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I think I can prove the closure of $A$, that is $\bar A$, is connected, as there are many other threads on this site. I am then just not sure how to make the jump to show formally that B is connected.

AdLibitum
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Daniel Ward
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    You could try to show that every map $B\to{0,1}$ is constant, using the fact that every map $A\to{0,1}$ is constant, and using that maps preserve the closure of sets. – Stefan Hamcke Mar 24 '15 at 13:50

2 Answers2

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If $B$ is not connected then non-empty sets $U,V$ exist with $U\cap\overline{V}=\varnothing=\overline{U}\cap V$ and $B=U\cup V$.

Then the sets $U\cap A$ and $V\cap A$ are disjoint, open in $A$ and are covering $A$.

So one of these sets must be empty since $A$ is connected.

If $U\cap A=\varnothing$ then $A\subseteq V$ and consequently $U\subseteq B\subseteq \overline{A}\subseteq\overline{V}$.

Then $\varnothing\neq U\subseteq\overline{V}$ contradicting that $U\cap\overline{V}=\varnothing$.

Likewise $V\cap A=\varnothing$ leads to a contradiction.

drhab
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If $B\subset U\cup V$ where $U$ and $V$ are disjoint, open sets, then the same thing can be said for $A$. But since $A$ is connected, then $U$ or $V$ is disjoint with $A$. Let's say WLOG that $A\cap U=\emptyset$.

Therefore, $A\subset \complement U$ and (since $\complement U$ is closed) $\bar A\subset \complement U$. This implies that $B\subset\complement U$, or equivalently, $B\cap U=\emptyset$.

ajotatxe
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