Let $(X, \mathcal{T})$ be a Hausdorff topological space.

Definition. $X$ has the finite intersection property (FIP), if for every family of closed subsets $(A_i)_{i \in I} \subset X$ with $\bigcap_{i \in I} A_i = \emptyset$ the exist indices $i_1, \ldots, i_k$, so that already $\bigcap_{\ell = 1}^{k} A_{i_{\ell}} = \emptyset$.

I've already proven that $X$ is compact iff $X$ has the FIP, which basically follows from deMorgans Principles and the fact that $(A_i^C)_{i \in I}$ are an open cover of $X$.

Let $(A_i)_{i \in I} \subset X$ be a compact family. Prove that if $\bigcap_{i \in I} A_i = \emptyset$ the existence of indices $i_1, \ldots, i_k$, so that $\bigcap_{\ell = 1}^{k} A_{i_{\ell}} = \emptyset$ already follows.

My ideas

I know that finite unions of compact sets and countable intersections are compact because $(X, \mathcal{T})$ is Hausdorff, but haven't been able to use that.

I also tried reaching a contradiction by assuming that for all finite sets of inidices $\{i_1, \ldots, i_k\}$ the intersection $\bigcap_{\ell = 1}^{k} A_{i_{\ell}}$ is nonempty but I haven't been able to show that then the infinite intersection is nonempty, which would by the desired contradiction.

All help is greatly appreciated.

Edit (my approach using the hint)

And $A_{i_0} \subset X$ is a compact subspace, so we have a family $(\tilde{A}_i := A_{i_0} \cap A_i)_{i \in I}$ in a compact space with $\bigcap_{i \in I} \tilde{A}_i = \emptyset$ and by FIP we know that finitely many indices are enough, so already $\bigcap_{\ell = 1}^{k} \tilde{A_{i_{\ell}}} = \emptyset$. Now we have for those finitely many indices we have $$ \emptyset = \bigcap_{\ell = 1}^{k} A_{i_0} \cap A_{i_{\ell}} = A_{i_0} \cap \bigcap_{\ell = 1}^{k} A_{i_{\ell}} $$ And since $A_{i_0} \neq \emptyset$, we have $\bigcap_{\ell = 1}^{k} A_{i_{\ell}}$ as required.

  • $\begingroup$ Normally not a set $X$ is said to have FIP, but a collection of subsets of $X$ is said to have FIP if this collection satisfies certain conditions. $\endgroup$ – drhab Mar 15 '19 at 14:21
  • $\begingroup$ I know, but this is our definition (our course is in German), and finite intersection property seemed to be the only apt translation. $\endgroup$ – Viktor Glombik Mar 15 '19 at 14:41
  • 1
    $\begingroup$ @drhab but the statement is equivalent to the more usual: every closed family with the FIP has non-empty intersection (it's the contrapositive statement), which is more well-known. $\endgroup$ – Henno Brandsma Mar 15 '19 at 16:44

Given some family of compact subsets $A_i, i \in I$ with empty intersection:

Fix some $A_{i_0}$ (which is compact), and define $B_i = A_{i_0} \cap A_i$.

Then some $B_i$ can be empty (and we are done, using the finite set $\{i, i_0\}$) or all $B_i$ are non-empty and $\bigcap_{i \in I} B_i = \bigcap_{i \in I} A_i =\emptyset$ and the already proved fact for the compact space $A_{i_0}$ (we use Hausdorffness in that all $B_i$ are closed in $A_{i_0}$ so that that theorem applies) implies that for a finite subset $J$ of $I$ we have that $\bigcap_{i \in J} B_i = \emptyset$ and then we have that for the finite subset $J'=J \cup \{i_0\}$ of $I$ that $$\bigcap_{i \in J'} A_i = \emptyset$$

as required.


$\textbf{Hint:}$ Fix $i_0$ and consider the family $(A_{i_0}\cap A_i)_{i\in I}$ which consists of closed subsets of $A_{i_0}$.

  • $\begingroup$ Do I need to use what I have shown previously? $\endgroup$ – Viktor Glombik Mar 15 '19 at 17:10
  • $\begingroup$ Yes, you can use FIP $\iff$ compact for the space $A_{i_0}$ with the subspace topology. Do you know how? $\endgroup$ – lulu Mar 15 '19 at 17:15
  • $\begingroup$ I edited, to include my approach. is it correct? $\endgroup$ – Viktor Glombik Mar 15 '19 at 17:53
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    $\begingroup$ Almost. You cannot conclude from $A_{i_0}\neq \emptyset $ that $\bigcap_{\ell = 1}^{k} A_{i_{\ell}}=\emptyset$. Instead notice that $A_{i_0} \cap \bigcap_{\ell = 1}^{k} A_{i_{\ell}}$ is already a finite intersection. Also to be precise you should convince yourself that the $A_{i_o}\cap A_i$ are closed in $A_{i_0}$. $\endgroup$ – lulu Mar 15 '19 at 18:11

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