4

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.

Fred Dashiell
  • 1,674

1 Answers1

8

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.

Stefan Geschke
  • 16,346
  • 2
  • 53
  • 80
  • 3
    Indeed there is: Alexander subbase theorem is equivalent to the Boolean prime ideal theorem, hence strictly weaker than AC. – godelian Sep 21 '12 at 08:23
  • 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 09:53
  • 2
    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
  • 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 00:24
  • There is a discussion in Marcel Erne, "Prime Ideal Theorems and Syatems of Finite Character", in CMUC 38,3 (1997), 513-536. The original proof seems to go back to Rubin and Scott in 1954, and a published proof by Parovicenko in 1969 (see the Erne paper for details). I am surprised it does not seem to be covered in any textbooks or monographs. – Fred Dashiell Sep 22 '12 at 01:30
  • I am going to check Stefan's "answer" to indicate the question has been answered, but of course the answer is a collaborative one. – Fred Dashiell Sep 22 '12 at 01:36
  • @Fred Dashiell: Are you sure that FIP implies the Alexander Subbase Theorem? Don't you have to pick, given an open cover $\mathcal U$ of a space $X$, for each point $x\in X$ some open set $O$ in the basis of the topology such that $x\in O$ and $O$ is a subset of an element of $\mathcal U$? Or can you avoid making these choices somehow? – Stefan Geschke Sep 22 '12 at 17:39
  • Every collection $\mathscr{U}$ of open sets defines a unique collection $\mathscr{V}_\mathscr{U}$ of elements of the base $\mathscr{B}$, which is just the union of all collections $\mathscr{V}_U \subset \mathscr{B}$, where for $U \in \mathscr{U}$, $\mathscr{V}_U$ is just the collection of all elements of the base contained in $U$. I don't think this uses any choice, does it? – Fred Dashiell Sep 23 '12 at 13:32
  • So an open cover $\mathscr{U}$ of $X$ defines a specific open cover $\mathscr{V}$ by elements of the base, such that each $V \in \mathscr{V}$ is contained in some element of $\mathscr{U}$.

    Taking a finite subcover from $\mathscr{V}$ gives a finite set of elements of $\mathscr{U}$ which also covers $X$. Here we have used a basic choice principle: Every finite collection of nonempty sets has a choice function. Isn't that a theorem in ZF?

     – Fred Dashiell Sep 23 '12 at 13:39
  • As for proving the Alexander subbase lemma (A) is equivalent to the Prime Ideal Theorem (PIT), note that the treatment by Bourbaki General Topology, Ex. I.9.1 shows that the ultrafilter theorem (UF) implies (A). And Howard-Rubin, "Consequences of the Axiom of Choice (1998), Note 34 proves (A) implies compactness of the generalized Cantor space $2^I$, which by Jech "Axiom of Choice" problem 17, p. 28 is equivalent to (PIT). Maybe there is a shorter chain, but these stepa are all pretty elementary, as opposed to the treatment given in Erne's paper. – Fred Dashiell Sep 23 '12 at 13:59
  • @Fred Dashiell: I missed that from an open cover you can just define an open cover consisting of basis elements: just take all the basic open sets that sit inside some element of the original cover. There is certainly no choice involved here. Everything is clear now. – Stefan Geschke Sep 23 '12 at 21:52