**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 $\overline{T} \subset T_1$. This implies (1) and (2).

Let $A$ be a nonempty set of connected subsets of $X$ such that $\Omega = \bigcap _{T \in A} T$ is nonempty. We claim $E = \bigcup _{T \in A} T$ is connected. Namely, $E$ is nonempty as it contains $\Omega $. Say $E = E_1 \amalg E_2$ with $E_ i$ closed in $E$. We may assume $E_1$ meets $\Omega $ (after renumbering). Then each $T \in A$ meets $E_1$ and hence must be contained in $E_1$ as $T$ is connected. Hence $E \subset E_1$ which proves the claim.

Let $W \subset X$ be a nonempty connected subset. If we apply the result of the previous paragraph to the set of all connected subsets of $X$ containing $W$, then we see that $E$ is a connected component of $X$. This implies existence and uniqueness in (3).

Let $x \in X$. Taking $W = \{ x\} $ in the previous paragraph we see that $x$ is contained in a unique connected component of $X$. Any two distinct connected components must be disjoint (by the result of the second paragraph).

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 (5)

Comment #3563 by Laurent Moret-Bailly on

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