# A formula that is upwards absolute but not downwards absolute

I know that any existential formula is upwards absolute (an any universal formula is downwards absolute) but I was looking for an example of a formula that is upwards absolute but not downwards absolute.

Ideally this would be an existential formula and the models would be transitive , if this is possible?

Thanks very much for any help.

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I think you should mention in the question that you're asking about a formula in the language of set theory, absolute for models of ZFC, which is probably what you meant (surmising from the tags). – tomasz Dec 2 '12 at 13:42

## 1 Answer

$$0^\#\text{ exists}$$

This statement is upwards absolute. If $0^\#$ exists then it exists in all larger [transitive] models. However it is not downwards absolute because $L\models\lnot\exists 0^\#$.

Equally if the universe is "sufficiently different" from some of its inner models then the formula $\exists f\colon\alpha\to\beta\text{ a bijection}$ can be true in some models and false in others.

For example if $\omega_1^L$ is countable in $M$ then $M\models\exists f\colon\omega\to\omega_1^L$, and any model extending $M$ also satisfies this formula, however $L$ itself doesn't. Note that $\omega$ is definable without parameters and $\omega_1^L$ is the least ordinal that has no constructible injection into $\omega$, and so it is also definable without parameters.

An even simpler variant of the previous example is simply writing the axiom $V\neq L$. Recall that there is a formula $\varphi(x)$ such that $\varphi(x)$ is true if and only if $x\in L$. Therefore $V=L$ can be expressed as $\forall x.\varphi(x)$.

Writing $\exists x.\lnot\varphi(x)$ is clearly not downwards absolute, because this statement would be false in $L$ of the model; but it is also clearly upwards absolute because if $M\subseteq N$ (both transitive with the same ordinals) and $M\models V\neq L$ then $N$ cannot have satisfy $V=L$.

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Hey thanks for the answer. Sorry if this is a stupid question but what do you mean by $0^{\#}$ ? – hmmmm Dec 1 '12 at 16:06
@hmmmm: $0^\#$ is a sort-of-large-cardinal axiom. It is in fact a real number which encodes the formulas which are true in $L$ in such way that we can decode them (not in $L$, of course, but in the universe) despite $L$ being a proper class. The second example should be much clearer though... Note that $\omega_1^L$ is definable in any universe [which is transitive and contains all the ordinals], and so in $\omega$. Therefore you can really write this without parameters. If we're at it, you can simply require $V\neq L$, and that would be upwards absolute. – Asaf Karagila Dec 1 '12 at 16:12
Ok thanks, I didn't quite follow that (the stuff about $0^{\#}$) but I will have a look at some stuff and have a think about it. Thanks for the answer :) – hmmmm Dec 1 '12 at 16:14
@hmmmm: Yes, it's a very nontrivial concept. Again, the second example and the last part of the previous comment should suffice, though. – Asaf Karagila Dec 1 '12 at 16:14
Yeah I understand that so thanks but thanks for the other part to, gives me something to think about! – hmmmm Dec 1 '12 at 16:15