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!

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.