# Tag Info

## New answers tagged large-cardinals

6

To complement Asaf's answer, let me point out that you don't need any large cardinals at all to get embeddings like this; they exist as long as there is an uncountable transitive model of ZFC+"there are no inaccessible cardinals". If $N$ is such an uncountable model we can get an embedding by taking a countable elementary substructure $X\prec N$ and ...

3

Note that in Asaf's answer, the elementary embedding $j: M\rightarrow N$ does not "live" (that is, is not definable in) $M$. By contrast, if we have an elementary embedding $j: M\rightarrow N$ which is definable in $M$ (from parameters in $M$), then $crit(j)$ is inaccessible, in fact measurable, in $M$: letting $\kappa=crit(j)$, we form the ultrafilter ...

5

No, not necessarily. Suppose $\kappa$ is measurable in $W$, let $M$ be $W[G]$ the model obtained after forcing with $\operatorname{Col}(\omega,<\kappa)$. If $j\colon W\to W'$ is an ultrapower embedding with critical point $\kappa$, then we can force over $M$ to obtain a model in which $j$ can be extended from $W$ to $M$. Another important example is the ...

0

Your argument is correct but the point is that you do not have to appeal to Godel's 2nd incompleteness theorem to prove this and Jech does that.

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