Lemma 5.7.3. Let $X$ be a topological space.

1. If $T \subset X$ is connected, then so is its closure.

2. Any connected component of $X$ is closed (but not necessarily open).

3. Every connected subset of $X$ is contained in a unique connected component of $X$.

4. Every point of $X$ is contained in a unique connected component, in other words, $X$ is the disjoint union of its connected components.

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. $\square$

Comment #3563 by Laurent Moret-Bailly on

In (3) and (4) there is no mention of uniqueness of the component in question. But (of course) this is used, e.g. in the statement of Lemma 08ZL.

Comment #7824 by Ryo Suzuki on

On the proof of (4), uniqueness is not proven. Also, (3) is not proven. I write proposed amendment:

Proof of (3). Let $E$ be connected. Consider the set $A$ of connected subsets containing $E$. Let $T = \bigcup A$. We claim $T$ is connected. Namely, supposed that $T = T_1 \coprod T_2$ is a disjoint union of two open and closed subsets of $T$. Because $E$ is connected and $E\subset T$, $E\subset T_1$ or $E\subset T_2$. Without loss of generality, we can suppose $E\subset T_1$. So $E'\cap T_1 \neq \varnothing$ for all $E'\in A$. By connectedness of $E'$, $E'\subset T_1$. Hence $T = T_1$. Immediately, $T$ is a unique connected components containing $E$.

Comment #7828 by Laurent Moret-Bailly on

I second Ryo Suzuki's comment, but here is a slightly more genral property that immediately implies (3) and (4): "If $\mathscr{T}$ is a nonempty set of connected subsets with nonempty intersection, then $E=\bigcup_{T\in\mathscr{T}}T$ is connected." Proof (like Ryo's proof, this does not use Zorn's lemma): Fix $x\in\bigcap_{T\in\mathscr{T}}T$. Then $x\in E$ since $\mathscr{T}\neq\emptyset$, hence $E\neq\emptyset$. Let $E_0\subset E$ be open, closed and nonempty. Then $E_0$ must meet some $T_0\in\mathscr{T}$, hence contain it since $E_0\cap T_0$ is nonempty, open and closed in $T_0$. Therefore $E_0$ contains $x$, hence meets (and contains) every $T$. QED

I also suggest to state explicitly that the first paragraph of the proof proves (1) and (2).

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• 10 comment(s) on Section 5.7: Connected components

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