I'm going through the proof that if $\kappa$ is inaccessible then $V_\kappa \vDash \mathrm{ZFC}$ and how thus we have $\mathrm{ZF} \nvdash \text{"There exist inaccessible cardinals"}$.

So the last part is by taking $\kappa$ to be the least inaccessible cardinal and then showing that $V_\kappa \vDash \text{"There is no inaccessible cardinal"}$.

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?

  • $\begingroup$ Yes, you should show it's absolute. It shouldn't be too hard, though maybe a bit tedious. Where did you get stuck? $\endgroup$ Jun 2, 2019 at 0:17
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    $\begingroup$ By the way, LaTeX formatting works here. You can type $V_\kappa$ to get $V_\kappa$. $\endgroup$ Jun 2, 2019 at 0:18
  • $\begingroup$ So first proving that being a cardinal is absolute; if $\lambda$ is an ordinal (less than $\kappa$), why can't there be an onto mapping from $\lambda$ onto $\kappa$ added after $V_\kappa$, i.e, in $V_{\kappa + i}$ or something (and thus $V_\kappa$ thinks $\lambda$ is smaller than $\kappa$)? I feel like I'm missing something very obvious here haha. $\endgroup$ Jun 2, 2019 at 0:22
  • $\begingroup$ Suppose $\lambda \in V_\kappa$ is an ordinal that is not really a cardinal. Then there would be a bijection $f$ from $\lambda$ to some smaller ordinal $\beta \in \lambda$. This map $f$ is a subset of $\lambda \times \beta \subset \lambda \times \lambda$, so its rank is something like 3 more than the rank of $\lambda$ (depending on exactly how you define ordered pairs). But $\kappa$ is a limit ordinal so the rank of $f$ is still less than $\kappa$. Therefore $V_\kappa$ knows that $\lambda$ isn't a cardinal because it can see the witness $f$. $\endgroup$ Jun 2, 2019 at 0:30
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    $\begingroup$ If I'm not mistaken, I think all those statements are absolute as soon as $\kappa$ is merely a limit ordinal larger than $\omega$. I don't think any of the relevant witnesses go up in rank by more than a small finite number. Of course, if $\kappa$ is pretty small, then this just boils down to "there are no inaccessibles in $V_\kappa$, and $V_\kappa$ knows that there aren't". $\endgroup$ Jun 2, 2019 at 0:51

1 Answer 1


Since this is really just a question about absoluteness, let me leave inaccessibility behind for the moment and focus on a simpler still-interesting example: cardinal-ness.

Suppose $\alpha$ is an ordinal which in $V$ is not a cardinal. Then in $V$ there is some smaller ordinal $\beta<\alpha$ and some surjection $f:\beta\rightarrow\alpha$. Now each "bit" of $f$ (that is, each pair $\langle x,y\rangle$ with $f(x)=y$) is an element of $V_\alpha$, so $f$ itself is a subset of $V_\alpha$. But this means that $f$ is an element of $V_{\alpha+1}$.

So if $\alpha$ is not a cardinal, then a "witness" to this already exists in $V_{\alpha+1}$. Conversely, anything $V_{\alpha+1}$ thinks is a witness to the non-cardinality of $\alpha$ actually is (since "being a surjective function from a smaller ordinal" is absolute).

As a corollary we get:

If $\lambda$ is a limit ordinal then $V_\lambda$ is "correct about cardinals:" for each $\alpha\in V_\lambda$ we have $V_\lambda\models$ "$\alpha$ is a cardinal" iff $\alpha$ is in fact a cardinal.

Proof: For $\alpha<\lambda$ we have $\alpha+1<\lambda$.

Regularity is similar: a witness to non-regularity is a cofinal map from a smaller ordinal, so just think about where such a thing has to live.

Do you see how to treat strong limit-ness this way as well?

Note that absoluteness in each of these cases is stemming from the fact that the failure of the property in question is witnessed by an object whose "witnessing-ness" is simple enough that we already know it's absolute. This "bootstrapping" doesn't get us everything, though, and with more complicated properties more care needs to be taken (since sometimes the absoluteness we want or expect actually fails!).


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