OPEN QUESTIONS

Q1. Measures and ultrafilters. (***)

Let us consider the two following statements:
UF (ultrafilter theorem): Every filter on a set is contained in a ultrafilter.
MUF (measured ultrafilter theorem): For every boolean algebra B, if there exists a non-trivial finitely additive measure on B, then there exists a ultrafilter on B.

It is well known that UF implies MUF.
[-] Reciprocally, does the statement MUF imply the ultrafilter theorem?

Remark. The statements UF and MUF are respectively equivalent to the following statements (see [Mo90] and [Mo91]):
GE (Gelfand Extremal) If A is a non-trivial Gelfand algebra, then the closed unit ball of its continuous dual has at least one extreme point which is not 0.
WGE (Weak Gelfand Extremal) If A is a Gelfand algebra, then the closed unit ball of its continuous dual has at least one extreme point.

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Q2. Extreme points in convex sets. (**)

[-]Does the following axiom imply the axiom of choice?
VKM: If E is a linear topological vector space which is Hausdorff and locally convex, then every non-empty convex-compact convex subset of E has an extreme point.

Here we say that a subset C of E is convex-compact iff for every family F of closed convex sets of C that satisfies the finite intersection property, the intersection of the family F is not empty.

This question was raised by Bell and Fremlin (1972), see [B-F72]: in their paper, the authors have shown that VKM + Hahn-Banach entails the axiom of choice. Hence, in order to prove that VKM implies the axiom of choice, it is sufficient to prove that VKM entails the Hahn-Banach axiom.

We have obtained some partial answers to this question:
The axiom VKM implies that every well-ordered family of non-empty sets has a non-empty product (see [Mo92] and [Fo-Mo]). In particular, the axiom VKM entails the axiom of dependent choices.

In the hope of solving the whole question, we look for various convex compact convex sets in Hausdorff locally convex topological vector spaces (these spaces must be built in ZF or at least in ZF+VKM! )

This question leads us to the continuous Hahn-Banach property (see question Q5).


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Q3. Baire spaces. (****)

In set theory with the axiom of choice, it is known that every compact Hausdorff space is a Baire space. But this statement is not provable in ZF. Nevertheless, the following statement is provable in ZF:
If a compact Hausdorff space has a well-orderable base of open sets, then it is a Baire space. In particular, for every ordinal n, the product space [0,1]^n (which is both compact and Hausdorff) is a Baire space.

[-] In this light, is the following statement provable in ZF?
Every compact Hausdorff space which is countable is a Baire space.


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Q4. Countable continua. (****)

[-]Are the two following statements provable in ZF?
-Every nonempty Hausdorff compact space which is countable has at least one isolated point.

-A compact Hausdorff space which is countable and which has at least two elements is not connected.


In other words, does there exist a model of ZF in which there is an infinite countable continuum?
Here, a continuum is a Hausdorff compact connected topological space.

The previous question has a philosophical flavour because it is a way of asking:
Does there exist a space (with at least two elements) which is both discrete and continuous?

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Q5. The continuous Hahn-Banach property. (*)

A vector space E over IR is said to satisfy the analytic Hahn-Banach property iff for every sublinear functional p:E-->IR, for every subspace F of E, and for every linear functional f:F-->IR such that f<=p, there exists a linear functional g:E-->IR such that g extends f and g<=p. A topological vector space E is said to satisfy the continuous Hahn-Banach property iff for every continuous sublinear functional p:E-->IR, for every subspace F of E, and for every linear functional f:F-->IR such that f<=p, there exists a linear functional g:E-->IR such that g extends f and g<=p.

The continuous Hahn-Banach property is equivalent to the classical geometrical forms of the Hahn-Banach property; it is also equivalent to the fact that for every set I, the normed space l1(I) satisfies the continuous Hahn-Banach property (see [Fo-Mo]).

If a normed space satisfies the continuous Hahn-Banach property, then the closed unit ball of its continuous dual is convex-compact in the weak* topology (this idea is in [Lu69]). Hence, proving the continuous Hahn-Banach property on various spaces is a way to obtain convex-compact sets in ZF (see question Q2).

For example Hilbert spaces satisfy the continuous Hahn-Banach property (see [Fo-Mo]).
More generally, uniformly convex Gâteaux-differentiable Banach spaces satisfy the continuous Hahn-Banach property, and, using the axiom of dependent choices, one can prove the continuous Hahn-Banach property for Gâteaux-differentiable Banach spaces (see [Do-Mo]).
One can also prove in ZF the continuous Hahn-Banach property for uniformly smooth Banach spaces (see [Al-Mo]).


[-] Is the continuous Hahn-Banach property for Frechet-differentiable Banach spaces provable in ZF?
[-] Is the continuous Hahn-Banach property for Gâteaux-differentiable Banach spaces provable in ZF?

[-] More generally, which normed spaces satisfy the continuous Hahn-Banach property in ZF?

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Q6. Super-reflexive spaces without choice.

Our search of Banach spaces which satisfy the continuous Hahn-Banach property leads us to the following remark: since uniformly convex Gâteaux differentiable Banach spaces satisfy the continuous Hahn-Banach property, every Banach space which is isomorphic with a uniformly convex Gâteaux differentiable Banach space also satisfies the continuous Hahn-Banach property.
In the literature, such spaces are called super-reflexive. There are many equivalent definitions of this notion, and the proofs of those equivalencies generally rely on some form of the axiom of choice because ultraproducts of Banach spaces (hence nontrivial ultrafilters) occur in these proofs.
[-] Among the many definitions of "super-reflexive space", which ones are equivalent (in ZF) to the following: "to be isomorphic with a uniformly convex Gâteaux differentiable Banach space".

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Q7. The finite tree property.

Among the various definitions which are equivalent (in set-theory with the axiom of choice) to super-reflexivity, there is one which is: "not to satisfy the finite tree property". It has been shown by Enflo, (see [En72]), that if a normed space E does not satisfy the finite tree property , then the space E has an equivalent norm which is uniformly convex. This theorem is provable in ZF . In set theory with the axiom of choice, it can be proved that uniformly smooth spaces do not satisfy the finite tree property: the classical proof (see [Bea85]) relies on a proof by contradiction, and the use of envelopes of Banach spaces (quotients of ultraproducts of Banach spaces). Hence the classical proof relies on the existence of nontrivial ultrafilters so it uses the axiom of choice... (see [Bl77])

[-] Is there an effective proof of the fact that uniformly smooth spaces do not satisfy the finite tree property?

More precisely:

[-] Is there a computable mapping f such that for every finite dimensional uniformly smooth normed space E , with modulus of smoothness r , the map f(r) is a witness of the fact that E does not satisfy the finite tree property.

Here we say that f(r) is a witness of the fact that E does not satisfy the finite tree property iff f(r) is a mapping from IR to IN such that, for every real number e>0 , if we put n=f(r)(e) then no (e,n)-tree is contained in the closed unit ball of E .

The same question holds for uniformly nonsquare spaces:

[-] Is there a computable mapping f such that for every finite dimensional normed space E , which is uniformly nonsquare for d>0 , the map f(d) is a witness of the fact that E does not satisfy the finite tree property.


Yet another question:
It is well known that if a normed space does not satisfy the finite tree property, then its continuous dual does not satisfy this property. Is there an effective proof of this result? In other words:
[-] Is there a computable mapping f such that for every finite dimensional normed space E , if t is a witness of the fact that E does not satisfy the finite tree property, then f(t) is a witness of the fact that the continuous dual E' does not satisfy the finite tree property.

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Q8. Weak compactness and reflexive spaces.

Say that a Banach space E is (simply) reflexive if the canonical mapping j_E from E to the continuous bidual E'' is isometric
and onto.  Using the axiom of choice, one can prove the following axiom:
RA: "The closed unit ball of a reflexive space is compact in the weak topology."
More precisely, the Tychonov axiom for compact Hausdorff spaces (which is equivalent to the Boolean Prime Ideal theorem, and also to the Alaoglu axiom), implies the axiom RA .

[-] Does the axiom RA imply the Tychonov axiom?
ANSWER: No!  see [Mo04]

Remark Consider the following statement:
RAh: "The closed unit ball of a Hilbert space is compact in the weak topology."
This statement is not provable in ZF (see [Fo-Mo]) but it is a consequence of the axiom of dependent choices (see [De-Mo]) hence RAh does not imply the Tychonov axiom.

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Q9. Reflexive spaces without choice.

There are many definitions of reflexivity: the equivalencies between these definitions generally rely on some form of the axiom of choice. It would be a good thing to know, among these equivalencies, which ones hold in ZF. Some of them (convex-reflexivity, simple-reflexivity, J-reflexivity, onto-reflexivity, omega-reflexivity) are studied in [Mo04].
Marianne Morillon