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Lemma: Suppose that $A,B \subseteq X$ are connected and $A \cap B \neq \emptyset$ , then $A \cup B$ is connected.

How would I go about proving this? I think I understand the consequences of the lemma and it seems sort of obvious why it would be true, but I can't figure out how to prove it.

Any help would be great.


marked as duplicate by freakish, Lee Mosher, GNUSupporter 8964民主女神 地下教會, Lee David Chung Lin, Lord Shark the Unknown Feb 12 at 4:24

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  • $\begingroup$ Suppose $A\cup B$ is a disjoint union of open sets $U \cup V$. GIven that $A$ and $B$ are connected, what can you say about $U$ and $V$? $\endgroup$ – Artur Araujo Feb 11 at 13:08
  • $\begingroup$ This is a similar question about metric spaces, but this is a question about topology. I'm aware of similarities but I'm not sure I can use quite the same reasoning. $\endgroup$ – Penguinking14 Feb 11 at 16:53
  • $\begingroup$ @Penguinking14 Solutions posted there only use topology. The "metric" assumption is irrelevant. $\endgroup$ – freakish Feb 11 at 20:16

Suppose $p \in A \cap B$. To show connectedness of $A \cup B$, write $A \cup B=U \cup V$, where $U$ and $V$ are open disjoint subsets of $A \cup B$. It suffices to show $U$ or $V$ is empty...

So where is $p$? It must be in $U$ or in $V$, say $p\in U$ for definiteness; by symmetry it doesn't matter (or rename letters in the following proof).

Then $A = (A \cap U) \cup (B \cap V)$ (simple set theory). Also, as $U \cap A$ is open in $A$ and $V \cap A$ is open in $A$ too (subspace topolgoy wrt a subspace topology is again the subspace topology). And these sets are still disjoint. And we know $A$ is connected, so both sets cannot be non-empty at the same time, and we already know $p \in U \cap A \neq \emptyset$, so $V \cap A = \emptyset$.

The exact same argument (mutatis mutandis) can be made for $B$ (also connected) as well so $V \cap B=\emptyset$.

But then $$V = V \cap (A \cup B) = (V \cap A) \cup (V \cap B) = \emptyset \cup \emptyset = \emptyset$$

and we are done: $A \cup B$ is connected.


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