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[UPDATE] The question has been updated and is mostly unrelated to the question above the line. The updated question is below the line.

The following argument supports a "yes" answer; is it convincing?

Let $\Omega$ be the least uncountable ordinal. A finite ($L_{\omega \omega}$) formula of signature $\{\in\}$ of one free variable $\varphi(x)$ defines an element of $\Omega$ if $$(\Omega,\in) \models \exists!_x \varphi(x).$$ Let $X$ be the set of all $\varphi(x)$ such that $\varphi$ defines an element of $\Omega$. Since $L_{\omega \omega}$ is countable, so is $X$. Let $F\colon X\rightarrow \Omega$ be such that $F(\varphi)$ is the element of $\Omega$ such that $\varphi(F(\varphi))$. Let $F[X]$ be the range of $F$. Since $F[X]$ is a countable subset of $\Omega$, and since no countable subset of $\Omega$ is cofinal in $\Omega$, there exists a least $\beta\in\Omega$ such that $F[X]\subseteq\beta$. Since the preceding remarks can be formalized in ZFC and define $\beta$ uniquely, there exists a formula $\psi(x)$ such that $$(\Omega,\in)\models\forall_x \psi(x) \iff x = \beta.$$ Since $\psi\in X$ and $F(\psi)=\beta$, we have $\beta\in F[X]\subseteq \beta$, hence $\beta\in\beta$.


Question: is ZFC inconsistent?

As above, the following argument supports a "yes" answer.

Let $\Omega$ be the least uncountable ordinal. A finite $(L_{\omega \omega})$ sentence $\varphi$ of signature $\{\in\}$ selects an infinite countable ordinal if there exists an infinite countable ordinal $\alpha$ and set $x$ of rank $\alpha$ such that $$(\alpha\cup x,\in)\models\varphi.$$ Let $C$ be the set of all $\varphi$ such that $\varphi$ selects an infinite countable ordinal. Since $L_{\omega \omega}$ is countable, so is $C$. Let $Q\colon C\rightarrow\Omega$ be such that $Q(\varphi)$ is the least infinite countable ordinal $\alpha$ such that there exists a set $x$ of rank $\alpha$ such that $(\alpha\cup x,\in)\models\varphi$. Let $Q[C]$ be the range of $Q$. Since $Q[C]$ is a countable subset of $\Omega$, and since no countable subset of $\Omega$ is cofinal in $\Omega$, there exists a least $\beta\in\Omega$ such that $Q[C]\subseteq\beta$. Is there a $\psi\in C$ such that $Q(\psi)=\beta$?

Case 1: yes. We have $Q(\psi)\in Q[C]\subseteq\beta$, hence $\beta\in\beta$, a contradiction.

Case 2: no. In other words, for all $\varphi\in C$, $Q(\varphi)\not=\beta$. If there exists a $\varphi\in C$ such that $Q(\varphi)>\beta$, then we would have $\beta\in Q(\varphi)\in Q[C]\subseteq\beta$, a contradiction. Therefore, for all $\varphi\in C$, $Q(\varphi)<\beta$. Let $\psi$ be the sentence, "there exists an infinite countable ordinal $\gamma$ and set $u$ such that for all $\varphi\in L_{\omega \omega}$, if $\varphi$ selects an infinite countable ordinal, then $(\gamma\cup u,\in)\models\varphi$. Claim: $Q(\psi)=\beta$, a contradiction.

[UPDATE] Replace the definition of $\psi$ with "there exists an infinite countable ordinal $\gamma$ such that for all $\varphi\in L_{\omega \omega}$, if $\varphi$ selects an infinite countable ordinal, then there exists a set $u$ such that $(\gamma\cup u,\in)\models\varphi$."

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    $\begingroup$ $\beta$ can be defined by a formula in ZFC, but not in the structure $(\Omega,{\in})$. $\endgroup$ Jun 7, 2019 at 14:40
  • $\begingroup$ I agree with this comment, but I don't see how it is relevant since the argument doesn't depend on $\beta$ being defined in $(\Omega,\in)$. $\endgroup$ Jun 7, 2019 at 14:48
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    $\begingroup$ The argument most certainly does depend on that: you claim “$(\Omega,{\in})\models\forall x\,(\psi(x)\iff x=\beta)$”, whereas what you get is “$(V,{\in})\models\forall x\,(\psi(x)\iff x=\beta)$” (modulo the fact that truth in $V$ is not actually definable inside $V$, so it cannot be written that way). $\endgroup$ Jun 7, 2019 at 15:07
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    $\begingroup$ (Incidentally, repeatedly changing a question is frowned upon. If you want to try another modification I'd ask it in a separate question - and I'd recommend math.stackexchange instead of mathoverflow.) Also, "As above, the following argument supports a "yes" answer." suggests that the above-the-line section actually does support the relevant claim, which it does not (per my answer). $\endgroup$ Jun 8, 2019 at 6:44
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    $\begingroup$ @DavidPokorny That has exactly the same problem - we can't use a single $\gamma$ for all $\varphi$s. And if you flip quantifiers so each $\varphi$ is allowed its own $\gamma$, then every ordinal satisfies that property since internally it amounts to "every set which selects an ordinal, selects an ordinal." No variation on this is going to work, and keeping moving the goalposts isn't helpful. Again, further variations should be asked in a separate question. Otherwise trying to answer this post is like trying to nail soup to a wall. $\endgroup$ Jun 9, 2019 at 5:44

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Note: The question looks at ordinals themselves as structures. However, they're far too weak by themselves to talk about anything as complicated as definability. Instead of looking at $\theta$, we want to look at a canonical transitive set of height $\theta$ - say, the $\theta$th level of the constructible universe. This is a point I ignored at first since it is tangential to the real issue, but I've decided I should clarify it (and correct my answer).


Is it convincing?

Absolutely not. It rests on two fundamental assumptions: that our notion of definability be "limited" (= only countably many definitions) and "self-contained" (= able to talk about itself to a certain extent), and these assumptions are in tension (to put it mildly). The argument in your post makes an unfixable error on the second point, the argument you suggest in a comment makes an unfixable error on the first point, and no variant I can imagine addresses both points simultaneously.


The argument of your post hinges on the set $X$ and the map $F$ being definable in $\omega_1$ (which is the standard notation for the least uncountable ordinal). Specifically, your claim

since the preceding remarks can be formalized in ZFC and define $\beta$ uniquely, there exists a formula $\psi(x)$ such that $$L_{\omega_1}\models\forall_x \psi(x) \iff x = \beta$$

conflates definability in $V$ (= the pre-comma clause) with definability in $L_{\omega_1}$ (= the post-comma clause).

All that you can actually conclude is that there is a formula $\psi_x$ such that

$$V\models\forall x(\psi(x)\iff x=\beta).$$ But that's not what you need.

Actually, you can improve that a bit: $\beta$ is in fact definable in $L_{\omega_1+n}$ (more generally for any ordinal $\alpha$ truth in $L_\alpha$ is definable in $L_{\alpha+n}$, and $\alpha$ is definable in $L_\alpha$) for a small finite $n$ - I think $n=2$ is enough. But that still doesn't help.

And of course $F$ and $X$ are not definable over $L_{\omega_1}$: they are definable only in a larger context (remember Tarski's undefinability theorem). Indeed, this can be taken as a proof of a special case of Tarski's result.


A response one might attempt at this point is to allow multiple "domains of definition." For example - and you suggest that in a comment - we could look at the set of countable ordinals which are definable in $L_\alpha$ for some uncountable $\alpha$. More generally, given any class-or-set $C$ of ordinals, we could consider the least ordinal $\beta$ not definable in $L_\alpha$ for any $\alpha\in C$.

This however doesn't save us. We still have to somehow argue that "definability in some element of $C$" is, well, definable in some element of $C$. The simplest way to get this is to have $C$ be large enough that some element $\theta\in C$ "sees" all the elements of $C$ necessary to define all the relevant ordinals $<\beta$ (e.g. the proposal in your comment); but this requires $C$ to be uncountable, at which point the counting argument showing $\beta<\omega_1$ breaks down since you have to count both formulas and "definability contexts" (= elements of $C$). Looking at your proposal in particular, every countable ordinal $\sigma$ is definable in $L_\alpha$ for some uncountable $\alpha$ (just take $\alpha=\omega_1+\sigma+1$ and consider the formula "$\omega_1+x$ exists and is the largest ordinal"). If $\beta=\omega_1$ we can't conclude anything.


In the comments you suggest yet another modification (and this will be the last modification I address - if you have another idea you should ask it in a separate question, and additionally I strongly feel that it would be more appropriate at MSE rather than MO); possibly two modifications if my reading of your response is correct.

  • Take $\beta$ to be the least ordinal such that $L_\beta$ satisfies "there is some $\gamma$ such that for every $\varphi$, if $L_\eta\models\varphi$ for some $\eta$ then $L_\gamma\models\varphi$." No such $\beta$ exists - consider $\varphi_0$ and $\varphi_1$ asserting that there is a largest ordinal and there is not a largest ordinal respectively.

  • Your comment in response seems garbled, since unless I'm misreading it it suggests the same thing. I think what you want to do is allow $\gamma$ to depend on $\varphi$. But now this is trivial: every ordinal satisfies "If $\varphi$ holds in some $L_\eta$, then $\varphi$ holds in some $L_\eta$." You might object that the second half of this sentence should add "with $\eta$ less than my height," but that's an "external" property: all the ordinals a level of $L$ sees are those less than its own height!


To sum up: in every nontrivial version of this question, you need to somehow express the "correctness" of $L_\beta$ internally to $L_\beta$, and you can't do this. Since it's the same mistake every time, I'll reiterate my original answer to your question:

Absolutely not.

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  • $\begingroup$ The question has been revised to reflect these remarks. $\endgroup$ Jun 8, 2019 at 6:23

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