Proving subset is not connected iff there exist open sets in X

Asked Oct 19 '13 at 23:49

Active Oct 20 '13 at 00:56

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Prove that E $\subseteq$ X is not connected if and only if there exist open sets $A, B \subseteq X$ such that $E \subseteq A ∪ B, A ∩ B$ = $\emptyset$ and $E ∩ A$ and $E ∩ B$ are both nonempty.

$X$ is a nonempty set equipped with a metric d.

Having a hard time proving the ⇒ part of the proof. Help!

edited Oct 20 '13 at 00:16

asked Oct 19 '13 at 23:49

akeenlogician

2

What is the definition of connectedness in this context? In some books, the right half is used as a definition. – Joel Oct 19 '13 at 23:51
1

I think you have the cap and cup the wrong way round as if $A\cup B=\emptyset$, then $A\cap B=\emptyset$. – Dan Rust Oct 20 '13 at 00:07
@Joel: Only definition I have is E is connected if E is not a union of two non-empty, separated sets. – akeenlogician Oct 20 '13 at 00:15
Dan, you are of course right...sorry for the typo. I've changed it. – akeenlogician Oct 20 '13 at 00:16
Well, if by 'separated' you meant 'disjoint', then this definition is not good, as then no space with $\ge 2$ points would be connected.. The exercise is basically the right definition. – Berci Oct 20 '13 at 00:20
What you are asked to prove is the most popular (and in my opinion, the best) definition of a disconnected set. I'm not sure how you can "prove" it! Your professor must be using a different (but equivalent) definition or else s/he would not have assigned this problem. What does "separated" mean in your comment above? – Stefan Smith Oct 20 '13 at 00:35
Ok, for separated, I have two subsets A and B of a metric space are separated if no point of A lies in the closure of B and no point of B lies in the closure of A, which seems to be saying that the two sets are disjoint. – akeenlogician Oct 20 '13 at 00:44
See also: Prove that $E$ is disconnected iff there exists two open disjoint sets $A$,$B$ in $X$ – Martin Sleziak Jun 18 '15 at 05:29

Proving subset is not connected iff there exist open sets in X

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