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 Theorem $\Rightarrow$ 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.

  • 74,358
  • 9
  • 254
  • 333
  • 11
    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
  • 4
    5 See also Carother,A Short Course on Approximation Theory (1998?), p.139-141, http://personal.bgsu.edu/~carother/Approx.html and mainly the remark in p.141. – vesszabo Jun 19 '12 at 17:59
  • 1
    @vesszabo: thanks a lot for this reference. – t.b. Jun 20 '12 at 04:18
  • 1
    Is this answered or would a bounty be of use? – Rudy the Reindeer Nov 30 '12 at 11:07
  • 14
    @Matt: This is **definitely** not a question for a bounty. Whoever answers this could probably write and publish a paper with this result, which is much more substantial than any bounty on this site. – Asaf Karagila Dec 10 '12 at 22:03
  • 1
    @MattN. Thank you very much, but I agree with Asaf's assessment. A bounty would not be of use in this case. I doubt the answer is known at all, that's why I didn't take the question to MO either. – t.b. Dec 10 '12 at 23:37
  • 1
    @AsafKaragila Just to be sure that I get OP: they are looking for a proof of Baire from open mapping without using any choice (i.e. in ZF). Correct? – Rudy the Reindeer Dec 13 '12 at 16:13
  • 4
    @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
  • 1
    @AsafKaragila What is a separating model? Also: In the first Fraenkel-Mostowski model DC fails. Why is it clear that there OMT also fails? – Rudy the Reindeer May 24 '13 at 08:17
  • 3
    @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 08:22
  • 1
    @AsafKaragila Meanwhile I'll think about how to prove that it holds. – Rudy the Reindeer May 24 '13 at 08:38
  • 1
    @AsafKaragila Though that wouldn't even settle the question: permutation models are not models of $\mathsf{ZF}$. – Rudy the Reindeer May 24 '13 at 10:22
  • 2
    @MattN. but one could presumably use Jech-Sochor since everything you need for OMT looks bounded to me. – theHigherGeometer May 27 '14 at 08:02
  • 1
    Hello there, it seems as though the proof given in Schechter uses the axiom of dependent choices in an essential way. Hence, my answer below would actually be invalid, lest one could somehow prove the CGT or the OMT from ZF+CC+UBP. – Cloudscape Nov 20 '16 at 18:42

1 Answers1


EDIT after more than a year:

At least, the uniform boundedness principle can be proven using only the axiom of countable choice, or $\mathbf{CC}$ for short. However, I have been unable to find a proof for the fact that any of the three facts

  • Every Banach space is barrelled
  • On a Banach space, a lower semi-continuous seminorm is always continuous
  • The uniform boundedness principle

implies either the closed graph theorem or the open mapping theorem (however, the two are equivalent in ZF). In particular, the argument in 27.37 of Schechter uses dependent choice, seemingly in an essential way.

Here is the sketch for $\mathbf{CC} \Rightarrow$ UBP:

We first note that for a linear operator $T$, $$ \max\{\|T(x - y)\|, \|T(x + y)\|\} \ge 1/2 (\|T(x - y)\| + \|T(x + y)\|) \ge \|T(y)\| $$ due to the triangle inequality $\|a - b\| \le \|a\| + \|b\|$.

Instead of applying the axiom of dependent choice, we first pick a sequence of operators $\|T_n\| \ge 4^n$ and $$ x_n \in \left\{x \in X \middle| \|x\| \le 1 \text{ and } \|T_n(x)\| \ge 2/3 \|T_n\|\right\} =: S_n $$ using countable choice. Then, since dependent choice with a function instead of a relation is a theorem of ZF, we define a function which maps $(y_n, n)$ to $(y_{n+1}, n+1)$ where $$ y_{n+1} = \begin{cases} y_n - 3^{-(n+1)} x_{n+1} & \|T_{n+1}(x_{n+1} - 3^{-(n+1)} x_{n+1})\| \ge \frac{2}{3} 3^{-(n+1)} \|T_{n+1}\| \\ y_n + 3^{-(n+1)} x_{n+1} & \text{otherwise}. \end{cases} $$ If we set $y_1 := 1/3 x_1$, then we may define a sequence from the repeated application of that function. The key lemma of Sokal's ensures that $\|T_n(y_n)\| \ge \frac{2}{3} 3^{-n} \|T_n\|$ for all $n \in \mathbb N$. From the triangle inequality follows if $k > n$ $$ \|y_n - y_k\| \le \sum_{j=n}^\infty \|y_j - y_{j+1}\| \le \sum_{j=n}^\infty \frac{2}{3} 3^{-(j+1)} \le \frac{1}{2} 3^{-n} $$

Thus, $(y_l)_{l \in \mathbb N}$ is Cauchy (with limit $y$, say), and it also follows that

$$ \|T_n(y)\| \ge \left| \|T_n(y_n)\| - \|T_n(y - y_n)\| \right| \ge 1/6 (4/3)^n. $$

  • 2,308
  • 12
  • 22
  • 3
    I've taken a quick look, and I think it looks correct! A couple comments: (a) I guess you want to take $y_0 = 0$ or something like that, to get the induction started. (2) The notation $y_{n+1} \in 3^{-(n+1)} S_{n+1}$ is a little hard to sort out. I guess a simpler way of stating the desired statement is simply that $\|T_n y_n\| \ge 2 \cdot 3^{-(n+1)} \|T_n\|$ for all $n$. – Nate Eldredge Sep 04 '15 at 17:01
  • Answer updated according to your request; I started with $y_1$ since that's what I'm used to. If there is a reason to do otherwise, let me know, I'll change it then. – Cloudscape Sep 04 '15 at 17:08
  • Awesome. Thank you very much! – t.b. Sep 12 '15 at 10:56
  • 2
    @t.b.: It's nice to see attention for such an ancient question. It's nice to see your presence in any case! – robjohn Sep 12 '15 at 13:06
  • @t.b.: What did you need this for? – Cloudscape Sep 12 '15 at 13:36
  • 3
    To satisfy my curiosity: I was wondering whether the theorem of Baire is actually needed for the fundamental principles of functional analysis. I'm still amazed that this isn't the case. [A few years ago there were a number of questions that were related to weak forms of the AoC and functional analytic principles, probably due to Asaf's influence]. @robjohn: Thanks a lot. I merely wanted to give credit where credit is due, and this amazing answer deserved proper acceptance and more attention. Unfortunately, I still don't have much time these days... See you! – t.b. Sep 18 '15 at 14:04
  • Was this paper submitted for publication? – Asaf Karagila Jan 03 '16 at 22:11
  • Yes it is submitted, but it is still in review. The website of the journal is available here: http://logicandanalysis.org/index.php/jla/index – Cloudscape Jan 04 '16 at 14:51
  • Have you found the missing brick in UBP => CC? I.e. a method of extracting a cardinality-bounded subsequence of subsets? If so, one should perhaps consider to add you (as an author) and your proof into the thing, to have it available at one place. – Cloudscape Jan 04 '16 at 15:03
  • 1
    No. But I am writing a final paper about functional analysis and the axiom of choice in a functional analysis course. I will survey your proof. And while I don't expect to solve any open problems, I will probably invariably think about them. Ping me again in a month, after I've submitted the thing (you should also use `@Asaf` so I'll be notified of your comment). – Asaf Karagila Jan 04 '16 at 21:38
  • @AsafKaragila I should add though that some of the referees seem to be quite serious mathematicians, and to be honest I could imagine that one of them cracks the problem. Hence, it might be advisable to only seriously work on this problem after they have stopped the review process. – Cloudscape Jan 05 '16 at 08:21
  • @Asaf this comment serves the purpose of making sure you are notified – Cloudscape Jan 05 '16 at 08:22
  • 2
    I know, but I am not going to copy-paste the result. I'll read it, and if it's relatively short I will just re-digest it (and of course, I'll let you know if I find mistakes, even though I can assure you that I am not the referee!) and write it in my final paper. If it seems out of place, I will just refer to the preprint and write something not inaccurate like "a proof has been proposed that ..." (And the comment notification worked with both of them, I was notified of *two* comments). – Asaf Karagila Jan 05 '16 at 09:17
  • @AsafKaragila Ping (sorry for forgetting to ping earlier) – Cloudscape Feb 20 '16 at 10:25
  • Does that mean that the paper is now accepted for publication? – Asaf Karagila Nov 20 '16 at 21:07
  • @AsafKaragila On condition that I make certain things more precise, it is; I hope that a strongly revised version, which I am preparing at the moment, will convince the referees (who were a little sceptical due to my, well, a bit imprecise writing style, which hopefully I was able to improve). The revised version will also appear on the arxiv, as a new version of the current paper. – Cloudscape Nov 20 '16 at 22:18
  • Thanks! Do keep me posted! – Asaf Karagila Nov 20 '16 at 22:19
  • Does your theorem imply that Banach spaces are barreled without DC? – Yai0Phah Feb 02 '21 at 19:57