From Jech, Set Theory:

Let $\kappa<\lambda$ be cardinals and suppose there exists an elementary embedding $j:V\longrightarrow M$ s.t.

(i) $j(\gamma)=\gamma$ for all $\gamma< \kappa$

(ii) $j(\kappa)>\lambda$

(iii)$M^\lambda\subset M$

Then there exists a fine normal measure $U$ on $P_{\kappa}(\lambda)=\{X\in P(\lambda)\mid |X|<\kappa\}$.

Proof:By (iii) the set $\{j(\gamma)\mid\gamma<\lambda\}$ belongs to $M$ and so the following defines an ultrafilter over $P_{\kappa}(\lambda)$: $$X\in U\iff\{j(\gamma)\mid \gamma<\lambda\}\in j(X)$$

A standard argument shows that $U$ is a $\kappa$-complete ultrafilter. $U$ is a fine measure because for every $\alpha\in \lambda, \{P\mid \alpha\in P\}$ is in $U$.

Now why is the "bold-type" sentence true? I suspect that the hypoteses (i)-(ii) are to be used, but how?

Thank you in advance.

  • $\begingroup$ Your definition of supercompactness is wrong in several places, e.g. I think you mean $\gamma < \kappa$ and $\kappa < \lambda < j(\kappa)$ and that this describes "$\kappa$ is $\lambda$-compact"... $\endgroup$ – Stefan Mesken Oct 25 '17 at 15:49

By elementarity of $j$, you have $j(\{P:\alpha\in P\})=\{P:j(\alpha)\in P\}$, and this set contains $\{j(\gamma):\gamma<\lambda\}$ (because $\alpha<\lambda$). So $\{P:\alpha\in P\}\in U$ by definition of $U$.

  • $\begingroup$ Thank you. Just why can we write $\{P:j(\alpha)\in P\}$ and not just $\{P:j(\alpha)\in j(P)\}$? I think this is the actual important passage I am missing. $\endgroup$ – W. Rether Oct 25 '17 at 17:10
  • 2
    $\begingroup$ Let $X$ be the set $\{P:\alpha\in P\}$. Then $X$ and $\alpha$ satisfy in $V$ the formula $(\forall z)\,(z\in x\leftrightarrow y\in z)$ (with $X$ as the value of $x$ and $\alpha$ as the value of $y$). By elementarity, $j(X)$ (as the value of $x$) and $j(\alpha)$ (as the value of $y$) satisfy the same formula in $M$. $\endgroup$ – Andreas Blass Oct 25 '17 at 17:21

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