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I was thinking about constructible universe, and I had the following idea.

Suppose we have a predicate $\phi$ defined in a language of set theory. Suppose moreover, that the statement "there exists $X$ such that $\phi(X)$; and for any $X, Y$ if $\phi(X)$ and $\phi(Y)$, then $X=Y$" is provable in ZFC (essentially, that the formula $\phi$ determines a unique set). Then that set $X$ must belong to some level $L_\alpha$ of the constructible hierarchy.

The argument is as follows: since the existence and uniqueness is provable in ZFC, it must be true in the constructible universe, which is a model of ZFC. But since constructible universe is an inner model of ZFC, that particular element of the constructible universe that serves as a witness of the truth of the formula in the constructible universe model of ZFC must also be the unique $X$ in the "whole" ZFC$-$we cannot have a different one, because there is already a set that satisfies $X$ in the constructible hierarchy.

Is this argument sound? It feels solid and completely obvious to me, and vacuous at the same time.

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The argument is clearly wrong. $\mathcal P(\omega)$ is definable without parameters, and yet it is not necessarily constructible, as shown by Cohen.

The thing to remember is that the formula $\phi$ is not absolute between models. So $\mathcal P(\omega)$ and $\mathcal P(\omega)^L$ might be different.

In some cases, there are sets whose definition is very robust and unique, but they cannot even exist in $L$. One example of this kind is $0^\#$, which has a parameter free definition, and can be represented as a fairly canonical set of integers. But nevertheless, this set cannot exist in $L$. Other examples would be any real, since you can code it into the continuum pattern below $\aleph_\omega$ (for example), and even much more than that.

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  • $\begingroup$ Ah, so the point here is that even though $P(\omega)$ is uniquely defined without parameters, then even though we might have $\omega = \omega^L$, we have that $P(\omega)^L$ is the set of constructible subsets of $\omega$, and not necessarily all subsets in the larger model sense, correct? Also, can you give me a reference for the fact proved by Cohen you mention? $\endgroup$ – xyzzyz Aug 10 '18 at 22:24
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    $\begingroup$ Correct. The reference? Well, forcing. Literally every text about forcing starts with Cohen forcing, the original and simplest example of forcing. Force over a model of $V=L$, and you get a model of $V\neq L$ which has non-constructible reals. $\endgroup$ – Asaf Karagila♦ Aug 10 '18 at 22:25

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