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A nice observation by C.E. Blair1, 2, 3 shows that the Baire category theorem for complete metric spaces is equivalent to the axiom of (countable) dependent choice.

On the other hand, the three classical consequences of the Baire category theorem in basic functional analysis — the open mapping theorem, the closed graph theorem and the uniform boundedness principle (as well as Zabreiko's lemma) — are equivalent to each other in Zermelo–Fraenkel set theory without choice: that is to say, if one is added as an axiom to ZF then the others follow4.

Each of these results has a more or less direct proof from the Baire category theorem and all the proofs “avoiding Baire” I'm aware of5 involve dependent choice in a way that doesn't seem to be replaceable by weaker forms of choice.

Hence I'm asking about the converse:

Does the open mapping theorem imply the Baire category theorem?

If not, is it at least true that the open mapping theorem implies the axiom of dependent choice for subsets of the reals?

I imagine that applying any of the above results to a judiciously chosen space and/or operator(s) might yield the desired conclusion, similarly to what happens in Bell's and Fremlin's geometric version of the axiom of choice6. Unfortunately, I couldn't find a promising place to start.

Needless to say that I checked numerous things on the web form of Howard and Rubin's book Consequences of the Axiom of Choice, but without much success: The only articles that I found this way are J.D. Maitland Wright's articles7.


Footnotes and References:

1 Charles E. Blair, The Baire category theorem implies the principle of dependent choices, Bull. Acad. Polon. Sci. Sér. Sci. Math. Astronom. Phys. 25 (1977), no. 10, 933–934.

2 Since Blair's article is hard to find, the proof can be found in the notes to chapter 9, page 95 of John C. Oxtoby, Measure and Category, Springer GTM 2, Second Edition, 1980.

3 Here's the idea of Blair's argument for the implication Baire Category Thereom $\Rightarrow$ Dependent Choice: let $S$ be a set and let $R \subset S \times S$ be a relation such that for all $s \in S$ there exists $t \in S$ such that $(s,t) \in R$. Equip $S^{\mathbb{N}}$ with the complete metric $d(f,g) = 2^{-\min\{n\,:\,f(n)\neq g(n)\}}$, put $$ U_n = \bigcup_{m = n+1}^{\infty} \bigcup_{(s,t) \in R} \{f \in S^{\mathbb{N}}\,:\,f(n) = s, \,f(m)=t\}, $$ observe that $U_n$ is open and dense and use $f \in \bigcap_{n=1}^\infty U_n$ and the well-order on $\mathbb{N}$ to find a strictly increasing sequence $k_1 \lt k_2 \lt \cdots$ such that the sequence $(x_n)_{n=1}^\infty$ given by $x_n = f(k_n)$ satisfies $(x_n, x_{n+1}) \in R$ for all $n \in \mathbb{N}$.

4 See e.g. E. Schechter, Handbook of Analysis and its foundations, 27.27, pp. 734ff.

5 A good example for this is Sokal's A Really Simple Elementary Proof of the Uniform Boundedness Theorem, The American Mathematical Monthly Vol. 118, No. 5 (May 2011), pp. 450–452, ArXiV Version. While admittedly it is beautifully simple and elementary, it involves a plain application of dependent choice in the main argument.

6 Bell and Fremlin, A Geometric Form of the Axiom of Choice, Fund. Math. vol. 77 (1972), 167–170.

7 The full list of relevant articles can be obtained with this ZBlatt query two of which appeared in rather obscure proceedings, so I couldn't get my hands on them, yet. The third article is J. D. Maitland Wright, All operators on a Hilbert space are bounded, Bull. Amer. Math. Soc. 79 (1973), 1247–1250.

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t.b.: I have the paper Functional analysis for the practical man from you Zentralblatt query. Feel free to e-mail me if you don't get the paper elsewhere. (You can find my email address easily from my website - the link is given in my profile.) –  Martin Sleziak May 21 '12 at 14:12
    
Thank you very much! I've sent you an email. –  t.b. May 21 '12 at 14:18
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5 See also Carother,A Short Course on Approximation Theory (1998?), p.139-141, personal.bgsu.edu/~carother/Approx.html and mainly the remark in p.141. –  vesszabo Jun 19 '12 at 17:59
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@Matt: Correct. Equivalently of DC from OMT; or alternatively a separating model in which DC fails but OMT holds. –  Asaf Karagila Dec 13 '12 at 16:17
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@Matt: Separating models are models where one statement is true, and the other is false. It's not entirely obvious, but it's likely. Perhaps in some vector space which is generated from the amorphous set (the set of atoms), or so. I'll need to think about it some more. –  Asaf Karagila May 24 '13 at 8:22
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It's important to remark that the closed graph theorem and the open mapping theorem hold for a class of locally convex topological vector spaces which is strictly larger than the class of Baire spaces (that is, the ones for which the Baire theorem holds). For instance, if $E$ and $F$ are strict inductive limits of sequences $(E_n)_{n\in\mathbb{N}},(F_n)_{n\in\mathbb{N}}$ of Baire spaces such that $E_n\subsetneq E_{n+1}$ and $F_n\subsetneq F_{n+1}$ for all $n$ (e.g. $E=F=\mathscr{D}(U)$, the space of smooth real- or complex-valued functions with compact support on an open subset $U\subset\mathbb{R}^n$, if Baire's theorem holds true for complete metrizable locally convex spaces), then $E$ and $F$ are not Baire spaces but the closed graph theorem and the open mapping theorem hold for them (see, for instance, Chapter 5 of H. Jarchow, "Locally Convex Spaces" (B. G. Teubner, 1981) for a fairly thorough discussion on these matters). I'm not sure enough of the details of the proofs to see whether it needs only ZF or it needs additional assumptions such as AC or DC, which would kill this would-be counterexample.

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PLEASE read the question. t.b.'s questions are very thorough and very complete. This is not some half-question posted from notes or homework sheets. This is a very complete question about the consistency of a certain implication in the absence of the axiom of choice. It is not about "If space X has property P then it has property Q", which is what you have written about in your answer. –  Asaf Karagila Dec 10 '12 at 22:16
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Because the question states the following: The BCT implies DC in ZF. BCT also implies OMT in ZF. Does OMT imply BCT in ZF? –  Asaf Karagila Dec 10 '12 at 22:18
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I now see what you are trying to say. Thanks! At the same time, Pedro has been trying to answer a question. He may have misunderstood, but I guess he probably deserves a less offensive treatment. –  Sanchez Dec 10 '12 at 22:42
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I suppose you're right and I have been quite hostile. Pedro, I apologize if I've been rude to you. –  Asaf Karagila Dec 10 '12 at 22:47
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Caps lock, bold face and multiple punctuation -- is it Mathemagic12345 or a broken keyboard? We will never know. –  Matt N. Dec 12 '12 at 14:55
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