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The Banach-Alaoglu theorem is well-known. It states that the closed unit ball in the dual space of a normed space is $\text{wk}^*$-compact. The proof relies heavily on Tychonoff's theorem.

As I have recently figured out thanks to the nice guys on the chat belonging to this website is that Tychonoff's theorem is equivalent to the Axiom of Choice.

Can we prove Tychonoff (or something else equivalent) from Alaoglu?

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1  
It isn't mentioned in this source, but in general a great source for AC matters is consequences.emich.edu/conseq.htm . –  Qiaochu Yuan May 14 '12 at 23:38
    
@Asaf Excellent. You're one of the cool guys now. –  Jonas Teuwen May 15 '12 at 0:08
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I replaced my comment with an answer because I wanted to add a bit more on what I wrote, and also to correct the title of the paper I cited. Two birds, one Stone-Cech compactification. –  Asaf Karagila May 15 '12 at 0:29
    
Just a small remark: You don't need Tychonoff's theorem to prove Alaoglu's theorem: A topological space is compact if and only if every universal net converges. Take a universal net $\varphi_i$ in the unit ball of $X^\ast$. For every $x \in X$ you have that $\varphi_i(x)$ converges because it is a universal net in some closed disk in $\mathbb{C}$; call the limit $\varphi(x)$. A direct check shows that $\varphi \in X^\ast$ and it lies in the unit ball. By definition $\varphi_i \to \varphi$ in the weak$^\ast$-topology. –  t.b. May 15 '12 at 9:11
    
@t.b.: based on their relationship to filters, I would guess that the machinery needed to set up the notion of universal nets is the same as the machinery needed to set up the notion of ultrafilters (namely one needs some form of the ultrafilter lemma). So Tychonoff's theorem for compact Hausdorff spaces still appears. –  Qiaochu Yuan May 15 '12 at 21:23

4 Answers 4

up vote 18 down vote accepted

I decided to outline the proof of the equivalence of some of the results mentioned in the other answers since it is quite easy.

The following statements are equivalent in ZF:

  1. Banach–Alaoglu: If $X$ is a normed space then the unit ball in $X^\ast$ is weak$^\ast$-compact.
  2. The ultrafilter lemma: Every filter is contained in an ultrafilter.
  3. Tikhonov for Hausdorff spaces: An arbitrary product of compact Hausdorff spaces is compact.
  4. Tikhonov for the unit interval: An arbitrary product of copies of $[0,1]$ is compact.

Remark. It is a theorem of Halpern–Levy, The Boolean prime ideal theorem does not imply the axiom of choice, Axiomatic Set Theory Part 1, Proc. Symp. Pure Math. Vol. 13 (1971), 83–134, that these equivalent statements are strictly weaker than the full axiom of choice. See also Jech's book The Axiom of Choice, chapter 7.

The implications 2. $\Rightarrow$ 3. and 4. $\Rightarrow$ 1. are standard (the first one is obtained by inspection of the usual proof of Tikhonov's theorem while the other is one of the usual proofs of the Banach-Alaoglu theorem, as explained by M. Turgeon here) and 3. $\Rightarrow$ 4. is trivial.

It remains to prove that the Banach–Alaoglu theorem implies the ultrafilter lemma.

  1. Recall that an ultrafilter on a set $S$ is the same thing as a $\{0,1\}$-valued finitely additive probability measure defined on the entire power set $P(S)$: For an ultrafilter $\mathfrak{U}$ and $A \subset S$ set $$ \mu_{\mathfrak{U}}(A) = \begin{cases} 1, & \text{if } A \in \mathfrak{U}, \\ 0, & \text{otherwise} \end{cases} $$ to get a finitely additive measure $\mu_\mathfrak{U}$. Conversely, given a finitely additive $\{0,1\}$-valued probability measure $\mu$, define $\mathfrak{U}_\mu = \{A \subset S\,:\,\mu(A) = 1\}$ then the the mutually exclusive possibilities $A \in \mathfrak{U}_\mu$ or $S \smallsetminus A \in\mathfrak{U}_\mu$ imply that $\mathfrak{U}_\mu$ is an ultrafilter.

  2. Observe that $\operatorname{ba}(S) = \ell^{\infty}(S)^\ast$ is the same as the Banach space of bounded (and signed) finitely additive measures on $P(S)$ with the total variation norm. The identification is straightforward: since $P(S) = \{0,1\}^S \subset \ell^{\infty}(S)$, every bounded linear functional on $\ell^{\infty}(S)$ gives a finitely additive measure. For the reverse direction, use the fact that the $\mathbb{Q}$-linear span of $P(S)$ is norm-dense in $\ell^{\infty}(S)$. The identification of the norm follows by direct inspection of the definitions.

  3. Let $B$ be the unit ball of $\operatorname{ba}(S)$, equipped with the weak$^\ast$-topology, so that it is compact by Banach-Alaoglu. The “subset of ultrafilters” $$ U = \{\mu \in B\,:\,\mu(A)\in\{0,1\} \text{ for all } A \subsetneqq S\text{ and }\mu(S) = 1\} \subset B $$ is weak$^\ast$-closed and we have a map $\delta: S \to U$ sending $s \in S$ to the corresponding Dirac measure (= principal ultrafilter).

    Given a filter $\mathfrak{F}$ on $S$, the collection $\mathcal{F} = \left\{\overline{\delta(F)}\right\}_{F \in \mathfrak{F}}$ of closed subsets of $U$ has the finite intersection property, so by compactness of $U$ the intersection $\bigcap \mathcal{F}$ is non-empty. Let $\mu$ be an element of that intersection. Notice that $\mu$ is $\{0,1\}$-valued and $\mu(F) = 1$ for all $F \in \mathfrak{F}$, so we're done, because the ultrafilter $\mathfrak{U}_\mu$ contains $\mathfrak{F}$.

Remark: Note that the idea is to implicitly work with the Stone–Čech compactification $\beta S$ by identifying it with the weak$^\ast$-closure $U$ of $\delta(S) \subset \operatorname{ba}(S)$ via the Gel'fand isomorphism $C(\beta S) = \ell^\infty(S)$, but putting it this way again requires relying on an equivalent of the ultrafilter lemma.


I second Asaf's recommendation to read the short article by Bell and Fremlin, A Geometric Form of the Axiom of Choice, Fund. Math. vol. 77 (1972), 167–170, showing various implications between functional analytic principles and choice. Especially the fact that the axiom of choice follows from the existence of an extreme point in the unit ball of a dual Banach space is a beautiful observation in the geometry of Banach spaces.

It would be tempting to put the upshot of Bell and Fremlin as “Hahn–Banach and Kreĭn–Mil'man imply the axiom of choice” (as one can sometimes read), however, the situation is slightly more subtle than that. More details in Bell–Fremlin and a bit more at the end of this answer.


Let me add a few points to Asaf's answer:

The Tikhonov theorem implies the Hahn–Banach theorem by an argument of Łoś–Ryll-Nardzewski, On the application of Tychonoff's theorem in mathematical proofs, Studia Math. 38 (1951), 233–237.

The idea is this: Let $U$ be a subspace of the vector space $V$ and let $f: U \to \mathbb{R}$ be a functional dominated by a sublinear functional $p: V \to [0,\infty)$. If we want to extend $f$ to $v \in V \smallsetminus U$ then we can only choose $\tilde{f}(v) \in [-p(-v),p(v)]$ if we want the extension $\tilde{f}$ to be dominated by $p$. For each finite subset of $V \smallsetminus U$ we can choose from a finite product of compact intervals plus we have linearity constraints to fulfill — which amounts to saying that finding a Hahn–Banach extension is the same as finding an element in a gigantic projective limit of compact spaces. That this projective limit isn't empty follows from Tikhonov's theorem. More details are in section 5 of loc. cit.

A direct proof of the Hahn–Banach theorem from the ultrafilter lemma, using the language of non-standard analysis, was given by Luxemburg in Two applications of the method of construction by ultrapowers to analysis, Bull. Amer. Math. Soc. 68 (1962), 416–419, see also his article Reduced powers of the real number system and equivalents of the Hahn- Banach extension theorem, Appl. Model Theory Algebra, Anal., Probab., Proc. Int. Sympos. Calif. Inst. Technol. 1967, 123-137 (1969) ZBlatt review.

In this latter article Luxemburg proves among many other things that Hahn–Banach is equivalent to the statement that the unit ball in the dual of a normed space is convex-compact: every cover by open and convex sets has a finite subcover.

This and various other facts seem to indicate that Hahn–Banach might imply Banach-Alaoglu (or the ultrafilter lemma) — at least that's the sentiment expressed by Luxemburg and others in various places.

However, surprisingly enough, this turns out to be wrong: D. Pincus, The strength of the Hahn-Banach theorem, Lecture Notes in Mathematics Volume 369 (1974), 203–248 constructs a model in which Hahn-Banach holds, but both the ultrafilter lemma and Kreĭn–Mil'man fail, so they are in fact independent of ZF+HB. In the same paper Pincus also established that the axiom of choice is independent of Hahn–Banach and Kreĭn–Mil'man (at least in ZFA). See also the announcement Independence of the prime ideal theorem from the Hahn Banach theorem, Bull. Amer. Math. Soc. 78 (1972), 766-770.

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3  
Have I told you that you're crazy? But it's good crazy... :-) –  Asaf Karagila May 15 '12 at 21:36
    
Wow, tb, you're one of the crazy bros. Best answer so far! I should study the proofs in detail. –  Jonas Teuwen May 15 '12 at 21:37
    
I wonder why you prefer the printed version over the one I linked which is easier to read... –  Asaf Karagila May 16 '12 at 13:04
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@Asaf: It's a persistent link... Yours truly, Mr crazy, but good crazy library man. –  t.b. May 16 '12 at 13:14
    
Great answer! And thanks to you, my answer to a previous question circulates! –  M Turgeon May 21 '12 at 15:39

The Banach-Alaoglu principle is equivalent to, among others, following theorems: the Ultrafilter Lemma, the Boolean prime ideal theorem, Stone' representation theorem, the Tychonoff theorem for compact Hausdorff spaces, the completeness theorem for first order logic$\ldots$

Proofs of the equivalence of these, and more, theorems can be found in the "Handbook of Analysis and Its Foundations" by Eric Schechter. All these results have an extremely non-constructive flavor.

Edit: I removed a reference to a paper attempting to show that the Banach-Alaoglu theorem is equivalent to the Tychonoff theorem for compact Hausdorff spaces, since t.b. raised some legitimate concerns about the proof. The equivalence is, however, correct.

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Another equivalent which may be of interest to this topic (at least more than FOL completeness theorem) is the general Arzela-Ascoli theorem. –  Asaf Karagila May 15 '12 at 0:13
    
I'm a bit puzzled about the use and claims about Urysohn's lemma in the proof of Theorem 7 in Rossi's preprint. It's not true that Urysohn's lemma is independent of choice and that the proof is "by construction". In Läuchli's second model Urysohn breaks down (even for locally compact spaces), see $\mathcal{N}8$ in Howard-Rubin. Now I'm sure that the argument can be repaired, but this looks like a gap to me (also the choice of the functions $f_\lambda$ in that proof looks rather fishy given the claims that AC isn't used). –  t.b. May 15 '12 at 9:33
    
Lauchli and his models... sheer fun! He's my hero for shattering choice as hard as humanly possible (the rationals have two non-isomorphic algebraic closures). –  Asaf Karagila May 15 '12 at 10:11
    
@t.b.: You are right. The standard "constructive" proof of Urysohn's lemma for normal spaces seems to involve something like DC. The $f_\lambda$ are constructed by the same approach. –  Michael Greinecker May 15 '12 at 10:13

No. The compact sets whose product you want to be compact in Banach-Alaoglu are also Hausdorff. The statement that a product of compact Hausdorff spaces is compact (Hausdorff) is equivalent to the ultrafilter lemma, which is known to be strictly weaker than the axiom of choice.

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So, am I right if I say that if we have a normed space (which is Hausdorff) we only need the ultrafilter lemma to prove Alaoglu? –  Jonas Teuwen May 14 '12 at 23:43
    
@Jonas: Yes, but the fact that the space is Hausdorff has nothing to do with that. I mean, it follows, but the claim "Every product of compact Hasudorff spaces is compact" is enough to prove Alaoglu as a theorem. Similarly saying that "Every filter can be extended to an ultrafilter" is enough to prove that. –  Asaf Karagila May 15 '12 at 0:34

To add on the previous answers, Bell and Fremlin proved in [1] that a slightly weaker version of Alaoglu's theorem is equivalent to Hahn-Banach, and that together with a modified version of the Krein-Milman theorem (SKM) it applies the axiom of choice in full.

It is worth noting two interesting things on this:

  1. Hahn-Banach is strictly weaker than the Boolean Prime Ideal theorem (mentioned by Qiaochu and Michael).
  2. The Boolean Prime Ideal theorem with the ordinary Krein-Milman theorem imply the axiom of choice in full.

So there is some sort of a tipping point here, we can slightly weaken Banach-Alaoglu or slightly weaken SKM.


Bibliography:

  1. J. Bell and D. Fremlin, A Geometric Form of the Axiom of Choice. Fund. Math. vol. 77, 1972. (Official scan from the journal page here and the abstract page of the Polish Virtual Library)
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