Add totally disconnected

Thanks to Laurent Moret-Bailly
https://stacks.math.columbia.edu/tag/0901#comment-4224

topology.tex

@@ -549,6 +549,21 @@ \section{Connected components}
 which are not open.
 \end{proof}
 
+\begin{remark}
+\label{remark-quasi-components}
+\begin{reference}
+\cite[Example 6.1.24]{Engelking}
+\end{reference}
+Let $X$ be a topological space and $x \in X$. Let $Z \subset X$ be the
+connected component of $X$ passing through $x$. Consider the intersection
+$E$ of all open and closed subsets of $X$ containing $x$. It is clear that
+$Z \subset E$. In general $Z \not = E$. For example, let
+$X = \{x, y, z_1, z_2, \ldots\}$ with the topology with the following
+basis of opens, $\{z_n\}$, $\{x, z_n, z_{n + 1}, \ldots\}$, and
+$\{y, z_n, z_{n + 1}, \ldots\}$ for all $n$. Then $Z = \{x\}$ and
+$E = \{x, y\}$. We omit the details.
+\end{remark}
+
 \begin{lemma}
 \label{lemma-connected-fibres-quotient-topology-connected-components}
 Let $f : X \to Y$ be a continuous map of topological spaces.
@@ -3905,13 +3920,20 @@ \section{Spectral spaces}
 \begin{lemma}
 \label{lemma-constructible-hausdorff-quasi-compact}
 Let $X$ be a spectral space. The constructible topology is
-Hausdorff and quasi-compact.
+Hausdorff, totally disconnected, and quasi-compact.
 \end{lemma}
 
 \begin{proof}
-Since the collection of all quasi-compact opens forms a basis for the
-topology on $X$ and $X$ is sober, it is clear that $X$ is Hausdorff in
-the constructible topology.
+Let $x, y \in X$ with $x \not = y$. Since $X$ is sober, there
+is an open subset $U$ containing exactly one of the two points $x, y$.
+Say $x \in U$. We may replace $U$ by a quasi-compact open
+neighbourhood of $x$ contained in $U$. Then $U$ and $U^c$ are open and
+closed in the constructible topology. Hence $X$ is Hausdorff in the
+constructible topology because $x \in U$ and $y \in U^c$ are
+disjoint opens in the constructible topology. The existence of $U$
+also implies $x$ and $y$ are in distinct connected components in the
+constructible topology, whence $X$ is totally disconnected in the
+constructible topology.
 
 \medskip\noindent
 Let $\mathcal{B}$ be the collection of subsets $B \subset X$