Lemma 5.7.3 (004T)—The Stacks project

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

If $T \subset X$ is connected, then so is its closure.
Any connected component of $X$ is closed (but not necessarily open).
Every connected subset of $X$ is contained in a unique connected component of $X$.
Every point of $X$ is contained in a unique connected component, in other words, $X$ is the 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 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 (4)

Comment #3563 by Laurent Moret-Bailly on September 03, 2018 at 15:26

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 #3687 by Johan on October 22, 2018 at 15:22

Thanks and fixed here.

Comment #7824 by Ryo Suzuki on October 08, 2022 at 15:07

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 October 09, 2022 at 03:33

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).

There are also:

10 comment(s) on Section 5.7: Connected components

Comments (4)

Post a comment