# Why is Axiom of Choice "a convenient and safe labour-saving device"?

In Terence Tao's Analysis, he states

The axiom is almost universally accepted by mathematicians. One reason for this conﬁdence is a theorem due to the great logician Kurt Godel, who showed that a result proven using the axiom of choice will never contradict a result proven without the axiom of choice. More precisely, Godel demonstrated that the axiom of choice is undecidable; it can neither be proved nor disproved from the other axioms of set theory, so long as those axioms are themselves consistent.

Then he writes

In practice, this means that any “real-life” application of analysis (more precisely, any application involving only “decidable” questions) which can be rigorously supported using the axiom of choice, can also be rigorously supported without the axiom of choice, though in many cases it would take a much more complicated and lengthier argument to do so if one were not allowed to use the axiom of choice. Thus one can view the axiom of choice as a convenient and safe labour-saving device in analysis.

In ZF, if a proposition is "decidable", and if we prove the proposition in ZFC, then it is true in ZF. I understand it. If we prove sth is true in ZFC, then it is undecidable or true in ZF. but I think the hardest part is to demonstrate a proposition is "decidable". How can we do that? Is there anyway to prove a proposition "decidable"? So I think the axiom of choice is not safe. I don't understand Tao's words. Also, I don't understand why he asserts "real-life" application is always "decidable".

• It would probably help, for those of us without access to the book, if you gave a little bit more context, so that we know what the "this" in the first sentence refers to. Commented Jun 9, 2018 at 8:02
• @MeesdeVries Thank you for your advice. Commented Jun 9, 2018 at 8:24
• The main reason why the axiom of choice is "safe" is that for most people it represents our intuition of sets, and Gödel showed that if there was any inconsistency with it; there was one with the rest of the axioms. Commented Jun 9, 2018 at 9:33
• @Max it is true that this is how we can first look at it, but would you say so about the well ordering principle?(ofc, without knowing they are equivalent), the move to think about AC it may be intuitive but the more you study it AC become less and less intuitive. I would say a big reason it is "safe" is because it is useful to get desired result(like the existence of ordinal for every cardinal), and, ofc, I can't argue about that that Gödel's proof of the consistency of AC in ZF helps a lot
– Holo
Commented Jun 25, 2018 at 18:55
• @Holo to be honest I don't find the well ordering principle unintuitive at all; on the contrary the usual proof is super intuitive: pick a first element, then a second one, etc. at some point you reach a limit, but don't worry, you have infinite time so you can go on. Going on like this, at some point you reach the end: you've picked every element; and thus you have your well-ordering. This feels super intuitive to me. I would argue that most "weird" looking statements following from (or equivalent to) AC aren't weird at all; and that we should expect mathematics to be surprising at times Commented Jun 25, 2018 at 19:01

When you want to prove that a particular and very hard to define sequence converges, it is sometimes easier to just prove "Every monotone sequence with an upper bound is convergent."

Of course, that means you need to verify that your sequence is monotone and bounded from above. But the general theorem is simpler.

When you move forward in analysis, you run into all sort of things that have general theories. Measure, Baire category, to name two main examples.

These theories are sensitive to the axiom of choice being present or removed. For example, it could be that the real numbers are a countable union of countable sets, which would destroy all theory of measure. Or it could be that all sets are Lebesgue measurable, or have the Baire property, which again changes the way these theories behave.

All these theories, when you try to apply them to stuff that "normally comes up", you find out that they can be proved by hand in the usual cases. It requires you, however, to fuss about $\varepsilon$s and $\delta$s, or work harder and produce actual computations of things, whereas the general theories just guarantee you certain things exist.

This is why sometimes it is just easier to work with choice, when you want to talk about the theoretical parts. This is what Tao means.

But the axiom of choice comes with a terrible price of giving you all sort of weird sets, like Vitali sets, Bernstein sets, and so on. So some people would argue that this is a reason to reject choice. That it is inconsistent, or at least incompatible with our intuition. Gödel proved, however, that if the rest of $\sf ZF$ is consistent, then $\sf ZFC$ is also consistent. So adding the axiom of choice will not cause real contradictions, only odd paradoxes.

This means that accepted or rejecting the axiom of choice is not about consequences in reality, but about simplifying proofs.

Let me just point out that Gödel only proved that the axiom of choice cannot be disproved from $\sf ZF$. It was Cohen who later showed that it cannot be proved either.

• I think it may be good to emphasize that for suitably definable sets (of reals, say) one can prove that there are suitably definable choice functions (the proof taking place in set theory without choice, or with only very modest forms of choice being assumed). But these results are far from being routine. The study of these so-called selectors plays an important role in descriptive set theory and in other areas, such as optimization in analysis, see for instance here. Commented Jun 9, 2018 at 13:12
• Thank you very much, but I'm still kind of confused. Could you explain more about "All these theories, when you try to apply them to stuff that "normally comes up", you find out that they can be proved by hand in the usual cases."? What is stuff that "normally comes up"? And How can I find out that "they can be proved by hand in the usual cases"? Is there any example? Thank you so much. Commented Jun 11, 2018 at 5:19
• @Eric: Let me use an example from real life. Someone I once knew was studying mathematics and physics. He was taking some physics course, and the professor always assumed that any harmonic oscillator is continuous. But, having studied some mathematics, the someone I knew was familiar with the idea of discontinuous harmonic oscillators. So he asked the professor, why do you always assume that the oscillators are continuous? And the professor replied, can you show me a discontinuous harmonic oscillator? Commented Jun 11, 2018 at 6:54
• @Eric: The lesson from that is that physicists assume things are continuous, because the things we can observe are generally described by continuous things (at least on the lower levels of physics). I am not an analyst, I can't tell you if someone is naturally occurring or not. I can just tell you my experience with questions my colleagues asked me about choice. And I can only tell you that as with many others similar vague-explanations, the best advice I can give is to put this in the back of your head, and move forward with your work. At some point, with more experience, you'll understand. Commented Jun 11, 2018 at 6:58
• @Daniel: Yes, but arguably, these don't really pop up in analysis all that much (with the exception of $\ell^2$ and the likes of it, but you don't really use the existence of a Hamel basis for that space in analysis all that much). Commented Jun 24, 2018 at 10:20

There is Shoenfield's absoluteness theorem, which could be what Terence Tao is referring to. At a simpler level, observe the following:

As usual, we use the standard interpretation of arithmetical sentences in ZF (as sentences about $ω$). Then for every axiom $A$ of ZFC, we have that ZF proves that the constructible universe $L$ satisfies $A$. Now take any arithmetical sentence $Q$ such that ZFC proves $Q$. Then ZF proves that $L$ satisfies $Q$, and hence also that $Q$ is true (because $ω$ is the same in $L$). We can hence observe (in a suitable metasystem MS) that any proof of any arithmetical sentence $Q$ within ZFC can be transformed into a proof of $Q$ within ZF. This weaker absoluteness theorem can be summed up as:

If you can prove some arithmetical sentence within ZFC, you can prove it within ZF alone.

This already implies that any theorem of ZFC about just the natural numbers does not depend on the axiom of choice. This could also be what he means by "decidable" (and anyway he did put it in scare-quotes). We can state a precise version: For any sentence whose truth value can be decided by a program that uses some finite Turing jump, if it can be proven within ZFC then it can be proven within ZF.

• MS can be much weaker than ZF, so this absoluteness itself does not depend on anything close to ZF. So if you want to use ZFC to prove some arithmetical sentence, the biggest 'leap-of-faith' would not be in any use of the axiom of choice but rather in the 'faith' in ZF itself. Commented Jun 16, 2018 at 15:24
• I know of some number theorists that object to this sort of argument because they find it too "metamathematical". So I think it is worth pointing out that rather than a metamathematical trick, the proof of any instance of Shoenfield absoluteness is formalizable in a low level version of analysis, and consists of a fairly explicit construction that from an explicit, say, arithmetic description of a set, allows us to build the desired choice function. Commented Jun 16, 2018 at 17:29
• @AndrésE.Caicedo: Hmm I see. I'm not sure what you mean by "arithmetic description" in general though. Say for example a proof of a strengthened Ramsey theorem via an ultrafilter? =P Commented Jun 16, 2018 at 17:37
• What I mean is this: A $\Sigma^1_2$ statement asserts that certain $\Pi^1_1$ set is nonempty. What Shoenfield absoluteness gives us is an "explicit" witness, that is, it describes a "procedure" that gives us an element of the $\Pi^1_1$ set. Commented Jun 16, 2018 at 17:59