The Stacks project

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 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

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 be connected. Consider the set of connected subsets containing . Let . We claim is connected. Namely, supposed that is a disjoint union of two open and closed subsets of . Because is connected and , or . Without loss of generality, we can suppose . So for all . By connectedness of , . Hence . Immediately, is a unique connected components containing .

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 is a nonempty set of connected subsets with nonempty intersection, then is connected." Proof (like Ryo's proof, this does not use Zorn's lemma): Fix . Then since , hence . Let be open, closed and nonempty. Then must meet some , hence contain it since is nonempty, open and closed in . Therefore contains , hence meets (and contains) every . 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

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