# Check My Proof: Finite Intersection Property

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.

• 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. – drhab Mar 15 '19 at 14:21
• I know, but this is our definition (our course is in German), and finite intersection property seemed to be the only apt translation. – Viktor Glombik Mar 15 '19 at 14:41
• @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. – 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 $$A_{i_0}$$.

• Do I need to use what I have shown previously? – Viktor Glombik Mar 15 '19 at 17:10
• Yes, you can use FIP $\iff$ compact for the space $A_{i_0}$ with the subspace topology. Do you know how? – lulu Mar 15 '19 at 17:15
• I edited, to include my approach. is it correct? – Viktor Glombik Mar 15 '19 at 17:53
• 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 $A_{i_0}$. – lulu Mar 15 '19 at 18:11