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I’ve read somewhere on this site (I believe from a JDH comment) that an argument in favor of AC (I believe from Asaf Karagila) is that without AC, there exists a real number which is not definable from a countable sequence of ordinal parameters.

I’d like a reference to a paper/post on this fun fact or a proof of it.

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2 Answers 2

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The argument, given to me by Hugh Woodin over a drink in Barcelona in September 2016, is a nice retort to "AC is obviously false since there cannot be a well-ordering of the real numbers". It is arguable whether or not it is philosophically sound, but as a retort to a claim that is definitely not sound by itself, it works absolutely great.

If there is no well-ordering of the real numbers, then $\Bbb R\nsubseteq\rm HOD$. Namely, there is a real number which does not belong to $\rm HOD$. Namely, there is a real number which is not definable from ordinal parameters. This is because the class $\rm HOD$ can be well-ordered in a very definable and very concrete way already in $\sf ZF$.

So, if $\Bbb R\setminus\rm HOD\neq\varnothing$, that means that we can instantiate this fact and pick such a real number.

So, which one is it? If you want to claim that it is impossible to exhibit a well-ordering of the reals, and therefore the Axiom of Choice must be false, you should be able to exhibit a real number that is not ordinal definable. Since otherwise, by failing to find a witness for a non-definable real, you either agree that having a definable witness is not necessary for existential statements or else $\Bbb R\subseteq\rm HOD$ after all, in which case it can be well-ordered sans the Axiom of Choice (and one can bootstrap this argument through the von Neumann hierarchy to argue that $V=\rm HOD$).

Is that a good philosophical argument? Probably not. Is it a good comeback to a bad philosophical argument? Absolutely.

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    $\begingroup$ I like this argument quite a lot. $\endgroup$ Commented Jun 25 at 16:13
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    $\begingroup$ 😄 ${{}}{{}}{}$ $\endgroup$ Commented Jun 25 at 17:53
  • $\begingroup$ How would this bootstrapping work? Do you mean “replace all occurrences of $\Bbb R$ with an arbitrary set $X$ and use the same argument” or that $\Bbb R\subseteq HOD \to V=HOD$? $\endgroup$
    – Lave Cave
    Commented Jun 26 at 0:55
  • $\begingroup$ @LaveCave: Recall that $\Bbb R$ is really just $V_{\omega+1}$ in disguise; once that's established to be a part of $\rm HOD$, we can continue by induction. But alternatively, just use induction to show that $\cal P(\alpha)\subseteq\rm HOD$ for all ordinals. $\endgroup$
    – Asaf Karagila
    Commented Jun 26 at 6:19
  • $\begingroup$ Is ZFC strong enough to prove anything about what measure $\Bbb R \cap HOD$ is? Clearly if $V = HOD$ this is 1; is it always 1 or can it be something else? Basically, I am curious if "a random real number in $[0,1]$ would be expected to have the property above. $\endgroup$ Commented Jun 29 at 2:29
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Unfortunately, the claim you have stated is not true. Regardless of the axiom of choice, every real is definable from a countable sequence of ordinal parameters, since the real is definable from the sequence of its digits, for example, or from the numbers encoding the rationals below it, or from the numbers encoding a sequence of rational numbers converging to it.

However, perhaps you are thinking of the following:

Theorem. If there is no well-ordering of the real numbers, then not every real is definable from finitely many ordinal parameters.

Proof. The reason is that if every real is ordinal definable in that way, then they can be well-ordered by the OD order. Namely, this is the order that places $x<y$ if both are definable from ordinal parameters, but $x$ is definable in this way inside some $V_\theta$ where $\theta$ is smaller than any ordinal working this way for $y$, or they have the same smallest such $\theta$, but $x$ is definable by a formula with a smaller Gödel code, or they have the same such smallest formula, but $x$ is definable by that formula using lexically earlier ordinal parameters. This would be a well order of the reals if every real were definable in that way. $\Box$

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