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Let $M$ be a transitive model of ZFC. For convenience let assume that $M$ is countable. Now let us consider the least undefinable ordinal $\vartheta_M$ which is not definable from elements in $M$ and $M$ itself as parameters. For example, if $\alpha$ is the height of $M$ then $\alpha+\alpha$ and $\alpha^2$ are definable from $M$, so they are (strictly) less than $\vartheta_M$. Since there are only countable possible formulas and $|M|$ possible parameters we only have $|M|$ definable ordinals from $M$, so $\vartheta_M<|M|^+$.

My question is whether forcing preserves the size of $\vartheta_M$. That is, if $M[G]$ is a generic extension of $M$ then $\vartheta_{M[G]} = \vartheta_M$? I guess that $\vartheta_M$ only depends on the height of $M$. Thanks for any help.


I should provide more precise notion of definability. The definition of $\vartheta_M$ I imagine is: for a transitive model $M$, a definable class $C\subseteq M$ (that is, we have a formula $\ulcorner\phi\urcorner$ and parameters $a_1,\cdots, a_n\in M$ such that $x\in C\iff M\models \ulcorner\phi\urcorner(x, a_1,\cdots, a_n)$) and a definable ordering $\prec$ over $C$ well-ordered if for every definable $X\subseteq C$ either $X$ is empty or $X$ have the $\prec$-minimum.

$(C, R)$ might not be well-ordered in $V$. However if it is well-ordered then we can find the ordinal isomorphic to $(C, R)$ and we consider the least ordinal $\vartheta_M$ not isomorphic to any $(C, R)$. In that sense I can argue that $\vartheta_M < |M|^+$.

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    $\begingroup$ It may be the case that every set in our background universe is definable without parameters - see here. So you need to tell us more about your background universe. $\endgroup$ – Stefan Mesken Mar 6 '16 at 7:43
  • $\begingroup$ Note also that - even in case $\vartheta_m$ is defined - it doesn't have to be the case that $\vartheta_M < \mid M \mid^+$. And since $M[G]$ is definable from $M$ and $G$, changing $M$ to $M[G]$ gives us exactly on additional parameter, namely $G$ which may without loss of generality be taken as a subset of $\omega_1$. $\endgroup$ – Stefan Mesken Mar 6 '16 at 7:53
  • $\begingroup$ @Stefan I assume the OP is assuming $V$ is the "real" universe, so that doesn't happen. :) Less Platonistically, I assume the OP is interested in a background universe which satisfies something slightly stronger than ZFC, e.g. enough to make this sort of argument go through. $\endgroup$ – Noah Schweber Mar 6 '16 at 8:01
  • $\begingroup$ @Stefan I am confusing the notion of definability. I should formalize what the definability I imagine is and ask it is right. My question is motivated following question: what is the possible order-type that can be encoded into the universe? We can encode $\mathrm{On+On}$ or $\mathrm{On}^{<\omega}$ unless every transitive class is isomorphic to none of them. $\endgroup$ – Hanul Jeon Mar 6 '16 at 8:26
  • $\begingroup$ @Stefan I make my question more clearly. I do not certain such specification avoids technical problems you mentioned. $\endgroup$ – Hanul Jeon Mar 6 '16 at 9:17
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To clarify, I think that what the OP is asking is:

For $M\in V$ a countable transitive model of ZFC, let $\alpha(M)$ be the supremum of the ordinals in $V$ such that $\alpha(M)$ is not definable in $V$ by a first-order formula with parameters from $M\cup\{M\}$.

(Note that I write this slightly differently from the OP: the OP asks for the least "undefinable" ordinal, but I think they are tacitly assuming that the "definable" ordinals are closed downwards, which is not at all obvious to me.)

Then the question is, if (according to $V$) $N$ is a generic extension of $M$, is $\alpha(N)=\alpha(M)$?

(Note that since $M\not\in M[G]$, it's not even obvious that $\alpha(M)$ is "increasing" in $M$! In fact, by modifying the below argument I think we can show that it's not.)

If I'm interpreting the question correctly, the answer is no: forcing can definitely change what ordinals are definable in this sense. For example, for $N$ a set model of ZFC (inside $V$), let $\alpha_N$ be the minimum of the ordinals $\alpha$ such that the continuum pattern $$\{i: 2^{\aleph_{\omega_\alpha+i}}=\aleph_{\omega_\alpha+i+1}\}$$ is in $N$. Then there's no reason we can't have a model $M$ such that $\alpha_M=0$, but a forcing extension $M[G]\in V$ such that $\alpha_{M[G]}=\theta_M$ (maybe the continuum patterns in $V$ look Cohen over $M$, and we happen to pick $G$ to match the relevant pattern exactly). Of course, the obvious way to do this involves a terrible $V$, but there's no reason it can't happen.

Note that it is crucial in this argument that $G$ be generic over $M$, but not $V$. Indeed, it's not hard to show the following:

Let $M$ be a countable transitive model in $V$, $\mathbb{P}\in M$ a forcing notion, and $G$ $\mathbb{P}$-generic over $V$. Then $\alpha^{V[G]}(M[G])=\alpha^V(M)$.

Of course, we have to be a bit careful defining "$\alpha^{V[G]}$," but it's not hard.

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  • $\begingroup$ Note, however, that the original definition can still go through if the ground model is definable. This is certainly the case in set forcing extensions. $\endgroup$ – Asaf Karagila Mar 6 '16 at 9:24
  • $\begingroup$ @AsafKaragila I don't quite understand your comment, can you elaborate? (Keep in mind that I'm assuming $G\in V$, that is, $G$ is only generic over $M$.) $\endgroup$ – Noah Schweber Mar 6 '16 at 9:29
  • $\begingroup$ Since $M$ is definable in $M[G]$, internally and so certainly externally, anything you define in $M$ can be defined in $M[G]$. $\endgroup$ – Asaf Karagila Mar 6 '16 at 9:35
  • $\begingroup$ @Asaf I'm still puzzled by your comment. What is the "original definition" that you are referring to? Is it the one OP gave before editing his post? If so, what do you mean by saying that it still can go through? Sorry, I'm obviously missing your point. $\endgroup$ – Stefan Mesken Mar 7 '16 at 5:03
  • $\begingroup$ @AsafKaragila I assumed the OP was asking about ordinals being defined in $V$ using $M$ and parameters from $M$ - so I still don't see what this has to do with my answer. $\endgroup$ – Noah Schweber Mar 7 '16 at 5:43

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