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Let us say that a family $R$ of sets has the Finite Subcovering Property --- FSP --- if any subfamily of $R$ which covers the union $\cup B: B \in R$ has itself a finite subfamily which also covers. For example, take for $R$ the family of open balls in a compact metric space $M$. Clearly the family of finite intersections of open balls also has the FSP.

We say the Finite Intersection Principle --- FIP --- is the statement: If $R$ has the FSP then the family of all finite intersections of members of $R$ also has the FSP.

It is easily proved that FIP is a theorem in ZFC. The question is: Does ZF + FIP imply the Axiom of Choice?

I looked in Herrlich, "Axiom of Choice", and I did not see this, but I may have missed it. This must certainly be known: the question was raised by J. L. Kelley in Fund. Math. 37 (1950), p. 76.

Note that in the absence of the Axiom of Choice, we need to specify what "finite" should mean. For our purposes here, let's say a set is finite if it may be ordered so that every non-void subset has both a first element and a last element in the ordering. See Herrlich Section 4.1 for equivalent formulations.

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up vote 8 down vote accepted

This is not really an answer but too long for a comment:

Let me point out that there might be a relation to the Alexander Subbase Theorem here:
A topological space is compact iff the topology has a subbase with the FSP.

What FIP gives you is this:

If $R$ has the FSP and is a subbase for the topology of a space $X=\bigcup R$, then $X$ has a basis (namely all finite intersections of elements of $R$) with the FSP.

Unfortunately it seems that getting compactness from the existence of a basis with the FSP also requires some form of choice. So FIP could be strictly weaker than the Alexander Subbase Theorem. But an upper bound on how much choice is needed for the subbase theorem would also be an upper bound for FIP.

Unfortunately I don't have access to any Axiom of Choice book right now. But it seems likely that there is some information about the amount of choice needed for the subbasis theorem.

Edit: By godelian's comment below, this actually answers the question: FIP follows from Alexander's Subbase Theorem which follows from the Boolean prime ideal theorem which is known to be strictly weaker than the full Axiom of Choice. So no, FIP does not imply AC.

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Indeed there is: Alexander subbase theorem is equivalent to the Boolean prime ideal theorem, hence strictly weaker than AC. – godelian Sep 21 '12 at 8:23
This is interesting. However, there might still be a catch here. Is the general subbase theorem or the version for Hausdorff spaces equivalent to the Boolean prime ideal theorem? With the Tychonov theorem there is exactly this problem: The version for Hausdorff spaces is equivalent to the Boolean prime ideal theorem, the general version (without assuming Hausdorffness) is equivalent to the full Axiom of Choice. I would not at all be surprised if each version of the Alexander Subbase Theorem was equivalent to the respective version of Tychonov's theorem. – Stefan Geschke Sep 21 '12 at 9:53
Hi Stefan, it is the general version of Alexander subbase theorem the one which is equivalent to BPI, i.e., the version asserting that a topological space is compact iff there is a subbase with the FSP. This also implies that any proof of Tychonoff for the general case relying on alexander theorem must also make use of another choice principle (this is indeed the case, if one takes a close look at such proofs). – godelian Sep 21 '12 at 11:57
Excellent. Thank you. – Stefan Geschke Sep 21 '12 at 13:54
It seems clear to me that, conversely, FIP implies the Alexander subbase theorem. So they are equivalent. What we need to finish this discussion is some reference to the proof that the Alexander subbase theorem is equivalent to the Boolen Prime Ideal Theorem (or one of its many better-known equivalents). – Fred Dashiell Sep 22 '12 at 0:24

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