**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.

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 $T = T_2$.

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.
$\square$

## Comments (2)

Comment #3563 by Laurent Moret-Bailly on

Comment #3687 by Johan on

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