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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.

Asaf Karagila
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Penguinking14
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  • 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 $V$? – Artur Araujo Feb 11 '19 at 13:08
  • 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. – Penguinking14 Feb 11 '19 at 16:53
  • @Penguinking14 Solutions posted there only use topology. The "metric" assumption is irrelevant. – freakish Feb 11 '19 at 20:16

1 Answers1

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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.

Henno Brandsma
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