closed quasi-compact
neighbourhood. A scheme has the property that every point has a fundamental
system of open quasi-compact neighbourhoods.} if every
point has a fundamental system of quasi-compact neighbourhoods.
\end{definition}
\noindent
The term {\it locally compact space} in the literature often refers
to a space as in the following lemma.
\begin{lemma}
\label{lemma-locally-quasi-compact-Hausdorff}
A Hausdorff space is locally quasi-compact if and only if every point
has a quasi-compact neighbourhood.
\end{lemma}
\begin{proof}
Let $X$ be a Hausdorff space. Let $x \in X$ and let $x \in E \subset X$
be a quasi-compact neighbourhood. Then $E$ is closed by
Lemma \ref{lemma-quasi-compact-in-Hausdorff}.
Suppose that $x \in U \subset X$ is an open neighbourhood of $x$.
Then $Z = E \setminus U$ is a closed subset of $E$ not containing $x$.
Hence we can find a pair of disjoint open subsets $W, V \subset E$
of $E$ such that $x \in V$ and $Z \subset W$, see
Lemma \ref{lemma-quasi-compact-in-Hausdorff}.
It follows that $\overline{V} \subset E$ is a closed neighbourhood
of $x$ contained in $E \cap U$. Also $\overline{V}$ is quasi-compact
as a closed subset of $E$ (Lemma \ref{lemma-closed-in-quasi-compact}).
In this way we obtain a fundamental system of quasi-compact neighbourhoods
of $x$.
\end{proof}
\begin{lemma}[Baire category theorem]
\label{lemma-baire-category-locally-compact}
Let $X$ be a locally quasi-compact Hausdorff space.
Let $U_n \subset X$, $n \geq 1$ be dense open subsets. Then
$\bigcap_{n \geq 1} U_n$ is dense in $X$.
\end{lemma}
\begin{proof}
After replacing $U_n$ by $\bigcap_{i = 1, \ldots, n} U_i$
we may assume that $U_1 \supset U_2 \supset \ldots$.
Let $x \in X$. We will show that $x$ is in the closure of
$\bigcap_{n \geq 1} U_n$. Thus let $E$ be a neighbourhood of $x$.
To show that $E \cap \bigcap_{n \geq 1} U_n$ is nonempty we
may replace $E$ by a smaller neighbourhood. After replacing
$E$ by a smaller neighbourhood, we may assume that $E$ is quasi-compact.
\medskip\noindent
Set $x_0 = x$ and $E_0 = E$. Below, we will inductively choose
a point $x_i \in E_{i - 1} \cap U_i$ and a quasi-compact
neighbourhood $E_i$ of $x_i$ with $E_i \subset E_{i - 1} \cap U_i$.
Because $X$ is Hausdorff, the subsets $E_i \subset X$ are closed
(Lemma \ref{lemma-quasi-compact-in-Hausdorff}).
Since the $E_i$ are also nonempty we conclude that
$\bigcap_{i \geq 1} E_i$ is nonempty
(Lemma \ref{lemma-intersection-closed-in-quasi-compact}).
Since $\bigcap_{i \geq 1} E_i \subset E \cap \bigcap_{n \geq 1} U_n$
this proves the lemma.
\medskip\noindent
The base case $i = 0$ we have done above. Induction step.
Since $E_{i - 1}$ is a neighbourhood of $x_{i - 1}$ we can
find an open $x_{i - 1} \in W \subset E_{i - 1}$.
Since $U_i$ is dense in $X$
we see that $W \cap U_i$ is nonempty.
Pick any $x_i \in W \cap U_i$.
By definition of locally quasi-compact spaces we can
find a quasi-compact neighbourhood $E_i$ of $x_i$
contained in $W \cap U_i$. Then $E_i \subset E_{i - 1} \cap U_i$
as desired.
\end{proof}
\begin{lemma}
\label{lemma-relatively-compact-refinement}
Let $X$ be a Hausdorff and quasi-compact space.
Let $X = \bigcup_{i \in I} U_i$ be an open covering.
Then there exists an open covering $X = \bigcup_{i \in I} V_i$
such that $\overline{V_i} \subset U_i$ for all $i$.
\end{lemma}
\begin{proof}
Let $x \in X$. Choose an $i(x) \in I$ such that $x \in U_{i(x)}$.
Since $X \setminus U_{i(x)}$ and $\{x\}$ are disjoint closed
subsets of $X$, by Lemmas \ref{lemma-closed-in-quasi-compact} and
\ref{lemma-quasi-compact-in-Hausdorff}
there exists an open neighbourhood $U_x$ of $x$
whose closure is disjoint from $X \setminus U_{i(x)}$.
Thus $\overline{U_x} \subset U_{i(x)}$. Since $X$ is quasi-compact,
there is a finite list of points $x_1, \ldots, x_m$ such that
$X = U_{x_1} \cup \ldots \cup U_{x_m}$. Setting
$V_i = \bigcup_{i = i(x_j)} U_{x_j}$ the proof is finished.
\end{proof}
\begin{lemma}
\label{lemma-refine-covering}
Let $X$ be a Hausdorff and quasi-compact space.
Let $X = \bigcup_{i \in I} U_i$ be an open covering.
Suppose given an integer $p \geq 0$ and for every $(p + 1)$-tuple
$i_0, \ldots, i_p$ of $I$ an open covering
$U_{i_0} \cap \ldots \cap U_{i_p} = \bigcup W_{i_0 \ldots i_p, k}$.
Then there exists an open covering $X = \bigcup_{j \in J} V_j$
and a map $\alpha : J \to I$ such that $\overline{V_j} \subset U_{\alpha(j)}$
and such that each $V_{j_0} \cap \ldots \cap V_{j_p}$
is contained in $W_{\alpha(j_0) \ldots \alpha(j_p), k}$
for some $k$.
\end{lemma}
\begin{proof}
Since $X$ is quasi-compact, there is a reduction to the
case where $I$ is finite (details omitted).
We prove the result for $I$ finite by induction on $p$.
The base case $p = 0$ is immediate by taking a covering as in
Lemma \ref{lemma-relatively-compact-refinement}
refining the open covering $X = \bigcup W_{i_0, k}$.
\medskip\noindent
Induction step. Assume the lemma proven for $p - 1$.
For all $p$-tuples $i'_0, \ldots, i'_{p - 1}$ of $I$ let
$U_{i'_0} \cap \ldots \cap U_{i'_{p - 1}} =
\bigcup W_{i'_0 \ldots i'_{p - 1}, k}$
be a common refinement of the coverings
$U_{i_0} \cap \ldots \cap U_{i_p} = \bigcup W_{i_0 \ldots i_p, k}$
for those $(p + 1)$-tuples such that
$\{i'_0, \ldots, i'_{p - 1}\} = \{i_0, \ldots, i_p\}$ (equality of sets).
(There are finitely many of these as $I$ is finite.)
By induction there exists a solution for these opens, say
$X = \bigcup V_j$ and $\alpha : J \to I$.
At this point the covering $X = \bigcup_{j \in J} V_j$
and $\alpha$ satisfy $\overline{V_j} \subset U_{\alpha(j)}$
and each $V_{j_0} \cap \ldots \cap V_{j_p}$
is contained in $W_{\alpha(j_0) \ldots \alpha(j_p), k}$
for some $k$ if there is a repetition in $\alpha(j_0), \ldots, \alpha(j_p)$.
Of course, we may and do assume that $J$ is finite.
\medskip\noindent
Fix $i_0, \ldots, i_p \in I$ pairwise distinct. Consider $(p + 1)$-tuples
$j_0, \ldots, j_p \in J$ with $i_0 = \alpha(j_0), \ldots, i_p = \alpha(j_p)$
such that $V_{j_0} \cap \ldots \cap V_{j_p}$
is {\bf not} contained in $W_{\alpha(j_0) \ldots \alpha(j_p), k}$ for any $k$.
Let $N$ be the number of such $(p + 1)$-tuples. We will show how to decrease
$N$. Since
$$
\overline{V_{j_0}} \cap \ldots \cap \overline{V_{j_p}} \subset
U_{i_0} \cap \ldots \cap U_{i_p} = \bigcup W_{i_0 \ldots i_p, k}
$$
we find a finite set $K = \{k_1, \ldots, k_t\}$ such that the LHS
is contained in $\bigcup_{k \in K} W_{i_0 \ldots i_p, k}$.
Then we consider the open covering
$$
V_{j_0} =
(V_{j_0} \setminus (\overline{V_{j_1}} \cap \ldots \cap \overline{V_{j_p}}))
\cup (\bigcup\nolimits_{k \in K} V_{j_0} \cap W_{i_0 \ldots i_p, k})
$$
The first open on the RHS intersects $V_{j_1} \cap \ldots \cap V_{j_p}$
in the empty set and the other opens $V_{j_0, k}$ of the RHS
satisfy $V_{j_0, k} \cap V_{j_1} \ldots \cap V_{j_p} \subset
W_{\alpha(j_0) \ldots \alpha(j_p), k}$.
Set $J' = J \amalg K$. For $j \in J$ set $V'_j = V_j$ if $j \not = j_0$
and set $V'_{j_0} =
V_{j_0} \setminus (\overline{V_{j_1}} \cap \ldots \cap \overline{V_{j_p}})$.
For $k \in K$ set $V'_k = V_{j_0, k}$. Finally, the map $\alpha' : J' \to I$
is given by $\alpha$ on $J$ and maps every element of $K$ to $i_0$.
A simple check shows that $N$ has decreased by one under this replacement.
Repeating this procedure $N$ times we arrive at the situation where
$N = 0$.
\medskip\noindent
To finish the proof we argue by induction on the number $M$ of $(p + 1)$-tuples
$i_0, \ldots, i_p \in I$ with pairwise distinct entries for which there exists
a $(p + 1)$-tuple $j_0, \ldots, j_p \in J$ with
$i_0 = \alpha(j_0), \ldots, i_p = \alpha(j_p)$ such that
$V_{j_0} \cap \ldots \cap V_{j_p}$ is {\bf not} contained in
$W_{\alpha(j_0) \ldots \alpha(j_p), k}$ for any $k$. To do this, we
claim that the operation performed in the previous paragraph does not
increase $M$. This follows formally from the fact that the map
$\alpha' : J' \to I$ factors through a map $\beta : J' \to J$
such that $V'_{j'} \subset V_{\beta(j')}$.
\end{proof}
\begin{lemma}
\label{lemma-lift-covering-of-a-closed}
Let $X$ be a Hausdorff and locally quasi-compact space.
Let $Z \subset X$ be a quasi-compact (hence closed) subset.
Suppose given an integer $p \geq 0$, a set $I$, for every $i \in I$
an open $U_i \subset X$, and for every $(p + 1)$-tuple
$i_0, \ldots, i_p$ of $I$ an open
$W_{i_0 \ldots i_p} \subset U_{i_0} \cap \ldots \cap U_{i_p}$
such that
\begin{enumerate}
\item $Z \subset \bigcup U_i$, and
\item for every $i_0, \ldots, i_p$ we have
$W_{i_0 \ldots i_p} \cap Z = U_{i_0} \cap \ldots \cap U_{i_p} \cap Z$.
\end{enumerate}
Then there exist opens $V_i$ of $X$ such that
we have $Z \subset \bigcup V_i$,
for all $i$ we have $\overline{V_i} \subset U_i$, and
we have $V_{i_0} \cap \ldots \cap V_{i_p} \subset W_{i_0 \ldots i_p}$
for all $(p + 1)$-tuples $i_0, \ldots, i_p$.
\end{lemma}
\begin{proof}
Since $Z$ is quasi-compact, there is a reduction to the
case where $I$ is finite (details omitted).
Because $X$ is locally quasi-compact and $Z$ is quasi-compact,
we can find a neighbourhood $Z \subset E$ which is quasi-compact,
i.e., $E$ is quasi-compact and contains an open neighbourhood
of $Z$ in $X$. If we prove the result after replacing $X$ by $E$,
then the result follows. Hence we may assume $X$ is quasi-compact.
\medskip\noindent
We prove the result in case $I$ is finite and $X$ is quasi-compact
by induction on $p$. The base case is $p = 0$. In this case we have
$X = (X \setminus Z) \cup \bigcup W_i$. By
Lemma \ref{lemma-relatively-compact-refinement}
we can find a covering $X = V \cup \bigcup V_i$ by
opens $V_i \subset W_i$ and $V \subset X \setminus Z$
with $\overline{V_i} \subset W_i$ for all $i$. Then we see that
we obtain a solution of the problem posed by the lemma.
\medskip\noindent
Induction step. Assume the lemma proven for $p - 1$.
Set $W_{j_0 \ldots j_{p - 1}}$ equal to the intersection of
all $W_{i_0 \ldots i_p}$ with
$\{j_0, \ldots, j_{p - 1}\} = \{i_0, \ldots, i_p\}$ (equality of sets).
By induction there exists a solution for these opens, say
$V_i \subset U_i$.
It follows from our choice of $W_{j_0 \ldots j_{p - 1}}$ that we have
$V_{i_0} \cap \ldots \cap V_{i_p} \subset W_{i_0 \ldots i_p}$
for all $(p + 1)$-tuples $i_0, \ldots, i_p$ where $i_a = i_b$ for
some $0 \leq a < b \leq p$.
Thus we only need to modify our choice of
$V_i$ if $V_{i_0} \cap \ldots \cap V_{i_p} \not \subset W_{i_0 \ldots i_p}$
for some $(p + 1)$-tuple $i_0, \ldots, i_p$ with pairwise distinct elements.
In this case we have
$$
T =
\overline{V_{i_0} \cap \ldots \cap V_{i_p} \setminus W_{i_0 \ldots i_p}}
\subset
\overline{V_{i_0}} \cap \ldots \cap \overline{V_{i_p}} \setminus
W_{i_0 \ldots i_p}
$$
is a closed subset of $X$ contained in $U_{i_0} \cap \ldots \cap U_{i_p}$
not meeting $Z$. Hence we can replace $V_{i_0}$ by $V_{i_0} \setminus T$
to ``fix'' the problem. After repeating this finitely many times for each
of the problem tuples, the lemma is proven.
\end{proof}
\begin{lemma}
\label{lemma-lift-covering-of-quasi-compact-hausdorff-subset}
Let $X$ be a topological space. Let $Z \subset X$ be a quasi-compact subset
such that any two points of $Z$ have disjoint open neighbourhoods in $X$.
Suppose given an integer $p \geq 0$, a set $I$, for every $i \in I$
an open $U_i \subset X$, and for every $(p + 1)$-tuple
$i_0, \ldots, i_p$ of $I$ an open
$W_{i_0 \ldots i_p} \subset U_{i_0} \cap \ldots \cap U_{i_p}$
such that
\begin{enumerate}
\item $Z \subset \bigcup U_i$, and
\item for every $i_0, \ldots, i_p$ we have
$W_{i_0 \ldots i_p} \cap Z = U_{i_0} \cap \ldots \cap U_{i_p} \cap Z$.
\end{enumerate}
Then there exist opens $V_i$ of $X$ such that
\begin{enumerate}
\item $Z \subset \bigcup V_i$,
\item $V_i \subset U_i$ for all $i$,
\item $\overline{V_i} \cap Z \subset U_i$ for all $i$, and
\item $V_{i_0} \cap \ldots \cap V_{i_p} \subset W_{i_0 \ldots i_p}$
for all $(p + 1)$-tuples $i_0, \ldots, i_p$.
\end{enumerate}
\end{lemma}
\begin{proof}
Since $Z$ is quasi-compact, there is a reduction to the
case where $I$ is finite (details omitted).
We prove the result in case $I$ is finite by induction on $p$.
\medskip\noindent
The base case is $p = 0$.
For $z \in Z \cap U_i$ and $z' \in Z \setminus U_i$ there exist
disjoint opens $z \in V_{z, z'}$ and $z' \in W_{z, z'}$ of $X$.
Since $Z \setminus U_i$ is quasi-compact
(Lemma \ref{lemma-closed-in-quasi-compact}),
we can choose a finite nunber $z'_1, \ldots, z'_r$ such that
$Z \setminus U_i \subset W_{z, z'_1} \cup \ldots \cup W_{z, z'_r}$.
Then we see that
$V_z = U_{z, z'_1} \cap \ldots \cap U_{z, z'_r} \cap U_i$
is an open neighbourhood of $z$ contained in $U_i$
with the property that $\overline{V_z} \cap Z \subset U_i$.
Since $z$ and $i$ where arbitrary and since $Z$ is quasi-compact
we can find a finite list $z_1, i_1, \ldots, z_t, i_t$
and opens $V_{z_j} \subset U_{i_j}$ with
$\overline{V_{z_j}} \cap Z \subset U_{i_j}$
and $Z \subset \bigcup V_{z_j}$.
Then we can set $V_i = W_i \cap (\bigcup_{j : i = i_j} V_{z_j})$
to solve the problem for $p = 0$.
\medskip\noindent
Induction step. Assume the lemma proven for $p - 1$.
Set $W_{j_0 \ldots j_{p - 1}}$ equal to the intersection of
all $W_{i_0 \ldots i_p}$ with
$\{j_0, \ldots, j_{p - 1}\} = \{i_0, \ldots, i_p\}$ (equality of sets).
By induction there exists a solution for these opens, say
$V_i \subset U_i$.
It follows from our choice of $W_{j_0 \ldots j_{p - 1}}$ that we have
$V_{i_0} \cap \ldots \cap V_{i_p} \subset W_{i_0 \ldots i_p}$
for all $(p + 1)$-tuples $i_0, \ldots, i_p$ where $i_a = i_b$ for
some $0 \leq a < b \leq p$.
Thus we only need to modify our choice of
$V_i$ if $V_{i_0} \cap \ldots \cap V_{i_p} \not \subset W_{i_0 \ldots i_p}$
for some $(p + 1)$-tuple $i_0, \ldots, i_p$ with pairwise distinct elements.
In this case we have
$$
T =
\overline{V_{i_0} \cap \ldots \cap V_{i_p} \setminus W_{i_0 \ldots i_p}}
\subset
\overline{V_{i_0}} \cap \ldots \cap \overline{V_{i_p}} \setminus
W_{i_0 \ldots i_p}
$$
is a closed subset of $X$ not meeting $Z$ by our property (3) of the
opens $V_i$. Hence we can replace $V_{i_0}$ by $V_{i_0} \setminus T$
to ``fix'' the problem. After repeating this finitely many times for each
of the problem tuples, the lemma is proven.
\end{proof}
\section{Limits of spaces}
\label{section-limits}
\noindent
The category of topological spaces has products. Namely, if $I$ is a set
and for $i \in I$ we are given a topological space $X_i$ then we endow
$\prod_{i \in I} X_i$ with the {\it product topology}. As a basis for
the topology we use sets of the form $\prod U_i$ where $U_i \subset X_i$
is open and $U_i = X_i$ for almost all $i$.
\medskip\noindent
The category of topological spaces has equalizers. Namely, if
$a, b : X \to Y$ are morphisms of topological spaces, then the
equalizer of $a$ and $b$ is the subset $\{x \in X \mid a(x) = b(x)\} \subset X$
endowed with the induced topology.
\begin{lemma}
\label{lemma-limits}
The category of topological spaces has limits and the forgetful functor
to sets commutes with them.
\end{lemma}
\begin{proof}
This follows from the discussion above and
Categories, Lemma \ref{categories-lemma-limits-products-equalizers}.
It follows from the description above that the forgetful functor
commutes with limits. Another way to see this is to use
Categories, Lemma \ref{categories-lemma-adjoint-exact} and use that
the forgetful functor has a left adjoint, namely the functor which
assigns to a set the corresponding discrete topological space.
\end{proof}
\begin{lemma}
\label{lemma-describe-limits}
Let $\mathcal{I}$ be a cofiltered category. Let $i \mapsto X_i$ be a diagram
of topological spaces over $\mathcal{I}$. Let $X = \lim X_i$ be the limit
with projection maps $f_i : X \to X_i$.
\begin{enumerate}
\item Any open of $X$ is of the form $\bigcup_{j \in J} f_j^{-1}(U_j)$
for some subset $J \subset I$ and opens $U_j \subset X_j$.
\item Any quasi-compact open of $X$ is of the form
$f_i^{-1}(U_i)$ for some $i$ and some $U_i \subset X_i$ open.
\end{enumerate}
\end{lemma}
\begin{proof}
The construction of the limit given above shows that $X \subset \prod X_i$
with the induced topology. A basis for the topology of $\prod X_i$ are
the opens $\prod U_i$ where $U_i \subset X_i$ is open and $U_i = X_i$
for almost all $i$. Say $i_1, \ldots, i_n \in \Ob(\mathcal{I})$ are the
objects such that $U_{i_j} \not = X_{i_j}$. Then
$$
X \cap \prod U_i = f_{i_1}^{-1}(U_{i_1}) \cap \ldots \cap f_{i_n}^{-1}(U_{i_n})
$$
For a general limit of topological spaces these form a basis for the
topology on $X$. However, if $\mathcal{I}$ is cofiltered as in the statement
of the lemma, then we can pick a $j \in \Ob(\mathcal{I})$ and morphisms
$j \to i_l$, $l = 1, \ldots, n$. Let
$$
U_j =
(X_j \to X_{i_1})^{-1}(U_{i_1}) \cap \ldots \cap
(X_j \to X_{i_n})^{-1}(U_{i_n})
$$
Then it is clear that $X \cap \prod U_i = f_j^{-1}(U_j)$. Thus for any open
$W$ of $X$ there is a set $A$ and a map $\alpha : A \to \Ob(\mathcal{I})$ and
opens $U_a \subset X_{\alpha(a)}$ such that
$W = \bigcup f_{\alpha(a)}^{-1}(U_a)$. Set $J = \Im(\alpha)$ and
for $j \in J$ set $U_j = \bigcup_{\alpha(a) = j} U_a$ to see that
$W = \bigcup_{j \in J} f_j^{-1}(U_j)$.
This proves (1).
\medskip\noindent
To see (2) suppose that $\bigcup_{j \in J} f_j^{-1}(U_j)$ is quasi-compact.
Then it is equal to
$f_{j_1}^{-1}(U_{j_1}) \cup \ldots \cup f_{j_m}^{-1}(U_{j_m})$
for some $j_1, \ldots, j_m \in J$. Since $\mathcal{I}$ is cofiltered,
we can pick a $i \in \Ob(\mathcal{I})$ and morphisms
$i \to j_l$, $l = 1, \ldots, m$. Let
$$
U_i =
(X_i \to X_{j_1})^{-1}(U_{j_1}) \cup \ldots \cup
(X_i \to X_{j_m})^{-1}(U_{j_m})
$$
Then our open equals $f_i^{-1}(U_i)$ as desired.
\end{proof}
\begin{lemma}
\label{lemma-characterize-limit}
Let $\mathcal{I}$ be a cofiltered category. Let $i \mapsto X_i$ be a diagram
of topological spaces over $\mathcal{I}$. Let $X$ be a topological
space such that
\begin{enumerate}
\item $X = \lim X_i$ as a set (denote $f_i$ the projection maps),
\item the sets $f_i^{-1}(U_i)$ for $i \in \Ob(\mathcal{I})$ and
$U_i \subset X_i$ open form a basis for the topology of $X$.
\end{enumerate}
Then $X$ is the limit of the $X_i$ as a topological space.
\end{lemma}
\begin{proof}
Follows from the description of the limit topology
in Lemma \ref{lemma-describe-limits}.
\end{proof}
\begin{theorem}[Tychonov]
\label{theorem-tychonov}
A product of quasi-compact spaces is quasi-compact.
\end{theorem}
\begin{proof}
Let $I$ be a set and for $i \in I$ let $X_i$ be a quasi-compact topological
space. Set $X = \prod X_i$. Let $\mathcal{B}$ be the set of subsets of $X$
of the form $U_i \times \prod_{i' \in I, i' \not = i} X_{i'}$ where
$U_i \subset X_i$ is open. By construction this family is a subbasis
for the topology on $X$. By Lemma \ref{lemma-subbase-theorem} it
suffices to show that any covering $X = \bigcup_{j \in J} B_j$
by elements $B_j$ of $\mathcal{B}$ has a finite refinement.
We can decompose $J = \coprod J_i$ so that if $j \in J_i$, then
$B_j = U_j \times \prod_{i' \not = i} X_{i'}$ with $U_j \subset X_i$
open. If $X_i = \bigcup_{j \in J_i} U_j$, then there is a finite
refinement and we conclude that $X = \bigcup_{j \in J} B_j$
has a finite refinement. If this is not the case, then for every $i$
we can choose an point $x_i \in X_i$ which is not in
$\bigcup_{j \in J_i} U_j$. But then the point $x = (x_i)_{i \in I}$
is an element of $X$ not contained in $\bigcup_{j \in J} B_j$, a
contradiction.
\end{proof}
\noindent
The following lemma does not hold if one drops the assumption that
the spaces $X_i$ are Hausdorff, see
Examples, Section \ref{examples-section-lim-not-quasi-compact}.
\begin{lemma}
\label{lemma-inverse-limit-quasi-compact}
Let $\mathcal{I}$ be a category and let $i \mapsto X_i$
be a diagram over $\mathcal{I}$ in the category of topological
spaces. If each $X_i$ is quasi-compact and Hausdorff, then
$\lim X_i$ is quasi-compact.
\end{lemma}
\begin{proof}
Recall that $\lim X_i$ is a subspace of $\prod X_i$. By
Theorem \ref{theorem-tychonov} this product is quasi-compact. Hence it
suffices to show that $\lim X_i$ is a closed subspace of $\prod X_i$
(Lemma \ref{lemma-closed-in-quasi-compact}).
If $\varphi : j \to k$ is a morphism of $\mathcal{I}$, then
let $\Gamma_\varphi \subset X_j \times X_k$ denote the graph
of the corresponding continuous map $X_j \to X_k$. By
Lemma \ref{lemma-graph-closed} this graph is closed.
It is clear that $\lim X_i$ is the intersection of the
closed subsets
$$
\Gamma_\varphi \times \prod\nolimits_{l \not = j, k} X_l
\subset \prod X_i
$$
Thus the result follows.
\end{proof}
\noindent
The following lemma generalizes
Categories, Lemma \ref{categories-lemma-nonempty-limit}
and partially generalizes
Lemma \ref{lemma-intersection-closed-in-quasi-compact}.
\begin{lemma}
\label{lemma-nonempty-limit}
Let $\mathcal{I}$ be a cofiltered category and let $i \mapsto X_i$
be a diagram over $\mathcal{I}$ in the category of topological
spaces. If each $X_i$ is quasi-compact, Hausdorff, and nonempty, then
$\lim X_i$ is nonempty.
\end{lemma}
\begin{proof}
In the proof of Lemma \ref{lemma-inverse-limit-quasi-compact}
we have seen that $X = \lim X_i$ is the intersection of the
closed subsets
$$
Z_\varphi = \Gamma_\varphi \times \prod\nolimits_{l \not = j, k} X_l
$$
inside the quasi-compact space $\prod X_i$ where $\varphi : j \to k$
is a morphism of $\mathcal{I}$ and $\Gamma_\varphi \subset X_j \times X_k$
is the graph of the corresponding morphism $X_j \to X_k$. Hence by
Lemma \ref{lemma-intersection-closed-in-quasi-compact}
it suffices to show any finite intersection of these subsets is nonempty.
Assume $\varphi_t : j_t \to k_t$, $t = 1, \ldots, n$ is a finite collection
of morphisms of $\mathcal{I}$. Since $\mathcal{I}$ is cofiltered, we can
pick an object $j$ and a morphism $\psi_t : j \to j_t$ for each $t$.
For each pair $t, t'$ such that either (a) $j_t = j_{t'}$, or
(b) $j_t = k_{t'}$, or (c) $k_t = k_{t'}$ we obtain two morphisms
$j \to l$ with $l = j_t$ in case (a), (b) or $l = k_t$ in case (c).
Because $\mathcal{I}$ is cofiltered and since there are finitely
many pairs $(t, t')$ we may choose a map $j' \to j$ which equalizes
these two morphisms for all such pairs $(t, t')$. Pick an element
$x \in X_{j'}$ and for each $t$ let $x_{j_t}$, resp.\ $x_{k_t}$
be the image of $x$ under the morphism $X_{j'} \to X_j \to X_{j_t}$,
resp.\ $X_{j'} \to X_j \to X_{j_t} \to X_{k_t}$.
For any index $l \in \Ob(\mathcal{I})$ which is not equal to
$j_t$ or $k_t$ for some $t$ we pick an arbitrary element $x_l \in X_l$
(using the axiom of choice). Then $(x_i)_{i \in \Ob(\mathcal{I})}$
is in the intersection
$$
Z_{\varphi_1} \cap \ldots \cap Z_{\varphi_n}
$$
by construction and the proof is complete.
\end{proof}
\section{Constructible sets}
\label{section-constructible}
\begin{definition}
\label{definition-constructible}
Let $X$ be a topological space. Let $E \subset X$ be a subset of $X$.
\begin{enumerate}
\item We say $E$ is {\it constructible}\footnote{In the second edition
of EGA I \cite{EGA1-second} this was called a ``globally constructible''
set and a the terminology ``constructible'' was used for what we call a locally
constructible set.}
in $X$ if $E$ is a finite union
of subsets of the form $U \cap V^c$ where $U, V \subset X$ are open and
retrocompact in $X$.
\item We say $E$ is {\it locally constructible} in $X$ if there exists an open
covering $X = \bigcup V_i$ such that each $E \cap V_i$ is constructible
in $V_i$.
\end{enumerate}
\end{definition}
\begin{lemma}
\label{lemma-constructible}
The collection of constructible sets is closed under
finite intersections, finite unions and complements.
\end{lemma}
\begin{proof}
Note that if $U_1$, $U_2$ are open and retrocompact in $X$
then so is $U_1 \cup U_2$ because the union of two quasi-compact
subsets of $X$ is quasi-compact. It is also true that
$U_1 \cap U_2$ is retrocompact. Namely, suppose $U \subset X$
is quasi-compact open, then $U_2 \cap U$ is quasi-compact because
$U_2$ is retrocompact in $X$, and then we conclude
$U_1 \cap (U_2 \cap U)$ is quasi-compact because $U_1$ is
retrocompact in $X$. From this it is formal to show that
the complement of a constructible set is constructible,
that finite unions of constructibles are constructible, and
that finite intersections of constructibles are constructible.
\end{proof}
\begin{lemma}
\label{lemma-inverse-images-constructibles}
Let $f : X \to Y$ be a continuous map of topological spaces.
If the inverse image of every retrocompact open subset of $Y$
is retrocompact in $X$, then inverse images of constructible
sets are constructible.
\end{lemma}
\begin{proof}
This is true because $f^{-1}(U \cap V^c) = f^{-1}(U) \cap f^{-1}(V)^c$,
combined with the definition of constructible sets.
\end{proof}
\begin{lemma}
\label{lemma-open-immersion-constructible-inverse-image}
Let $U \subset X$ be open. For a constructible set
$E \subset X$ the intersection $E \cap U$ is constructible
in $U$.
\end{lemma}
\begin{proof}
Suppose that $V \subset X$ is retrocompact open in $X$.
It suffices to show that $V \cap U$ is retrocompact in $U$
by Lemma \ref{lemma-inverse-images-constructibles}. To show this
let $W \subset U$ be open and quasi-compact. Then $W$
is open and quasi-compact in $X$. Hence $V \cap W = V \cap U \cap W$
is quasi-compact as $V$ is retrocompact in $X$.
\end{proof}
\begin{lemma}
\label{lemma-quasi-compact-open-immersion-constructible-image}
Let $U \subset X$ be a retrocompact open. Let $E \subset U$.
If $E$ is constructible in $U$, then $E$ is constructible in $X$.
\end{lemma}
\begin{proof}
Suppose that $V, W \subset U$ are retrocompact open in $U$.
Then $V, W$ are retrocompact open in $X$
(Lemma \ref{lemma-composition-quasi-compact}).
Hence $V \cap (U \setminus W) = V \cap (X \setminus W)$
is constructible in $X$. We conclude since every constructible subset of $U$
is a finite union of subsets of the form $V \cap (U \setminus W)$.
\end{proof}
\begin{lemma}
\label{lemma-collate-constructible}
Let $X$ be a topological space. Let $E \subset X$ be a subset.
Let $X = V_1 \cup \ldots \cup V_m$ be a finite covering by
retrocompact opens.
Then $E$ is constructible in $X$ if and only if $E \cap V_j$
is constructible in $V_j$ for each $j = 1, \ldots, m$.
\end{lemma}
\begin{proof}
If $E$ is constructible in $X$, then by
Lemma \ref{lemma-open-immersion-constructible-inverse-image}
we see that $E \cap V_j$ is constructible in $V_j$ for all $j$.
Conversely, suppose that $E \cap V_j$
is constructible in $V_j$ for each $j = 1, \ldots, m$.
Then $E = \bigcup E \cap V_j$ is a finite union of
constructible sets by
Lemma \ref{lemma-quasi-compact-open-immersion-constructible-image}
and hence constructible.
\end{proof}
\begin{lemma}
\label{lemma-intersect-constructible-with-closed}
Let $X$ be a topological space. Let $Z \subset X$ be a closed
subset such that $X \setminus Z$ is quasi-compact.
Then for a constructible set $E \subset X$ the intersection
$E \cap Z$ is constructible in $Z$.
\end{lemma}
\begin{proof}
Suppose that $V \subset X$ is retrocompact open in $X$.
It suffices to show that $V \cap Z$ is retrocompact in $Z$
by Lemma \ref{lemma-inverse-images-constructibles}. To show this
let $W \subset Z$ be open and quasi-compact. The subset
$W' = W \cup (X \setminus Z)$ is quasi-compact, open, and $W = Z \cap W'$.
Hence $V \cap Z \cap W = V \cap Z \cap W'$
is a closed subset of the quasi-compact open $V \cap W'$
as $V$ is retrocompact in $X$. Thus $V \cap Z \cap W$ is quasi-compact
by Lemma \ref{lemma-closed-in-quasi-compact}.
\end{proof}
\begin{lemma}
\label{lemma-intersect-constructible-with-retrocompact}
Let $X$ be a topological space. Let $T \subset X$ be a subset. Suppose
\begin{enumerate}
\item $T$ is retrocompact in $X$,
\item quasi-compact opens form a basis for the topology on $X$.
\end{enumerate}
Then for a constructible set $E \subset X$ the intersection $E \cap T$ is
constructible in $T$.
\end{lemma}
\begin{proof}
Suppose that $V \subset X$ is retrocompact open in $X$.
It suffices to show that $V \cap T$ is retrocompact in $T$
by Lemma \ref{lemma-inverse-images-constructibles}. To show this
let $W \subset T$ be open and quasi-compact. By assumption (2)
we can find a quasi-compact open $W' \subset X$
such that $W = T \cap W'$ (details omitted).
Hence $V \cap T \cap W = V \cap T \cap W'$
is the intersection of $T$ with the quasi-compact open $V \cap W'$
as $V$ is retrocompact in $X$. Thus $V \cap T \cap W$ is quasi-compact.
\end{proof}
\begin{lemma}
\label{lemma-closed-constructible-image}
Let $Z \subset X$ be a closed subset whose complement is retrocompact open.
Let $E \subset Z$. If $E$ is constructible in $Z$, then $E$ is constructible
in $X$.
\end{lemma}
\begin{proof}
Suppose that $V \subset Z$ is retrocompact open in $Z$. Consider the open
subset $\tilde V = V \cup (X \setminus Z)$ of $X$. Let $W \subset X$ be
quasi-compact open. Then
$$
W \cap \tilde V =
\left(V \cap W\right) \cup \left((X \setminus Z) \cap W\right).
$$
The first part is quasi-compact as $V \cap W = V \cap (Z \cap W)$ and
$(Z \cap W)$ is quasi-compact open in $Z$
(Lemma \ref{lemma-closed-in-quasi-compact}) and $V$ is retrocompact in $Z$.
The second part is quasi-compact as $(X \setminus Z)$ is retrocompact in $X$.
In this way we see that $\tilde V$ is retrocompact in $X$.
Thus if $V_1, V_2 \subset Z$ are retrocompact open, then
$$
V_1 \cap (Z \setminus V_2) = \tilde V_1 \cap (X \setminus \tilde V_2)
$$
is constructible in $X$. We conclude since every constructible subset of $Z$
is a finite union of subsets of the form $V_1 \cap (Z \setminus V_2)$.
\end{proof}
\begin{lemma}
\label{lemma-constructible-is-retrocompact}
Let $X$ be a topological space. Every constructible
subset of $X$ is retrocompact.
\end{lemma}
\begin{proof}
Let $E = \bigcup_{i = 1, \ldots, n} U_i \cap V_i^c$ with $U_i, V_i$
retrocompact open in $X$. Let $W \subset X$ be quasi-compact open.
Then $E \cap W = \bigcup_{i = 1, \ldots, n} U_i \cap V_i^c \cap W$.
Thus it suffices to show that $U \cap V^c \cap W$ is quasi-compact
if $U, V$ are retrocompact open and $W$ is quasi-compact
open. This is true because $U \cap V^c \cap W$ is a closed
subset of the quasi-compact $U \cap W$ so
Lemma \ref{lemma-closed-in-quasi-compact}
applies.
\end{proof}
\noindent
Question: Does the following lemma also hold if we assume $X$ is a
quasi-compact topological space? Compare with
Lemma \ref{lemma-intersect-constructible-with-closed}.
\begin{lemma}
\label{lemma-intersect-constructible-with-constructible}
Let $X$ be a topological space. Assume
$X$ has a basis consisting of quasi-compact opens.
For $E, E'$ constructible in $X$, the intersection
$E \cap E'$ is constructible in $E$.
\end{lemma}
\begin{proof}
Combine Lemmas \ref{lemma-intersect-constructible-with-retrocompact} and
\ref{lemma-constructible-is-retrocompact}.
\end{proof}
\begin{lemma}
\label{lemma-constructible-in-constructible}
Let $X$ be a topological space. Assume
$X$ has a basis consisting of quasi-compact opens.
Let $E$ be constructible in $X$ and $F \subset E$ constructible in $E$.
Then $F$ is constructible in $X$.
\end{lemma}
\begin{proof}
Observe that any retrocompact subset $T$ of $X$ has a basis for the induced
topology consisting of quasi-compact opens. In particular this holds
for any constructible subset
(Lemma \ref{lemma-constructible-is-retrocompact}).
Write $E = E_1 \cup \ldots \cup E_n$ with $E_i = U_i \cap V_i^c$
where $U_i, V_i \subset X$ are retrocompact open.
Note that $E_i = E \cap E_i$ is constructible in $E$ by
Lemma \ref{lemma-intersect-constructible-with-constructible}.
Hence $F \cap E_i$ is constructible in $E_i$ by
Lemma \ref{lemma-intersect-constructible-with-constructible}.
Thus it suffices to prove the lemma in case $E = U \cap V^c$
where $U, V \subset X$ are retrocompact open.
In this case the inclusion $E \subset X$ is a composition
$$
E = U \cap V^c \to U \to X
$$
Then we can apply Lemma \ref{lemma-closed-constructible-image}
to the first inclusion and
Lemma \ref{lemma-quasi-compact-open-immersion-constructible-image}
to the second.
\end{proof}
\begin{lemma}
\label{lemma-collate-constructible-from-constructible}
Let $X$ be a topological space which has a basis for the topology
consisting of quasi-compact opens. Let $E \subset X$ be a subset.
Let $X = E_1 \cup \ldots \cup E_m$ be a finite covering by constructible
subsets. Then $E$ is constructible in $X$ if and only if $E \cap E_j$
is constructible in $E_j$ for each $j = 1, \ldots, m$.
\end{lemma}
\begin{proof}
Combine
Lemmas \ref{lemma-intersect-constructible-with-constructible} and
\ref{lemma-constructible-in-constructible}.
\end{proof}
\begin{lemma}
\label{lemma-generic-point-in-constructible}
Let $X$ be a topological space. Suppose that
$Z \subset X$ is irreducible. Let $E \subset X$
be a finite union of locally closed subsets (e.g.\ $E$
is constructible). The following are equivalent
\begin{enumerate}
\item The intersection $E \cap Z$ contains an open
dense subset of $Z$.
\item The intersection $E \cap Z$ is dense in $Z$.
\end{enumerate}
If $Z$ has a generic point $\xi$, then this is
also equivalent to
\begin{enumerate}
\item[(3)] We have $\xi \in E$.
\end{enumerate}
\end{lemma}
\begin{proof}
Write $E = \bigcup U_i \cap Z_i$ as the finite union of
intersections of open sets $U_i$ and closed sets $Z_i$.
Suppose that $E \cap Z$ is dense in $Z$. Note that
the closure of $E \cap Z$ is the union of the closures
of the intersections $U_i \cap Z_i \cap Z$. As $Z$ is irreducible we
conclude that the closure of $U_i \cap Z_i \cap Z$ is $Z$ for some $i$.
Fix such an $i$. It follows that $Z \subset Z_i$ since otherwise
the closed subset $Z \cap Z_i$ of $Z$ would not be dense in $Z$.
Then $U_i \cap Z_i \cap Z = U_i \cap Z$ is an open nonempty subset of $Z$.
Because $Z$ is irreducible, it is open dense. Hence $E \cap Z$
contains an open dense subset of $Z$.
The converse is obvious.
\medskip\noindent
Suppose that $\xi \in Z$ is a generic point. Of course if
(1) $\Leftrightarrow$ (2) holds, then $\xi \in E$. Conversely,
if $\xi \in E$, then $\xi \in U_i \cap Z_i$ for some $i = i_0$.
Clearly this implies $Z \subset Z_{i_0}$ and hence
$U_{i_0} \cap Z_{i_0} \cap Z = U_{i_0} \cap Z$ is an open
not empty subset of $Z$. We conclude as before.
\end{proof}
\section{Constructible sets and Noetherian spaces}
\label{section-constructible-Noetherian}
\begin{lemma}
\label{lemma-constructible-Noetherian-space}
Let $X$ be a Noetherian topological space.
The constructible sets in $X$ are precisely the finite unions
of locally closed subsets of $X$.
\end{lemma}
\begin{proof}
This follows immediately from
Lemma \ref{lemma-Noetherian-quasi-compact}.
\end{proof}
\begin{lemma}
\label{lemma-constructible-map-Noetherian}
Let $f : X \to Y$ be a continuous map of Noetherian topological spaces.
If $E \subset Y$ is constructible in $Y$, then $f^{-1}(E)$ is constructible
in $X$.
\end{lemma}
\begin{proof}
Follows immediately from
Lemma \ref{lemma-constructible-Noetherian-space}
and the definition of a continuous map.
\end{proof}
\begin{lemma}
\label{lemma-characterize-constructible-Noetherian}
Let $X$ be a Noetherian topological space.
Let $E \subset X$ be a subset.
The following are equivalent:
\begin{enumerate}
\item $E$ is constructible in $X$, and
\item for every irreducible closed $Z \subset X$ the intersection
$E \cap Z$ either contains a nonempty open of $Z$ or is not dense in $Z$.
\end{enumerate}
\end{lemma}
\begin{proof}
Assume $E$ is constructible and $Z \subset X$ irreducible closed.
Then $E \cap Z$ is constructible in $Z$ by
Lemma \ref{lemma-constructible-map-Noetherian}.
Hence $E \cap Z$ is a finite union of nonempty locally closed subsets
$T_i$ of $Z$. Clearly if none of the $T_i$ is open in $Z$, then
$E \cap Z$ is not dense in $Z$. In this way we see that (1) implies (2).
\medskip\noindent
Conversely, assume (2) holds. Consider the set $\mathcal{S}$ of closed
subsets $Y$ of $X$ such that $E \cap Y$ is not constructible in $Y$.
If $\mathcal{S} \not = \emptyset$, then it has a smallest element $Y$
as $X$ is Noetherian.
Let $Y = Y_1 \cup \ldots \cup Y_r$ be the decomposition of $Y$ into its
irreducible components, see
Lemma \ref{lemma-Noetherian}.
If $r > 1$, then each $Y_i \cap E$ is constructible in $Y_i$ and hence
a finite union of locally closed subsets of $Y_i$. Thus $E \cap Y$
is a finite union of locally closed subsets of $Y$ too and we conclude
that $E \cap Y$ is constructible in $Y$ by
Lemma \ref{lemma-constructible-Noetherian-space}.
This is a contradiction and so $r = 1$. If $r = 1$, then $Y$ is
irreducible, and by assumption (2) we see that $E \cap Y$ either
(a) contains an open $V$ of $Y$ or (b) is not dense in $Y$.
In case (a) we see, by minimality of $Y$, that $E \cap (Y \setminus V)$
is a finite union of locally closed subsets of $Y \setminus V$. Thus
$E \cap Y$ is a finite union of locally closed subsets of $Y$ and is
constructible by
Lemma \ref{lemma-constructible-Noetherian-space}.
This is a contradiction and so we must be in case (b).
In case (b) we see that $E \cap Y = E \cap Y'$ for some proper closed
subset $Y' \subset Y$. By minimality of $Y$ we see that
$E \cap Y'$ is a finite union of locally closed subsets of $Y'$ and
we see that $E \cap Y' = E \cap Y$ is a finite union of locally closed
subsets of $Y$ and is constructible by
Lemma \ref{lemma-constructible-Noetherian-space}.
This contradiction finishes the proof of the lemma.
\end{proof}
\begin{lemma}
\label{lemma-constructible-neighbourhood-Noetherian}
Let $X$ be a Noetherian topological space.
Let $x \in X$.
Let $E \subset X$ be constructible in $X$.
The following are equivalent:
\begin{enumerate}
\item $E$ is a neighbourhood of $x$, and
\item for every irreducible closed subset $Y$ of $X$ which contains
$x$ the intersection $E \cap Y$ is dense in $Y$.
\end{enumerate}
\end{lemma}
\begin{proof}
It is clear that (1) implies (2). Assume (2).
Consider the set $\mathcal{S}$ of closed subsets $Y$ of $X$ containing $x$
such that $E \cap Y$ is not a neighbourhood of $x$ in $Y$.
If $\mathcal{S} \not = \emptyset$, then it has a minimal element $Y$
as $X$ is Noetherian. Suppose $Y = Y_1 \cup Y_2$ with two smaller nonempty
closed subsets $Y_1$, $Y_2$. If $x \in Y_i$ for $i = 1, 2$, then $Y_i \cap E$
is a neighbourhood of $x$ in $Y_i$ and we conclude $Y \cap E$ is a
neighbourhood of $x$ in $Y$ which is a contradiction. If $x \in Y_1$ but
$x \not\in Y_2$ (say), then $Y_1 \cap E$ is a neighbourhood of $x$ in
$Y_1$ and hence also in $Y$, which is a contradiction as well.
We conclude that $Y$ is irreducible closed. By assumption (2) we see that
$E \cap Y$ is dense in $Y$. Thus $E \cap Y$ contains an open $V$ of $Y$, see
Lemma \ref{lemma-characterize-constructible-Noetherian}.
If $x \in V$ then $E \cap Y$ is a neighbourhood of $x$ in $Y$ which
is a contradiction. If $x \not \in V$, then $Y' = Y \setminus V$ is a
proper closed subset of $Y$ containing $x$. By minimality of $Y$
we see that $E \cap Y'$ contains an open neighbourhood $V' \subset Y'$
of $x$ in $Y'$. But then $V' \cup V$ is an open neighbourhood of $x$
in $Y$ contained in $E$, a contradiction.
This contradiction finishes the proof of the lemma.
\end{proof}
\begin{lemma}
\label{lemma-characterize-open-Noetherian}
Let $X$ be a Noetherian topological space.
Let $E \subset X$ be a subset.
The following are equivalent:
\begin{enumerate}
\item $E$ is open in $X$, and
\item for every irreducible closed subset $Y$ of $X$
the intersection $E \cap Y$ is either empty or
contains a nonempty open of $Y$.
\end{enumerate}
\end{lemma}
\begin{proof}
This follows formally from
Lemmas \ref{lemma-characterize-constructible-Noetherian} and
\ref{lemma-constructible-neighbourhood-Noetherian}.
\end{proof}
\section{Characterizing proper maps}
\label{section-proper}
\noindent
We include a section discussing the notion of a proper map in usual
topology. It turns out that in topology, the notion of being proper
is the same as the notion of being universally closed, in the sense
that any base change is a closed morphism (not just taking products
with spaces). The reason for doing this is that in algebraic geometry
we use this notion of universal closedness as the basis for our
definition of properness.
\begin{lemma}[Tube lemma]
\label{lemma-tube}
Let $X$ and $Y$ be topological spaces.
Let $A \subset X$ and $B \subset Y$ be quasi-compact subsets.
Let $A \times B \subset W \subset X \times Y$ with $W$
open in $X \times Y$. Then there exists opens $A \subset U \subset X$
and $B \subset V \subset Y$ such that $U \times V \subset W$.
\end{lemma}
\begin{proof}
For every $a \in A$ and $b \in B$ there exist opens
$U_{(a, b)}$ of $X$ and $V_{(a, b)}$ of $Y$ such that
$(a, b) \in U_{(a, b)} \times V_{(a, b)} \subset W$.
Fix $b$ and we see there exist a finite number $a_1, \ldots, a_n$
such that $A \subset U_{(a_1, b)} \cup \ldots \cup U_{(a_n, b)}$.
Hence
$$
A \times \{b\} \subset
(U_{(a_1, b)} \cup \ldots \cup U_{(a_n, b)}) \times
(V_{(a_1, b)} \cap \ldots \cap V_{(a_n, b)}) \subset W.
$$
Thus for every $b \in B$ there exists opens $U_b \subset X$ and
$V_b \subset Y$ such that $A \times \{b\} \subset U_b \times V_b \subset W$.
As above there exist a finite number $b_1, \ldots, b_m$ such
that $B \subset V_{b_1} \cup \ldots \cup V_{b_m}$.
Then we win because
$A \times B \subset
(U_{b_1} \cap \ldots \cap U_{b_m}) \times
(V_{b_1} \cup \ldots \cup V_{b_m})$.
\end{proof}
\noindent
The notation in the following definition may be slightly different
from what you are used to.
\begin{definition}
\label{definition-proper-map}
Let $f : X\to Y$ be a continuous map between topological spaces.
\begin{enumerate}
\item We say that the map $f$ is {\it closed}
iff the image of every closed subset is closed.
\item We say that the map $f$ is {\it proper}\footnote{This is the
terminology used in \cite{Bourbaki}. Usually this is what
is called ``universally closed'' in the literature. Thus our notion
of proper does not involve any separation conditions.} iff
the map $Z \times X\to Z \times Y$ is closed for any topological space
$Z$.
\item We say that the map $f$ is {\it quasi-proper} iff
the inverse image $f^{-1}(V)$ of every quasi-compact subset $V \subset Y$
is quasi-compact.
\item We say that $f$ is {\it universally closed} iff
the map $f': Z \times_Y X \to Z$ is closed for any map $g: Z \to Y$.
\end{enumerate}
\end{definition}
\noindent
The following lemma is useful later.
\begin{lemma}
\label{lemma-characterize-quasi-compact}
\begin{reference}
Combination of
\cite[I, p. 75, Lemme 1]{Bourbaki} and
\cite[I, p. 76, Corrolaire 1]{Bourbaki}.
\end{reference}
A topological space $X$ is quasi-compact if and only if the
projection map $Z \times X \to Z$ is closed for
any topological space $Z$.
\end{lemma}
\begin{proof}
(See also remark below.)
If $X$ is not quasi-compact, there exists an open covering
$X = \bigcup_{i \in I} U_i$ such that no finite
number of $U_i$ cover $X$.
Let $Z$ be the subset of the power set $\mathcal{P}(I)$ of $I$
consisting of $I$ and all nonempty finite subsets of $I$.
Define a topology on $Z$ with as a basis for the topology
the following sets:
\begin{enumerate}
\item All subsets of $Z\setminus\{I\}$.
\item For every finite subset $K$ of $I$ the set
$U_K := \{J\subset I \mid J \in Z, \ K\subset J \})$.
\end{enumerate}
It is left to the reader to verify this is the basis for a topology.
Consider the subset of $Z \times X$ defined by the formula
$$
M = \{(J, x) \mid J \in Z, \ x \in \bigcap\nolimits_{i \in J} U_i^c)\}
$$
If $(J, x) \not \in M$, then $x \in U_i$ for some $i \in J$.
Hence $U_{\{i\}} \times U_i \subset Z \times X$ is an open
subset containing $(J, x)$ and not intersecting $M$. Hence
$M$ is closed. The projection of $M$ to $Z$ is $Z-\{I\}$
which is not closed. Hence $Z \times X \to Z$ is not closed.
\medskip\noindent
Assume $X$ is quasi-compact. Let $Z$ be a topological space.
Let $M \subset Z \times X$ be closed. Let $z \in Z$ be a point
which is not in $\text{pr}_1(M)$. By the Tube Lemma \ref{lemma-tube}
there exists an open $U \subset Z$ such that $U \times X$ is
contained in the complement of $M$. Hence $\text{pr}_1(M)$ is closed.
\end{proof}
\begin{remark}
\label{remark-lemma-literature}
Lemma \ref{lemma-characterize-quasi-compact} is a combination of
\cite[I, p. 75, Lemme 1]{Bourbaki} and
\cite[I, p. 76, Corollaire 1]{Bourbaki}.
\end{remark}
\begin{theorem}
\label{theorem-characterize-proper}
\begin{reference}
In \cite[I, p. 75, Theorem 1]{Bourbaki} you can find:
(2) $\Leftrightarrow$ (4).
In \cite[I, p. 77, Proposition 6]{Bourbaki} you can find:
(2) $\Rightarrow$ (1).
\end{reference}
Let $f: X\to Y$ be a continuous map between
topological spaces. The following conditions are equivalent:
\begin{enumerate}
\item The map $f$ is quasi-proper and closed.
\item The map $f$ is proper.
\item The map $f$ is universally closed.
\item The map $f$ is closed and $f^{-1}(y)$ is quasi-compact for any
$y\in Y$.
\end{enumerate}
\end{theorem}
\begin{proof}
(See also the remark below.)
If the map $f$ satisfies (1), it automatically satisfies (4) because
any single point is quasi-compact.
\medskip\noindent
Assume map $f$ satisfies (4).
We will prove it is universally closed, i.e., (3) holds.
Let $g : Z \to Y$ be a continuous map of topological spaces
and consider the diagram
$$
\xymatrix{
Z \times_Y X \ar[r]_{g'} \ar[d]_{f'} & X \ar[d]^f \\
Z \ar[r]^g & Y
}
$$
During the proof we will use that $Z \times_Y X \to Z \times X$
is a homeomorphism onto its image, i.e., that we may identify
$Z \times_Y X$ with the corresponding subset of $Z \times X$ with
the induced topology.
The image of $f' : Z \times_Y X \to Z$ is
$\Im(f') = \{z : g(z) \in f(X)\}$.
Because $f(X)$ is closed, we see that
$\Im(f')$ is a closed subspace of $Z$.
Consider a closed subset $P \subset Z \times_Y X$.
Let $z \in Z$, $z \not \in f'(P)$.
If $z \not \in \Im(f')$, then $Z \setminus \Im(f')$
is an open neighbourhood which avoids $f'(P)$.
If $z$ is in $\Im(f')$
then $(f')^{-1}\{z\} = \{z\} \times f^{-1}\{g(z)\}$
and $f^{-1}\{g(z)\}$
is quasi-compact by assumption. Because $P$ is a closed
subset of $Z \times_Y X$, we have a closed $P'$ of $Z \times X$ such
that $P = P' \cap Z \times_Y X$.
Since $(f')^{-1}\{z\}$ is a subset of $P^c = P'^c \cup (Z \times_Y X)^c$,
and since $(f')^{-1}\{z\}$ is disjoint from $(Z \times_Y X)^c$
we see that $(f')^{-1}\{z\}$ is contained in $P'^c$.
We may apply the Tube Lemma \ref{lemma-tube} to
$(f')^{-1}\{z\} = \{z\} \times f^{-1}\{g(z)\}
\subset (P')^c \subset Z \times X$.
This gives $V \times U$ containing
$(f')^{-1}\{z\}$ where $U$ and $V$ are open sets in $X$ and $Z$
respectively and $V \times U$ has empty intersection with $P'$.
Then the set $V \cap g^{-1}(Y-f(U^c))$ is open in $Z$ since $f$
is closed, contains $z$, and has empty intersection with the image of $P$.
Thus $f'(P)$ is closed. In other words, the map $f$ is universally closed.
\medskip\noindent
The implication (3) $\Rightarrow$ (2) is trivial.
Namely, given any topological space $Z$ consider the projection
morphism $g : Z \times Y \to Y$. Then it is easy to see
that $f'$ is the map $Z \times X \to Z \times Y$, in other
words that $(Z \times Y) \times_Y X = Z \times X$. (This identification
is a purely categorical property having nothing to do with
topological spaces per se.)
\medskip\noindent
Assume $f$ satisfies (2). We will prove it satisfies (1).
Note that $f$ is closed as $f$ can be identified with the map
$\{pt\} \times X \to \{pt\} \times Y$ which is assumed closed.
Choose any quasi-compact subset $K \subset Y$.
Let $Z$ be any topological space.
Because $Z \times X \to Z \times Y$ is closed
we see the map $Z \times f^{-1}(K) \to Z \times K$
is closed (if $T$ is closed in $Z \times f^{-1}(K)$, write
$T = Z \times f^{-1}(K) \cap T'$ for some closed
$T' \subset Z \times X$). Because $K$ is quasi-compact,
$K \times Z\to Z$ is closed by Lemma \ref{lemma-characterize-quasi-compact}.
Hence the composition $Z \times f^{-1}(K)\to Z \times K \to Z$
is closed and therefore $f^{-1}(K)$ must be quasi-compact
by Lemma \ref{lemma-characterize-quasi-compact} again.
\end{proof}
\begin{remark}
\label{remark-proof-literature}
Here are some references to the literature.
In \cite[I, p. 75, Theorem 1]{Bourbaki} you can find:
(2) $\Leftrightarrow$ (4).
In \cite[I, p. 77, Proposition 6]{Bourbaki} you can find:
(2) $\Rightarrow$ (1).
Of course, trivially we have (1) $\Rightarrow$ (4).
Thus (1), (2) and (4) are equivalent.
Fan Zhou claimed and proved that (3) and (4) are equivalent;
let me know if you find a reference in the literature.
\end{remark}
\begin{lemma}
\label{lemma-closed-map}
\begin{slogan}
A map from a compact space to a Hausdorff space is a proper.
\end{slogan}
Let $f : X \to Y$ be a continuous map of topological spaces.
If $X$ is quasi-compact and $Y$ is Hausdorff, then $f$ is proper.
\end{lemma}
\begin{proof}
Since every point of $Y$ is closed, we see from
Lemma \ref{lemma-closed-in-quasi-compact}
that the closed subset $f^{-1}(y)$ of $X$ is quasi-compact for all $y \in Y$.
Thus, by Theorem \ref{theorem-characterize-proper}
it suffices to show that $f$ is closed.
If $E \subset X$ is closed, then it is quasi-compact
(Lemma \ref{lemma-closed-in-quasi-compact}),
hence $f(E) \subset Y$ is quasi-compact
(Lemma \ref{lemma-image-quasi-compact}),
hence $f(E)$ is closed in $Y$
(Lemma \ref{lemma-quasi-compact-in-Hausdorff}).
\end{proof}
\begin{lemma}
\label{lemma-bijective-map}
Let $f : X \to Y$ be a continuous map of topological spaces.
If $f$ is bijective, $X$ is quasi-compact, and $Y$ is Hausdorff,
then $f$ is a homeomorphism.
\end{lemma}
\begin{proof}
This follows immediately from Lemma \ref{lemma-closed-map}
which tells us that $f$ is closed, i.e., $f^{-1}$ is
continuous.
\end{proof}
\section{Jacobson spaces}
\label{section-space-jacobson}
\begin{definition}
\label{definition-space-jacobson}
Let $X$ be a topological space.
Let $X_0$ be the set of closed points of $X$.
We say that $X$ is {\it Jacobson} if every
closed subset $Z \subset X$ is the closure
of $Z \cap X_0$.
\end{definition}
\noindent
Note that a topological space $X$ is Jacobson if and only if
every nonempty locally closed subset of $X$
has a point closed in $X$.
\medskip\noindent
Let $X$ be a Jacobson space and let $X_0$ be the set
of closed points of $X$ with the induced topology.
Clearly, the definition implies that the morphism
$X_0 \to X$ induces a bijection between the closed
subsets of $X_0$ and the closed subsets of $X$.
Thus many properties of $X$ are inherited by $X_0$.
For example, the Krull dimensions of $X$ and $X_0$
are the same.
\begin{lemma}
\label{lemma-jacobson-check-irreducible-closed}
Let $X$ be a topological space. Let $X_0$ be the set
of closed points of $X$.
Suppose that for every point $x\in X$
the intersection $X_0 \cap \overline{\{x\}}$ is dense in $\overline{\{x\}}$.
Then $X$ is Jacobson.
\end{lemma}
\begin{proof}
Let $Z$ be closed subset of $X$
and $U$ be and open subset of $X$
such that $U\cap Z$ is nonempty.
Then for $x\in U\cap Z$ we have that $\overline{\{x\}}\cap U$ is a nonempty
subset of $Z\cap U$,
and by hypothesis it contains a point closed in $X$ as required.
\end{proof}
\begin{lemma}
\label{lemma-non-jacobson-Noetherian-characterize}
Let $X$ be a Kolmogorov topological space with a basis of quasi-compact
open sets.
If $X$ is not Jacobson, then there exists a non-closed point
$x \in X$ such that $\{x\}$ is locally closed.
\end{lemma}
\begin{proof}
As $X$ is not Jacobson there exists a closed set $Z$ and an open set $U$
in $X$ such that $Z \cap U$ is nonempty and does not contain points closed
in $X$. As $X$ has a basis of quasi-compact open sets we may replace $U$
by an open quasi-compact neighborhood of a point in $Z\cap U$ and so we may
assume that $U$ is quasi-compact open. By
Lemma \ref{lemma-quasi-compact-closed-point}, there exists a point
$x \in Z \cap U$ closed in $Z \cap U$,
and so $\{x\}$ is locally closed but not closed in $X$.
\end{proof}
\begin{lemma}
\label{lemma-jacobson-local}
Let $X$ be a topological space.
Let $X = \bigcup U_i$ be an open covering.
Then $X$ is Jacobson if and only if each $U_i$ is Jacobson.
Moreover, in this case $X_0 = \bigcup U_{i, 0}$.
\end{lemma}
\begin{proof}
Let $X$ be a topological space.
Let $X_0$ be the set of closed points of $X$.
Let $U_{i, 0}$ be the set of closed points of
$U_i$. Then $X_0 \cap U_i \subset U_{i, 0}$
but equality may not hold in general.
\medskip\noindent
First, assume that each $U_i$ is Jacobson.
We claim that in this case $X_0 \cap U_i = U_{i, 0}$.
Namely, suppose that $x \in U_{i, 0}$, i.e., $x$ is closed in
$U_i$. Let $\overline{\{x\}}$ be the closure
in $X$. Consider $\overline{\{x\}} \cap U_j$.
If $x \not \in U_j$, then $\overline{\{x\}} \cap U_j = \emptyset$.
If $x \in U_j$, then $U_i \cap U_j \subset U_j$
is an open subset of $U_j$ containing $x$.
Let $T' = U_j \setminus U_i \cap U_j$ and
$T = \{x\} \amalg T'$. Then $T$, $T'$
are closed subsets of $U_j$ and $T$ contains
$x$. As $U_j$ is Jacobson we see that the closed points of
$U_j$ are dense in $T$. Because $T = \{x\} \amalg T'$
this can only be the case if $x$ is closed in $U_j$.
Hence $\overline{\{x\}} \cap U_j = \{x\}$. We conclude
that $\overline{\{x\}} = \{ x \}$ as desired.
\medskip\noindent
Let $Z \subset X$ be a closed subset (still
assuming each $U_i$ is Jacobson).
Since now we know that $X_0 \cap Z \cap U_i
= U_{i, 0} \cap Z$ are dense in $Z \cap U_i$
it follows immediately that $X_0 \cap Z$ is
dense in $Z$.
\medskip\noindent
Conversely, assume that $X$ is Jacobson.
Let $Z \subset U_i$ be closed. Then
$X_0 \cap \overline{Z}$ is dense in $\overline{Z}$.
Hence also $X_0 \cap Z$ is dense in $Z$, because
$\overline{Z} \setminus Z$ is closed. As $X_0 \cap U_i
\subset U_{i, 0}$ we see that
$U_{i, 0} \cap Z$ is dense in $Z$.
Thus $U_i$ is Jacobson as desired.
\end{proof}
\begin{lemma}
\label{lemma-jacobson-inherited}
Let $X$ be Jacobson. The following types of subsets $T \subset X$
are Jacobson:
\begin{enumerate}
\item Open subspaces.
\item Closed subspaces.
\item Locally closed subspaces.
\item Unions of locally closed subspaces.
\item Constructible sets.
\item Any subset $T \subset X$ which locally on $X$
is a union of locally closed subsets.
\end{enumerate}
In each of these cases closed points of $T$ are
closed in $X$.
\end{lemma}
\begin{proof}
Let $X_0$ be the set of closed points of $X$. For any subset
$T \subset X$ we let $(*)$ denote the property:
\begin{itemize}
\item[$(*)$] Every nonempty locally closed subset of $T$ has a point
closed in $X$.
\end{itemize}
Note that always $X_0 \cap T \subset T_0$. Hence property $(*)$
implies that $T$ is Jacobson. In addition it clearly implies
that every closed point of $T$ is closed in $X$.
\medskip\noindent
Suppose that $T=\bigcup_i T_i$ with $T_i$ locally closed in $X$.
Take $A\subset T$ a locally closed nonempty subset in $T$,
then there exists a $T_i$ such that $A\cap T_i$ is nonempty, it is
locally closed in $T_i$ and so in $X$.
As $X$ is Jacobson $A$ has a point closed in $X$.
\end{proof}
\begin{lemma}
\label{lemma-finite-jacobson}
A finite Jacobson space is discrete.
\end{lemma}
\begin{proof}
If $X$ is finite Jacobson, $X_0 \subset X$ the subset of closed points,
then, on the one hand, $\overline{X_0} = X$. On the other hand, $X$,
and hence $X_0$ is finite, so
$X_0 =\{x_1, \ldots, x_n\} = \bigcup_{i = 1, \ldots, n} \{x_i\}$
is a finite union of closed sets, hence closed, so
$X = \overline{X_0} = X_0$. Every point is closed, and by
finiteness, every point is open.
\end{proof}
\begin{lemma}
\label{lemma-jacobson-equivalent-locally-closed}
\begin{slogan}
For Jacobson spaces, closed points see everything about the topology.
\end{slogan}
Suppose $X$ is a Jacobson topological space.
Let $X_0$ be the set of closed points of $X$.
There is a bijective, inclusion preserving correspondence
$$
\{\text{finite unions loc.\ closed subsets of } X\}
\leftrightarrow
\{\text{finite unions loc.\ closed subsets of } X_0\}
$$
given by $E \mapsto E \cap X_0$. This correspondence preserves
the subsets of locally closed, of open and of closed subsets.
\end{lemma}
\begin{proof}
We just prove that the correspondence $E \mapsto E \cap X_0$ is injective.
Indeed if $E\neq E'$ then without loss of generality $E\setminus E'$ is
nonempty, and it is a finite union of locally closed sets (details omitted).
As $X$ is Jacobson, we see that
$(E \setminus E') \cap X_0 = E \cap X_0 \setminus E' \cap X_0$ is not empty.
\end{proof}
\begin{lemma}
\label{lemma-jacobson-equivalent-constructible}
Suppose $X$ is a Jacobson topological space.
Let $X_0$ be the set of closed points of $X$.
There is a bijective, inclusion preserving correspondence
$$
\{\text{constructible subsets of } X\}
\leftrightarrow
\{\text{constructible subsets of } X_0\}
$$
given by $E \mapsto E \cap X_0$. This correspondence preserves
the subset of retrocompact open subsets, as well as complements
of these.
\end{lemma}
\begin{proof}
From Lemma \ref{lemma-jacobson-equivalent-locally-closed} above,
we just have to see that if $U$ is open in $X$ then $U\cap X_0$ is
retrocompact in $X_0$ if and only if $U$ is retrocompact in $X$.
This follows if we prove that for $U$ open in $X$ then $U\cap X_0$ is
quasi-compact if and only if $U$ is quasi-compact.
From Lemma \ref{lemma-jacobson-inherited} it follows that we may replace
$X$ by $U$ and assume that $U = X$.
Finally notice that any collection of opens $\mathcal{U}$ of $X$ cover
$X$ if and only if they cover $X_0$, using the Jacobson property
of $X$ in the closed $X\setminus \bigcup \mathcal{U}$ to find a point
in $X_0$ if it were nonempty.
\end{proof}
\section{Specialization}
\label{section-specialization}
\begin{definition}
\label{definition-specialization}
Let $X$ be a topological space.
\begin{enumerate}
\item If $x, x' \in X$ then we say $x$ is a {\it specialization} of $x'$,
or $x'$ is a {\it generalization} of $x$ if $x \in \overline{\{x'\}}$.
Notation: $x' \leadsto x$.
\item A subset $T \subset X$ is {\it stable under specialization}
if for all $x' \in T$ and every specialization $x' \leadsto x$ we have
$x \in T$.
\item A subset $T \subset X$ is {\it stable under generalization}
if for all $x \in T$ and every generalization $x' \leadsto x$ we have
$x' \in T$.
\end{enumerate}
\end{definition}
\begin{lemma}
\label{lemma-open-closed-specialization}
Let $X$ be a topological space.
\begin{enumerate}
\item Any closed subset of $X$ is stable under specialization.
\item Any open subset of $X$ is stable under generalization.
\item A subset $T \subset X$ is stable under specialization
if and only if
the complement $T^c$ is stable under generalization.
\end{enumerate}
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{definition}
\label{definition-lift-specializations}
Let $f : X \to Y$ be a continuous map of topological spaces.
\begin{enumerate}
\item We say that {\it specializations lift along $f$} or that $f$ is
{\it specializing} if given $y' \leadsto y$ in $Y$ and any $x'\in X$ with
$f(x') = y'$ there exists a specialization $x' \leadsto x$ of $x'$ in $X$ such
that $f(x) = y$.
\item We say that {\it generalizations lift along $f$} or that $f$ is
{\it generalizing} if given $y' \leadsto y$ in $Y$ and any $x\in X$ with
$f(x) = y$ there exists a generalization $x' \leadsto x$ of $x$ in $X$ such
that $f(x') = y'$.
\end{enumerate}
\end{definition}
\begin{lemma}
\label{lemma-lift-specialization-composition}
Suppose $f : X \to Y$ and $g : Y \to Z$ are continuous maps
of topological spaces. If specializations lift along both $f$ and $g$
then specializations lift along $g \circ f$</