# Does the open mapping theorem imply the Baire category theorem?

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.

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.

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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
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
@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
@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
@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

In my upload http://arxiv.org/abs/1509.01078 I tried to modify Sokal's idea in order to avoid the axiom of dependent choice and instead rely on the axiom of countable choice, which is strictly weaker than dependent choice.

EDIT (proof sketch(EDIT2) as wished for):

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 $\|a - b\| \le \|a\| + \|b\|$ (this is the key lemma in Sokal's proof).

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

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

EDIT 3: We may hence conclude that the open mapping theorem does not imply the Baire category theorem, since if that were true, we would have $$\text{Countable choice} \Rightarrow \text{UBP} \Rightarrow \text{Open mapping theorem} \Rightarrow \text{Baire} \Rightarrow \text{Dependent choice}.$$ But it is known that countable choice does NOT imply dependent choice (due to Jensen 1966; see, for example, A. Levy's Basic set theory, p. 169).

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 4 '15 at 17:01
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 4 at 21:38