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Assume that $E$ and $F$ are conneceted subsets of the metrix space X, such that $\bar{E} \cap F \neq \emptyset$. Prove that $E \cup F$ is conneceted as well.

When I draw A picture the statement appears pretty logic to me, but I don't know how I should prove it. Here are my attempts.

First of all, it's easy to show $\bar{E} \cup F$ is connected. Maybe I could somehow prove that this implies that $E \cup F$ is connected but I failed to do so. I know that we can make a sequence that converges to a point in F.

Can you please give me a hint to go on?

Koenraad van Duin
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  • I suppose the "it's easy to show..." means you know that the union of two connected sets with nonempty intersection is connected. So consider $E' = E \cup (\overline{E}\cap F)$. Note $E \cup F = E' \cup F$. – Daniel Fischer Oct 12 '13 at 19:42

1 Answers1

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HINT: Suppose that $U$ and $V$ are open sets in $X$ such that $E\cup F\subseteq U\cup V$, and $$U\cap(E\cup F)\ne\varnothing\ne V\cap(E\cup F)\;.$$ Without loss of generality we may assume that $U\cap E\ne\varnothing$.

  • Show that $E\subseteq U$.
  • Show that if $U\cap F\ne\varnothing$, then $F\subseteq U$, so $U\cap F=\varnothing$, and $F\subseteq V$.
  • Conclude that $U\cap V\ne\varnothing$.
Brian M. Scott
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