# Proving the Daisy Lemma [duplicate]

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 UnknownFeb 12 at 4:24

• 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 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 at 16:53
• @Penguinking14 Solutions posted there only use topology. The "metric" assumption is irrelevant. – 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.