12
$\begingroup$

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

Reset to default
21
$\begingroup$

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
    Commented Apr 13, 2020 at 0:49
  • 1
    $\begingroup$ @pmac (under the subspace topology) $S$ and $T$ above are both open and closed (because $A^c = B$) $\endgroup$
    – aidangallagher4
    Commented 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? $\endgroup$
    – Blue Tomato
    Commented Mar 12, 2022 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
    Commented 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
    Commented 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
    Commented Sep 16, 2013 at 12:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .