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 \input{preamble} % OK, start here. % \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 $g \circ f$</