2
$\begingroup$

Show that for $x_o\in X$, the connected component of $x_o$ is connected.

Attempt: So I'm trying to show that assuming that the union of connected sets that contain $x_o$ is not connected results in a contradiction.

Let $x_o\in X$ and let $A_1$ and $A_2$ connected such that $x_o\in A_1,A_2$. Suppose $A_1\cup A_2$ is not connected. Then $\exists U,V$ open that disconnect $A_1\cup A_2$. Without loss of generality, let $U\cap A_1\neq\emptyset$. Then, $V\cap A_2\neq\emptyset$. If $V\cap A_1\neq\emptyset$, $U\cap A_2=\emptyset$. But then $U,V$ do not disconnect $A_1\cup A_2$, contradiction.

  • 1
    You should not assume your connected component is the union of only two sets. See my answer for a way to address this issue.2012-02-29

2 Answers 2

2

Let $C_{x_0} = \bigcup_{\alpha \in \Gamma} A_\alpha$, where each $A_\alpha$ is a connected set containing $x_0$. Suppose $U$ and $V$ are open sets that disconnect $C_{x_0}$. Without loss of generality, let $x_0 \in U$, so that $A_\alpha \cap U \neq \emptyset$ for all $\alpha \in \Gamma$.

Now, there is $\beta \in \Gamma$ such $A_\beta \cap V \neq \emptyset$ (otherwise, $C_{x_0} \subseteq U$). In the relative topology on $A_\beta$, $U \cap A_\beta$ and $V \cap A_\beta$ are both open and so disconnect $A_\beta$ (a contradiction with the fact that $A_\beta$ is connected).

  • 0
    @Emir As Kannapan notes, $\Gamma$ is a set used to index the sets in the union. Since we don't know whether there are finitely-many, countably-many, or uncountably-many, it's best to index them in a generic way.2012-02-29
0

This follows from the theorem that states that $\bigcup F_\alpha$ is connected if each $F_\alpha$ is connected and $F_\alpha\cap F_\beta\neq\emptyset$ for each $\alpha,\beta.$ In particular, if $\{A_\alpha\}$ is the collection of connected sets containing $x_0$, then the connected component $\bigcup A_\alpha$ is connected since each $A_\alpha$ is connected and $x_0\in A_\alpha\cap A_\beta$ for each $\alpha,\beta.$