Can withdrawal of the Power set axiom enable absoluteness for set size statements?

86 Views Asked by At

If we look to models of $\small \sf ZFC$ we can have two transitive models $N,M$ such that for some formula $\phi$ in the language of $\small \sf ZFC$ we have:

$\forall \mathcal M ([\mathcal M\models ZFC] \to \mathcal M\models \exists!x (\phi))$

$M\models \exists x (\phi \land |x| = \kappa)$

$N\models \exists x (\phi \land |x| \neq \kappa)$,

for example left $\phi$ be $x=P(\omega)$ and let $\kappa= \omega_1$. I'll refer to the above as "size statements about some sets theorized to uniquely exist by $\small \sf ZFC$, are not absolute".

Now examine this theory, the question is:

Can we have the above situation raised with it also? In other words, can it theorize the unique existence of some sets the size statements about which are not absolute?

1

There are 1 best solutions below

0
On BEST ANSWER

Yes. Surely we can, under reasonable assumptions, find two transitive models $M, N$ of this theory, one of which (say $M$) has a powerset of $\omega$ and the other does not. Let $\varphi(x)$ be the formula which asserts that $x$ is $\mathcal P(\omega)$ if it exists and $\emptyset$ else. The given theory proves that there is a unique $x$ which satisfies $\varphi$. But now the respective witnesses in $M$ and $N$ have vastly different sizes. A variation of this trick works (essentially) whenever the given theory is incomplete (given that the question even makes sense for that theory).