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a) Let X be a topological space, and $\mathcal{A}$ a family of closed, compact subspaces. Then, if $\bigcap_{A \in \mathcal{A}}A \subseteq U$, and $U$ is open, then $\exists\ \mathcal{F} \subseteq \mathcal{A}$ such that $\mathcal{F}$ is finite and $\bigcap_{A \in \mathcal{F}}A \subseteq U$

b) Let X be a topological Hausdorff ("$T_2$" for some people) space, and $\mathcal{A}$ a family of compact subspaces such that $\bigcap_{A \in \mathcal{F}}A$ is connected for every finite $\mathcal{F}$. Prove that $\bigcap_{A \in \mathcal{A}}A$ is connected.

Proof: a) Suppose the contrary, then $\bigcap_{A \in \mathcal{F}}A \cap U^c \neq \emptyset$. By the Finite intersection Property, ($\bigcap_{A \in \mathcal{A}}A \cap U^c$ is compact), we arrive at a contradiction.

b) Here's where I get stuck... Suppose the contrary.

First, an observation: as $X$ is Hausdorff, so the subspaces are also closed.

Then, $\exists\ U, V$ open (there) such that $B:= \bigcap_{A \in \mathcal{A}}A = U \cup V$, and $U$ and $V$ are disjoint.

Then, $\exists\ U', V'$ open in $X$ such that $U = B \cap U'$, $V = B \cap V'$.

From there, I can get a finite $F$ such that $C:= \bigcap_{A \in \mathcal{F}}A \subset U' \cup V'$. I'd like to shrink the size of U' and V' so I disconnect C, but I couldn't so far. I think I haven't used "enough" the fact that X is Hausdorff.

Can anyone help me? Thanks!


marked as duplicate by Eric Wofsey, user91500, user296602, user223391, Fabian Jan 20 '16 at 9:00

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    $\begingroup$ It is better to use a decomposition into disjoint closed sets in b). $\endgroup$ – Daniel Fischer Oct 6 '15 at 20:41
  • $\begingroup$ Well, if we say that $U$ and $V$ are closed, then they are compact! $\endgroup$ – Guillermo Mosse Oct 6 '15 at 22:11
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    $\begingroup$ Right. Perhaps it's better to call them $D$ and $E$ then. Okay, so we have two disjoint compact sets is a Hausdorff space. Can we separate them by open sets? $\endgroup$ – Daniel Fischer Oct 6 '15 at 22:12
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    $\begingroup$ You have two open sets, $x \in U^x$ and $E \subset V^x$, with $U^x \cap V^x = \varnothing$. $\endgroup$ – Daniel Fischer Oct 6 '15 at 22:38
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    $\begingroup$ Bingo, that's it. $\endgroup$ – Daniel Fischer Oct 6 '15 at 22:50

Pick some $A_i\in \mathcal A$, suppose $\mathcal A=\{A_j:j\in J\}$. Then $A_i\subseteq U\cup \bigcup_{i\neq j} A_j$: if $x\in A_i$ is in every $A_j$; then $x\in U$, if there is some $j$ such that $x\notin A_j$; then $x\in A_j^c$. The $A_j$ are assumed to be closed so this open cover of $A_i$ must admit a finite subcover and we can assume it contains $U$, say $A_i\subseteq U\cup A_1^c\cdots \cup A_n^c$ (I've relabeled the $A_j$ in the finite subcover for ease of notation). Then evidently $\bigcap_{i=0}^n A_i\subseteq U$, where $A_0:= A_i$.

  • $\begingroup$ (Credits to Franco A. for the solutions.) $\endgroup$ – Pedro Tamaroff Oct 6 '15 at 22:36
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    $\begingroup$ It looks like you assume that $U$ and $V$ are disjoint. That needs an argument. So far, you can only say that either (without loss of generality) $L \cap U = \varnothing$, or $L \cap U \cap V \neq \varnothing$. $\endgroup$ – Daniel Fischer Oct 6 '15 at 22:41
  • $\begingroup$ @DanielFischer Let's try to fix that, then. $\endgroup$ – Pedro Tamaroff Oct 6 '15 at 22:48
  • $\begingroup$ @DanielFischer It seems the second point was already solved in the comments. $\endgroup$ – Pedro Tamaroff Oct 6 '15 at 23:15

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