# If $\{E_\alpha\}$ is connected, $\bigcap\limits_{\alpha\in A}E \neq \emptyset$, then $\bigcup\limits_{\alpha\in A}E$ is connected

If $\{E_\alpha\}_{\alpha\in A}$ is connected in $\mathbb{R}^n$, $\bigcap\limits_{\alpha\in A}E_\alpha \neq \emptyset$, then $\bigcup\limits_{\alpha\in A}E_\alpha$ is connected.

I have zero intuition on how to do a proof of this statement. Please help.

I started it by contradiction like this: suppose that $\bigcup\limits_{\alpha\in A}E_\alpha$ is not connected. Then there exist non-empty disjoint separating sets $U$ and $V$, such that $U$ and $V$ are relatively open in $\bigcup\limits_{\alpha\in A}E_\alpha$, and $U\cup V= \bigcup\limits_{\alpha\in A}E_\alpha$. But I have no idea what to do next.

• If $\bigcap E_\alpha\neq \emptyset$ then there must be something that is common to all sets, say $v$. Borrowing terminology from path-connectedness, for any two points $x$ and $y$ in $\bigcup E$, you could trace a path from $x$ to $v$ in some $E_\alpha$ where $x$ appears, and then from $v$ to $y$ in some $E_\alpha$ where $y$ appears since each $E_\alpha$ is itself connected. The general proof for ordinary connectedness should be similar. – JMoravitz Feb 3 '16 at 4:31
• If you understand the proof for when $A$ is a two element set, then you should have no problem with this. – Forever Mozart Feb 3 '16 at 4:33

## 2 Answers

Hint: Let $y\in \cap E_\alpha \subset \cup E_\alpha$. Then $y\in U$ or $y\in V$. **Fixing each $\alpha$, consider $U_\alpha = U\cap E_\alpha$ and $V_\alpha = V\cap E_\alpha$. They are open, disjoint and $U_\alpha \cup V_\alpha = E_\alpha$. Do you see how to use the connectivity now?

(Let's say, if $y\in U$. Then $y\in U_\alpha$ is nonempty. Thus $V_\alpha$ has to be empty as $E_\alpha$ is connected).

• John: Do you mean that $U_\alpha$ is open in $\mathbb{R}^n$? – sequence Feb 3 '16 at 5:34
• Unfortunately, I don't still see how to use connectivity. It is surprising for me, since I've done well in any other branch of mathematics, but topology seems completely out of touch with me. And this is just a relatively small part of an introductory real analysis course. – sequence Feb 3 '16 at 5:42
• I mean $U_\alpha, V_\alpha$ are open in $E_\alpha$. @sequence – user99914 Feb 3 '16 at 7:07
• Note that if $V_\alpha$ is empty, then $U_\alpha = E_\alpha$, this means $E_\alpha \subset U$. But as $\alpha$ is arbitrary... @sequence – user99914 Feb 3 '16 at 7:53
• @sequence. Yes. – user99914 Feb 3 '16 at 8:05

$X$ is connected iff the only nonempty clopen (closed and open) subset of $X$ is all of $X$.

$E:=\bigcup _{\alpha\in A}E_\alpha$ is connected: Let $U$ be a nonempty clopen subset of $E$. We show $U=E$. There exists $\beta\in A$ such that $U\cap E_\beta\neq\varnothing$. Since $E_\beta$ is connected, we have $E_\beta\subseteq U$. Now let $p\in \bigcap_{\alpha\in A}E_\alpha$. We have $p\in U$. So $U\cap E_\alpha\neq\varnothing$ for each $\alpha\in A$. Since each $E_\alpha$ is connected, we have $E_\alpha\subseteq U$ for each $\alpha\in A$. That is $E\subseteq U$. So $U=E$. Thus $E$ is connected.