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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!

Pedro
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Guillermo Mosse
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    It is better to use a decomposition into disjoint closed sets in b). – Daniel Fischer Oct 06 '15 at 20:41
  • Well, if we say that $U$ and $V$ are closed, then they are compact! – Guillermo Mosse Oct 06 '15 at 22:11
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    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? – Daniel Fischer Oct 06 '15 at 22:12
  • Yea, I was just thinking about that. In a Hausdorff space, you can separate points from compact sets with open sets. So, for each $x\ \in D$ consider $U^x$ that separates it from $E$. They cover $D$, which is compact, so we have a finite subcovering $U^{x_1},...,U^{x_n}$. I'll call the union of those, $V$. (To be continued) – Guillermo Mosse Oct 06 '15 at 22:17
  • Y could do the same for E and build another set $V'$, but $V$ and $V'$ could have an intersection....damm – Guillermo Mosse Oct 06 '15 at 22:37
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    You have two open sets, $x \in U^x$ and $E \subset V^x$, with $U^x \cap V^x = \varnothing$. – Daniel Fischer Oct 06 '15 at 22:38
  • Oh! I consider $V'$ the intersection of $V^{x_1},...,V^{x_n}$. $V$ and $V'$ are disjoint, and $V'$ contains $E$. My intersection $B$ is contained in $V \cup V'$, which is open. I can apply now $a)$, and say that my finite intersection is contained in only one of the open sets. As $B$ is contained in the finite intersection, I win. (I shouldn't have chosen the letters $V$ and $V'$, sorry). Thanks Daniel! – Guillermo Mosse Oct 06 '15 at 22:49
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    Bingo, that's it. – Daniel Fischer Oct 06 '15 at 22:50

1 Answers1

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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 $A_0:= A_i$.

Pedro
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