11
$\begingroup$

I'm starting with the basics of topology theory and I'm trying to show:

Let $A$ and $B$ be closed subsets of a topological space. If $A\cap B$ and $A\cup B$ are both connected, $A$ and $B$ are connected.

I've tried to prove it whit subsets operations and by contradiction, but I've failed with both strategies. Can you help me, please?

$\endgroup$

2 Answers 2

19
$\begingroup$

Suppose for a contradiction that $A$ is disconnected. Then there are closed, disjoint, non-empty subsets $S, T$ of $A$ such that $A=S\cup T$.
Since $A\cap B$ is connected, one of $B\cap S$ and $B\cap T$ must be empty, otherwise $A\cap B=(B\cap S)\cup (B\cap T)$. WLOG assume $B\cap S=\emptyset$.
Now, since $S$ and $T$ are closed in $A$ and $A$ is closed in $X$, $S$ and $T$ are closed in $X$. Therefore, $B\cup T$ and $S$ are closed non-empty subsets of $A\cup B$. Obviously, $(B\cup T)\cap S=\emptyset$ and $A\cup B=(B\cup T)\cup S$, so $A\cup B$ is disconnected, contradiction.

$\endgroup$
3
  • 1
    $\begingroup$ I know this is old but why did you phrase the partition in terms of closed sets? Aren't they defined to be open? $\endgroup$
    – ngc1300
    Apr 13, 2020 at 0:49
  • 1
    $\begingroup$ @pmac (under the subspace topology) $S$ and $T$ above are both open and closed (because $A^c = B$) $\endgroup$ Aug 27, 2020 at 12:36
  • $\begingroup$ If $A,B$ are open, then $S,T$ are open, disjoint and non-empty. Since $S,T$ are open in $A$ and $A$ is open in $X$, then $S,T$ are open in $X$. Therefore, $B\cup T$ and $S$ are open non-empty subsets of $A\cup B$. Is this correct? $\endgroup$ Mar 12 at 6:02
4
$\begingroup$

Let's use proof by contradiction. Without loss of generality, assume $A$ is not connected. Then (by definition) $A = A_1 \cup A_2$ such that $A_1 \subseteq U_1$, $A_2 \subseteq U_2$, $U_1 \cap U_2 = \emptyset$, and the sets $U_i$ are open. Can you proceed from there? Hint, you'll have to show one of $A \cap B$ or $A \cup B$ is in fact disconnected.

$\endgroup$
3
  • $\begingroup$ That's exactly the way I've proceeded, but I can't finish. Here's what I've got: If $A=A_1\cup A_2$, then $A\cap B=(A_1\cup A_2)\cap B=(A_1\cap B)\cup (A_2\cap B)$. It's clear that $(A_1\cap B)\cap (A_2\cap B)=\emptyset$ (because $(A_i\cap B)\subset A_i$), but are $A_i\cap B$ open? If they were open, then could we conclude that $A\cap B$ is not connected? And where do we use the other condition, that $A\cup B$ are connected? I'm not sure if I am right, thanks for your patience. $\endgroup$
    – Alejandro
    Sep 16, 2013 at 12:01
  • $\begingroup$ Yes, $A_i\cap B$ are open in $A\cap B$, but no you cannot conclude that $A\cap B$ is not connected, as $A_i\cap B$ might be empty. That is why you need that $A\cup B$ is connected and $A, B$ are closed. $\endgroup$
    – walcher
    Sep 16, 2013 at 12:07
  • $\begingroup$ I've had answered to Shaun before I've read your comment, @walcher. Now i've understood it perfectly. Thank you both. $\endgroup$
    – Alejandro
    Sep 16, 2013 at 12:16

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.