---
tag: 3Y504JQAY2
kind: proof
name: Proof that connected components are connected and closed
msc: [54D05]
proves: 0NRT0E2QT2
uses: [0R37DFJKFW, 0D939CMME7, 0D5YHQGDJY, 05GK32H72H, 0DRKT1QJR7, 0SHC6Z304Z, 0G5AV9Q4RR]
sources:
  - url: https://en.wikipedia.org/wiki/Connected_space
    locator: Connected components
---

## Proof

We show that $C(x)$ is connected, then that its closure is connected, so that the closure is not larger than $C(x)$.

1. Recall that $S \subseteq X$ is [[connected|0R37DFJKFW]] when it is a [[connected space|0D939CMME7]] in the [[subspace topology|0D5YHQGDJY]]. The open sets of $S$ are the sets $S \cap U$ with $U$ open in $X$.
2. A set $S$ is not connected exactly when there are open $U, V \subseteq X$ with $S \cap U$ and $S \cap V$ nonempty, $S \cap U \cap V = \varnothing$ and $S \subseteq U \cup V$.
3. The set $\{x\}$ is connected, since it has no two nonempty disjoint subsets. So $x \in C(x)$.
4. $C(x)$ is connected. Suppose not, and take $U, V$ as in step 2 for $S = C(x)$. As $x \in C(x) \subseteq U \cup V$, we may assume $x \in U$, after swapping $U$ and $V$.
5. Let $S$ be a connected subset of $X$ with $x \in S$. The sets $S \cap U$ and $S \cap V$ are open in $S$, disjoint, and cover $S$, because $S \subseteq C(x)$.
6. As $S$ is connected, one of them is empty. It is not $S \cap U$, which contains $x$. So $S \cap V = \varnothing$, and $S \subseteq U$.
7. $C(x)$ is the [[union|05GK32H72H]] of all such $S$, so $C(x) \subseteq U$ and $C(x) \cap V \subseteq C(x) \cap U \cap V = \varnothing$. This contradicts step 2. So $C(x)$ is connected.
8. Let $K = \overline{C(x)}$ be the [[closure|0DRKT1QJR7]]. Then $K$ is closed and $C(x) \subseteq K$, by [[this result|0SHC6Z304Z]].
9. $K$ is connected. Suppose not, and take $U, V$ as in step 2 for $S = K$.
10. Then $C(x) \cap U$ and $C(x) \cap V$ are open in $C(x)$, disjoint, and cover $C(x)$. As $C(x)$ is connected, one of them is empty. Say $C(x) \cap V = \varnothing$; the other case is the same with $U$ and $V$ swapped.
11. Then $C(x) \subseteq X \setminus V$, which is [[closed|0G5AV9Q4RR]]. By the minimality of the closure in [[this result|0SHC6Z304Z]], $K \subseteq X \setminus V$.
12. So $K \cap V = \varnothing$, which contradicts step 2. Thus $K$ is connected.
13. $K$ is connected and contains $x$, so $K$ is one of the sets in the union that defines $C(x)$. Thus $K \subseteq C(x)$.
14. With step 8, $C(x) = K$. So $C(x)$ is closed, and by step 7 it is connected.

## Idea

The union of connected sets through one point is connected, because any splitting would have to put each of them on the side of that point. The closure of a connected set is still connected. So the closure of the largest connected set through $x$ cannot be larger than it, which means that this set is closed.