# Bounded operators and axiom of choice

In the article below, it is shown that the proposition "Every linear operator defined on a whole Hilbert space is bounded" is consistent with the axioms of ZF + a weakened version of the axiom of choice (called DC).

So, if I want to prove that an operator A defined on a Hilbert space H is bounded, is it enough to just check that the axiom of choice was not used to define it?

And a related question: To show that a set is measurable, is it enough to check that the definition of this set didn't use the axiom of choice? (As similarly, the statement "all subsets of R" are Lebesgue measurable is consitent with ZF without the axiom of choice.)

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Do note that it is often hard to escape the axiom of choice. E.g. almost every time you say "complete $A$ to a basis" you use the axiom of choice. – Asaf Karagila Jun 23 '12 at 11:51

In the case of measurable sets, something like your statement is true, but your particular criteria is not literally correct. It is an over-simplification, since once one rises to a certain level of complexity, the issue isn't whether the definition of your set "uses the axiom of choice", but rather the issue is a matter of the properties of the ambient set theoretic universe in which you are defining your set.

The basic fact is that yes, sets of reals defined by particularly simple definitions are indeed automatically measurable. (One should assume at least a small amount of choice, say the DC principle, in order to have a satisfactory theory of Lebesgue measure.)

To illustrate the lowest level of this, you probably know that every Borel set is Lebesgue measurable. This is an instance of the definability claim, because the Borel sets are precisely those sets that have complexity $\Delta^1_1$ in the descriptive set-theoretic hierarchy, which means that they can be defined by a property involving quantification over finite objects plus a single universal quantifier over the reals, and equivalently by a definition with a single existential quantifier over the reals.

More generally, the $\Sigma^1_1$ sets are the projections of Borel sets, and these are also measurable. These are the sets that can be defined with a single existential real quantifier, followed by any number of quantifiers over a fixed countable realm. If one has Martin's axiom plus $\neg$CH, then this rises to the $\Sigma^1_2$ sets.

Above this, one has the very interesting phenomenon that the assertions that classes of sets are Lebesgue measurable begin to have large cardinal consistency strength. For example, Solovay proved that the assertion that every $\Sigma^1_3$ set is measurable--these are the sets definable with the quantifier structure $\exists x\forall y\exists z\ \varphi(\cdot,x,y,z)$, where $\varphi$ uses only arithmetic quantifiers---is equiconsistent with the existence of an inaccessible cardinal.

The existence of much stronger large cardinals have outright consequences for the measurability of projective sets. For example, the principle known as Projective Determinacy, which is equiconsistent with and implied by the existence of infinitely many Woodin cardinals, implies that every projective set is determined. Thus, under PD, any set of reals that can be defined by quantifying only over the reals and over the natural numbers will automatically be Lebesgue measurable.

Some of these issues were considered in other questions here on mathoverflow, such as ZFC plus every analytical set is measurable?.

Finally, to show that particular criterion you mention is not correct, observe that by forcing we can make any particular set of reals be the reals that are coded into a block of the GCH pattern on the cardinals $\aleph_\alpha$. Thus, I can define a set of reals by saying "the set of reals whose binary expansion occurs as a part of the GCH pattern". This definition does not use the axiom of choice, but it can be used to define any given set, which may or may not be measurable. There are numerous other ways to see a similar effect.

A similar observation applies the bounded operator case---in principle any set can be defined by reference to the GCH pattern, and these definitions do not in principle refer to the axiom of choice.

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This is the standard answer set theorists give to this type of question, and it is useful and interesting. Intuitively, it is a bit unsatisfying since one has the sense that one is never at risk of defining up an unmeasurable set by accident, but it is not always obvious that particular definitions are $\Sigma^1_1$. For that matter, one might even define a set by quantifying over sets of reals, but then there is no guarantee of measurability even given large cardinals, right? Also, just out of curiosity, is there any reason to think sets constructed using DC will be sometimes be measurable? – Marian Jun 23 '12 at 11:46
I agree that the general lesson of these results is that if you define a set projectively (quantifying only over reals and integers), then indeed we should expect it to be measurable (and the details of this involve large cardinals). And with practice, I think one does get a good sense of the complexity of the definitions one encounters. Descriptive set theorists are typically quite quick to say a particular definition is $\Sigma^1_2$ or $\Pi^1_4$ or whatever, and one gets the knack of it. As for your final question, to my way of thinking, definitions do not use axioms at all, ... – Joel David Hamkins Jun 23 '12 at 13:54
..but rather we use axioms when proving a theorem. So perhaps you have in mind a situation where you prove that there is a certain kind of set, and your proof uses DC but not AC. In this case, for the reasons I explain in the last paragraphs of my answer, we cannot deduce that it must or must not be measurable. – Joel David Hamkins Jun 23 '12 at 13:56
I don't agree with that. When you "use the axiom of choice" in a construction, what you are really doing is providing a construction relative to a particular choice function (or well-ordering or whatever). So you aren't defining a specific set, but rather defining a function from the choice-functions to the sets. The axiom of choice is needed in order to know that this is not a vacuous construction. It is consistent with ZFC that there are projectively definable well-orderings of $\mathbb{R}$, so in those universes, even when you use AC in this way, the resulting set is still simply definable. – Joel David Hamkins Jun 23 '12 at 14:19
Joel, you mentioned defining a set of reals in terms of the GCH pattern. If what you mean by GCH pattern is "for which $\alpha$ is $2^{\aleph_\alpha}=\aleph_{\alpha+1}$," then a good deal of choice is needed to ensure that this makes sense and does hat you want. Even if this is not what you meant by GCH pattern, it would seem to be a reasonable definition, but only in the presence of a good deal of choice. The bottom line is that uses of AC in defining sets of reals don't necessarily look like what you described in your comment (though they very often do). – Andreas Blass Jun 23 '12 at 16:19

Joel has explained very well the situation in ZF+DC and the effect of large cardinals. Let me add that a consistency result due to Solovay comes very close to achieving what the question hoped for, but (because it's a consistency result) only in certain models of set theory. Solovay's result says that the following theory is consistent (relative to ZFC plus an inaccessible cardinal): ZFC plus Lebesgue measurability of all sets of reals that are definable (in the language of set theory) with a countable sequence of ordinals as a parameter. (Note that the permitted parameters include reals and, via coding, countable sequences of reals.) I believe that this sort of definability is extensive enough to cover anything likely to arise in analysis (when no set theorists are involved).

I also believe that the analogous result holds, in the same model, for the part of the question about boundedness of Hilbert space operators, but I have not thought enough about this to make any guarantees.

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Andreas, automatic continuity holds in Solovay's model for Banach spaces and therefore for Hilbert spaces too. – Asaf Karagila Jun 23 '12 at 17:18

if I want to prove that an operator A defined on a Hilbert space H is bounded, is it enough to just check that the axiom of choice was not used to define it?

In a certain sense, the answer is "yes", taking into account the peculiarities that Joel notes. But if you actually have explicitly written down the definition of your operator on an explicit Banach space: in every known case it will be easy to simply apply the closed graph theorem to conclude continuity. Far easier than doing the meticulous checking that the Axiom of Choice has been avoided.

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If I recall correctly the common way of proving the closed graph theorem is through the Baire category theorem (in one way or another) which is equivalent to DC itself. Of course the question assumes DC, but it is worth mentioning that methinks. – Asaf Karagila Jun 23 '12 at 14:19