## Proving that being an inaccessible cardinal is absolute, for V_$\kappa$, where $\kappa$ is inaccessible?

I’m going through the proof that if $$\kappa$$ is inaccessible then V_$$\kappa$$ $$\vDash$$ ZFC and how thus we have ZF $$\nvdash$$ “There exist inaccessible cardinals”.

So the last part is by taking $$\kappa$$ to be the least inaccessible cardinal and then showing that V_$$\kappa$$ $$\vDash$$ “There is no inaccessible cardinal”

But for this to work, we must show that being an inaccessible cardinal is absolute for V_$$\kappa$$, yes? Otherwise it is possible V_$$\kappa$$ could just think it has inaccessible cardinals even though it doesn’t.

So how do we prove that being an inaccessible cardinlas is absolute? Jech simply leaves it to the reader, and I’m struggling to find it elsewhere. It must be easy, because no one even seems to bother proving it, or perhaps it’s not necessary at all? Am I missing something obvious? Is it not neccessary for this notion to be absolute?

## Are there uncountable cardinals $\kappa$ such that $|\kappa\cap\mathsf{Card}| = \kappa$?

All the cardinals $$\kappa\leq\aleph_0$$ have the property that there are precicely $$\kappa$$ cardinals less than $$\kappa$$. Of course, $$\aleph_1$$ lacks this property since there are only $$\aleph_0 +1= \aleph_0$$ cardinals less than it. And I suppose the same goes for all successor cardinals for the same reason.

But are there any limit cardinals for which the property returns? If so, are they in ZFC, or does one have to introduce large cardinals axioms to get them?

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## $\kappa$-ultrafilter on $\kappa$, creating a partition from $\bigcup_{\alpha < \lambda} X_\alpha \in$ filter

I’m reading some contents on set-theory for my own interest, and I stumbled upon some questions I cannot solve yet.

Let $$\mathcal{F}$$ be a $$\kappa$$-complete ultrafilter on $$\kappa$$, and $$\bigcup_{\alpha < \lambda} X_\alpha \in \mathcal{F}$$ with $$\lambda < \kappa$$. Show that $$\exists \ \alpha < \lambda$$ such that $$X_\alpha \in \mathcal{F}$$.

I proved that $$\mathcal{F}$$ is a $$\kappa$$-complete ultrafilter on $$\kappa$$ if and only if, for every partition $$\{Y_\alpha, \alpha<\lambda\}$$ of $$\kappa$$, there is $$\alpha < \lambda$$ such that $$Y_\alpha \in \mathcal{F}$$.

I guess I only have to “transform” my $$X_\alpha$$‘s into a partition of $$\kappa$$, but I don’t see how. Ideally I would create a partition of $$\kappa$$ from $$X_\alpha$$‘s and only “small elements”, i.e. elements not in $$\mathcal{F}$$. This will directly give me the desired conclusion. Any idea? Thanks

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## Does measurability of cardinal $\kappa$ imply measurability of $2^\kappa$?

A cardinal $$\kappa$$ is real-valued measurable if there is a $$\kappa$$-additive probability measure on $$2^\kappa$$ which vanishes on singletons. The existence of measurable $$\kappa$$ is independent of ZFC.

Question: if $$\kappa$$ is assumed to be real-valued measurable, does it necessarily follow that $$2^\kappa$$ is real-valued measurable?