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Note: assume that we are working in ZFC.

Let $\mathcal F$ be a nonempty set of nonempty finite sets with the finite intersection property. That is, for any finite $\mathcal G\subseteq\mathcal F$, the intersection of $\mathcal G$ is nonempty. Is it then the case that $\bigcap\mathcal F$ is nonempty?

I think it is - at least for $\lvert\mathcal F\rvert<$some large cardinal - but I can't explain why.

R. Burton
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    Since $\mathcal{F}$ is a nonempty set of nonempty finite sets with FIP, just select one such finite set $A$ and just check its elements one-by-one to conclude $\bigcap\mathcal{F}\neq\varnothing$. – user10354138 Oct 07 '23 at 21:42
  • @user10354138 I'm not sure that I understand. If $\bigcap \mathcal F$ is nonempty, then we can certainly choose $F\in\mathcal F$, and show $(\exists x\in F)(\forall G\in F)(x\in G)$. But how do we know that we can do this without assuming the conclusion? – R. Burton Oct 07 '23 at 21:47
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    @R.Burton Pick $A\in\mathcal{F}$. Let $A={a_1,...,a_n}$. By FIP, there must be some $i$ such that $a_i$ is in each $B\in\mathcal{F}$; otherwise, letting $B_i\in\mathcal{F}$ with $a_i\not\in B_i$ we would have $A\cap B_1\cap ...\cap B_n=\emptyset$, contradicting FIP. But then $a_i\in\bigcap\mathcal{F}$. – Noah Schweber Oct 07 '23 at 21:51
  • @user10354138 You should add that as an answer (I would have added my own comment as an answer but yours preceded it). – Noah Schweber Oct 07 '23 at 21:52
  • @NoahSchweber That's what I thought at first, but I don't know that we necessarily have such $a_i$. I am certain that $(\forall B\in\mathcal F)(\exists a\in A)(a\in B)$, but this is weaker than the required $(\exists a\in A)(\forall B\in\mathcal F)(a\in B)$. – R. Burton Oct 07 '23 at 21:56
  • @R.Burton We do have such an $a_i$. Keep reading my prior comment. If no $a_i$ fit the bill, we would get a contradiction with FIP (and note that choice has no role here since we're only making finitely many choices). – Noah Schweber Oct 07 '23 at 21:57

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Pick an $A\in\mathcal{F}$. We know $A$ is (nonempty) finite.

For every $a\in A$, either we have $a\in F$ for all $F\in\mathcal{F}$ (thus $a\in\bigcap\mathcal{F}$ and we are done), or we can find some $G_a\in\mathcal{F}$ with $a\notin G_a$.

Now if we can find $G_a$ for all $a\in A$ (note only finitely choices $a\in A\mapsto G_a\in\mathcal{F}$ here, so no AC is needed) then $\{A\}\cup\{G_a\colon a\in A\}$ is a finite subset of $\mathcal{F}$ with empty intersection, contradicting FIP.

user10354138
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  • In regard to the axiom of choice, if $\mathcal G_a={G\in\mathcal F:a\notin A}$, do we not need it in order to select one $G_a$ from each $\mathcal G_a$, since each $\mathcal G_a$ may be infinite? Or is choice only required for the selection of one element from each set of an infinite family? – R. Burton Oct 07 '23 at 22:18
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    @R.Burton Making finitely many choices does not need AC. You can just write out the function $f\colon a\in A\mapsto G_a$ by listing the pairs $(a,G_a)$ (or equivalent depending on how you define function), which is a wff. – user10354138 Oct 07 '23 at 22:21
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    See this answer if you are still having trouble proving finite choice in ZF (or just Z). – user10354138 Oct 07 '23 at 22:32