Absoluteness of inaccessible cardinals

Question

I'm studying large cardinals and I'm hoping to fully understand the proof that says ZFC is not able to prove the existence of inaccessibles (given ZFC is consistent, of course).

I've already fully understood and/or proven everything up to and including: if we had an inaccessible cardinal $\kappa$, then $V_\kappa$ is a (sub)model of ZFC. I also understood the main proof that takes the least inaccessible $\kappa$ and arrives at a contradiction by finding another inaccessible inside $V_\kappa$.

The only thing that's missing is the following Lemma:

Given a model $M$ of ZFC and an inaccessible $\kappa$, $\lambda$ is inaccessible in $M$ iff it is inaccessible in $V_\kappa\subset M$.

I feel like the proof of this lemma is highly technical and that I'm getting confused about the relativization of the axioms and the absoluteness of the functions and properties needed to define cardinals and inaccessibility.

I haven't found a proof anywhere. Does anyone know a good reference for this lemma?

Thanks in advance!

Answer 1

As discussed in the comments, what we want to prove is:

If $\kappa$ is inaccessible and $\lambda < \kappa$ then $V_\kappa\models \mbox{"$\lambda$ is inaccessible"},$ if and only if $\lambda$ is inaccessible.

Really we only need the forward direction to show $\sf ZFC\not\vdash \exists inaccessibles$, so that's the one we'll prove (the other direction is similar). We can just prove it for "regular" and "strong limit" separately. The high-level reason is that $V_\kappa$ "knows everything" that goes on below rank $\kappa$ and any potential witnesses contradicting the inaccessibility of $\lambda<\kappa$ would have rank $<\kappa,$ so $V_\kappa$ would know about them if they existed.

(This same idea works pretty much verbatim for any large cardinal property that is "witnessed locally", e.g. weak compactness, measurability. For instance, we can use a similar argument to show that if it's consistent that measurable cardinals exist, then it's consistent that there's only one and that it's the greatest inaccessible.)

If $\lambda$ is not regular, there is some sequence $\lambda_\alpha\uparrow \lambda$ for $\alpha < \beta$ where $\beta < \lambda.$ Well, then $(\lambda_\alpha: \alpha< \beta)\in V_\kappa$ and $V_\kappa\models \lambda_\alpha\uparrow \lambda,$ so $V_\kappa\models \mbox{"$\lambda$ is not regular"}.$

Likewise, if $\lambda$ is not a strong limit, there is an $\alpha < \lambda$ and a surjection $g:P(\alpha)\to \lambda$, but then $\lambda, \alpha, P(\alpha), g\in V_\kappa$ and $$V_\kappa\models \mbox{"$P(\alpha)$ is the power set of $\alpha<$, $\alpha < \lambda$ and $g:P(\alpha)\to \lambda$ is a surjection"}$$ so $V_\kappa\models \mbox{"$\lambda$ is not a strong limit."}$

Note we didn't use that $\kappa$ was inaccessible in any significant way (for this part - of course it's important for showing $V_\kappa\models \sf ZFC$), we just implicitly used that it was a limit ordinal in a couple places.