Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 2 down vote accepted

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

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 $V=L$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.