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I want to show that if A and B are two connected subsets of X, and if $Cl(A) \cap B \neq \emptyset$, then $A\cup B$ is connected.

I know that $X$ is separated (or not connected) if $A \neq \emptyset$ and $B \neq \emptyset$ and $\overline{A} \cap B = A \cap \overline{B} = \emptyset$ where $\overline{A}, \overline{B}$ are the closures of and $A$ and $B$ and if $X=A \cup B$. I know a set will be connected if it is not separated.

Intuitively, I feel that this statement makes sense because the only way A and B are connected but $A \cup B$ isn't is if there is some possible separate component of A and B. It must be that some part outside A (in this case, Cl(A)) allows a separation between A and B. I'm just not sure how to even begin proving this. Help would be much appreciated!

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  • $\begingroup$ This was listed among related questions (in the sidebar on the right): math.stackexchange.com/questions/1431179/… $\endgroup$ – Martin Sleziak Oct 20 '16 at 1:21
  • $\begingroup$ Thank you! Yes I have seen this, but I was wondering if there was a more general proof across all spaces (not just $\mathbb{R}$) without using continuity. $\endgroup$ – Nikitau Oct 20 '16 at 2:57
  • $\begingroup$ Maybe I am missing something, but neither the question nor the answer in the link I have given above say anything about $X=\mathbb R$. They are about arbitrary topological space $X$. $\endgroup$ – Martin Sleziak Oct 20 '16 at 4:58
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We prove this by contradiction. Assume $A\cup B$ is disconnected, then there are two open sets $U$ and $V$ such that

  • $A\cup B\subseteq U\cup V$ and $U\cap(A\cup B)\neq\emptyset, V\cap (A\cup B)\neq\emptyset$.
  • $U'\cap V'=\emptyset$ where $U'=U\cap(A\cup B), V'=V\cap (A\cup B)$. Note that $A\cup B=U'\cup V'$. Moreover $U'$ and $V'$ are open in $A\cup B$ under the subspace topology.

Since $A$ is connected, it follows that either $A\subseteq U$ or $A\subseteq V$, let's say $A\subseteq U$. Now, as $B$ is also connected, it follows that $B\subseteq U$ or $B\subseteq V$. Then we have $B\subseteq V$ because both $U$ and $V$ have non-empty intersections with $A\cup B$.

Now take $b\in Cl(A)\cap B$. Then $V$ is an open neighbourhood of $b$ thus $V\cap A\neq\emptyset$. However as $A\subseteq U$, it follows that $V\cap A\subseteq U$, thus $V'\cap U'\neq\emptyset$, which is a contradiction.

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  • $\begingroup$ It is incorrect to say that U and V are disjoint. What you should say is that $(U\cap (A\cup B))\cap (V\cap (A\cup B))=\phi.$ Now let $b\in B\cap \bar A.$ Then every nbhd of $b$ intersects $A,$ so $V$ intersects $A$ at a point $c.$ But then $c\in A\subset U\implies c\in U\cap (A\cup B) $ while $c\in A\cap V\implies c\in V\cap (A\cup B),$ a contradiction. $\endgroup$ – DanielWainfleet Oct 22 '16 at 7:46
  • $\begingroup$ @user254665 That's a good point. I've edit my answer. Thanks $\endgroup$ – Frank Lu Oct 22 '16 at 17:16

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