Tarski's Theorem on the undefinability of truth gives me a bit of a headache, and as a beginner I am still trying to grapple with its consequences. Here's a question.


Let $T$ be the set of Godel numbers of sentences $\sigma$ such that $V \models \sigma$. 


Now, I know that by Tarski's theorem, there is no formula that defines $T$ - that is to say, there is no first order formula $\tau (x)$ such that for each $n$, $n \in T \leftrightarrow V \models \tau (n)$. 


I can live with that, because obviously some sets are not definable. (Indeed, most are not definable.) But must $T$ exist somewhere in $V$, even if it not definable? The argument is not so obvious to me. Or will this depend on background assumptions? (Whether $V=L$, or whether this or that large cardinal exists, and so on ...) 



The set of Godel numbers of true sentences.
$\let\eq\leftrightarrow\def\god#1{\ulcorner#1\urcorner}$The question as such is not formally precise. The existence of a truth set is not expressible in the language of ZFC, so one cannot assert it from inside $V$, only from outside. My reading of the question is as follows. Let $\mathcal M=(M,E)$ be a model of ZFC, and $T^\mathcal M$ be the set of Gödel numbers of all sentences true in $\mathcal M$. Under which conditions does $\mathcal M$ contain $T^\mathcal M$, in the sense that there exists $t\in M$ such that $\{x\in M:(x,t)\in E\}=\{n^\mathcal M:n\in T\}$? I will write this as $\mathcal M\models T\in V$.


If $T^\mathcal M$ is in $\mathcal M$, then so is the set $\{n:\mathcal M\models n\in\omega\land\exists a\in T^\mathcal M\,n\le a\}$, i.e., the standard natural numbers. Thus, this can happen only if $\mathcal M$ has standard integers (i.e., it is an $\omega$-model).


In particular, every model of set theory has an elementary extension which satisfies $T\notin V$, so you cannot guarantee the existence of $T$ by any first-order axioms.


On the other hand, models of $T\in V$ do exist: for example, if $\kappa$ is an inaccessible cardinal, then $V\_\kappa$ contains the set of all sentences true in itself (as it contains every set of integers).


However:


**Proposition.** If an extension $S$ of ZFC has an $\omega$-model at all, then it also has an $\omega$-model satisfying $T\notin V$.


*Proof:* We may assume that $S$ is a complete theory, hence it is the theory of an $\omega$-model $\mathcal M\_0$. We need to show that there exists a model of $S$ which omits the types
$$\begin{align\*}
p(t)&=\{\phi\eq\god\phi\in t:\phi\text{ a sentence}\},\\\\
q(x)&=\{x\in\omega\}\cup\{x\ne n:n\in\omega\}.
\end{align\*}$$
By the omitting types theorem, it suffices to verify that neither type has a generator. If $q$ had a generator $\alpha(x)$, then $S$ proves $\alpha(x)\to x\in\omega$ and $\alpha(x)\to x\ne n$ for every natural number $n$, hence $\mathcal M\_0\models\forall x\,\neg\alpha(x)$. Since $S$ is complete, $\alpha$ is inconsistent with $S$, and thus cannot be a generator.


If $p$ had a generator $\alpha(t)$, then $S$ proves
$$\tag{$\*$}\alpha(t)\to(\phi\eq\god\phi\in t)$$
for every sentence $\phi$. Using self-reference, let $\phi$ be a sentence such that ZFC proves
$$\phi\eq\exists t\,(\alpha(t)\land\god\phi\notin t).$$
Then it is easy to see that $(\*)$ leads to $S\vdash\forall t\,\neg\alpha(t)$, hence $\alpha$ is no generator. QED


François Dorais suggested in the comments that a more relaxed reading of $\mathcal M\models T\in V$ could be that there is $t\in M$ such that $T^\mathcal M=\{n\in\omega:\mathcal M\models n\in t\}$. In other words, $\mathcal M$ satisfies $T\in V$ iff it realizes the type $p$ above. Under this reading, every consistent extension $S$ of ZFC has a model which satisfies $T\in V$, but is not an $\omega$-model (and moreover, every recursively saturated model of ZFC has this property). On the other hand, the same omitting types argument as above shows that every such $S$ also has a model which satisfies $T\notin V$.


