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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?

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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.

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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.

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  • $\begingroup$ 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 '13 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 '13 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 '13 at 12:16

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