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This is (a variation on) exercise 20.4 in Jech's "Set Theory." Let $j : V \to M$ witness $\lambda$-supercompactness of $\kappa$, and consider the normal measure $U$ on $P_{\kappa}(\lambda)$ consisting of those $x$ such that $j[\lambda] \in j(x)$. (How do you make the left quotation mark symbol to denote 'j-image-of-lambda'?)

We want to show that this measure extends the club filter.

This hint is as follows: Suppose $C$ is club. Then define $D = j[C]$. Then:

  1. $D$ is a directed subset of $j(C)$.
  2. $D$ has size $|C| \leq \lambda ^{< \kappa} < j(\kappa )$.
  3. Therefore $\bigcup D \in j(C)$.
  4. $\bigcup D = j[\lambda ]$

I'm fine with 1. I'm not sure about 2 - where is the argument taking place, in $V$ or in $M$, or both? For 3, it appears the underlying argument is this:

$V \vDash \forall E \subset C\ (E$ directed and $|E| < \kappa \Rightarrow \bigcup E \in C)$

and so

$M \vDash \forall E \subset j(C)\ (E$ directed and $|E| < j(\kappa) \Rightarrow \bigcup \in j(C))$

I can accept this assuming that 2 means "$D \in M$ and $M \vDash |D| < j(\kappa )$." I'm having trouble with 4 as well - I believe that $j[\lambda] \subseteq \bigcup D$, but why does the reverse inclusion hold, i.e. why is it that $x \in C, \beta \in j(x) \Rightarrow \beta \in j[\lambda]$?

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  • $\begingroup$ A partial answer for 2 and 3: As the ultrafilter $U$ is a fine measure we even have $M^{\lambda^{<\kappa}} \subset M$, and as $D \in M^{\lambda^{<\kappa}}$ we know that $D \in M$ (See Prop 22.11. in Kanamoris Book for a proof of this). Moreover as $M$ is transitive and $V$ thinks that the size of $C$ = size of $D$ $\le \lambda^{<\kappa} < j(\kappa)$, $M$ thinks that the size of $D$ is less than $j(\kappa)$. So both things that you wanted to accept 2 and 3 are done. $\endgroup$
    – Stefan Hoffelner
    Commented Apr 8, 2011 at 19:11
  • $\begingroup$ Hi oktan. Why does the fineness of $U$ give us said closure property of $M$? We're obtaining $U$ from $j : V \to M$, not vice versa. Also, sorry this may be a basic question, but how do we know $\lambda ^{<\kappa} < j(\kappa)$? $\endgroup$
    – Amit Kumar Gupta
    Commented Apr 8, 2011 at 21:04
  • $\begingroup$ The first question in the comment above is not so serious though; for the application I'm interested in I'm happy to assume $j : V \to M$ arises from some fine measure, say $U'$, on $P_{\kappa}(\lambda)$. $\endgroup$
    – Amit Kumar Gupta
    Commented Apr 8, 2011 at 21:06

1 Answer 1

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Suppose $\kappa$ is $\lambda$-supercompact for some $\lambda \geq \kappa$, and let $j: V \rightarrow M$ be an elementary embedding with critical point $\kappa$ such that $j(\kappa) > \lambda$ and $M^{\lambda} \subseteq M$ for some inner model $M$. First, observe that $V$ and $M$ agree on $P_{\kappa}\lambda$ because $M$ is closed under ${<}\kappa$ sequences. In particular, this means that $\lambda^{{<}\kappa} \leq (\lambda^{{<}\kappa})^M$ since $M \subseteq V$. But this then means that $j(\kappa) > (\lambda^{{<}\kappa})^M \geq \lambda^{{<}\kappa}$ because $j(\kappa)$ is inaccessible in $M$ and $j(\kappa)$ is greater than both $\lambda$ and $\kappa$. Next, note that any $x \in P_{\kappa}\lambda$ will be a subset of $\lambda$ having size less than the critical point $\kappa$ so that $j(x) = j''x \subseteq j''\lambda$.

[Specifically, if for some $\alpha < \kappa$, we have a bijection $f: \alpha \rightarrow x$, then $j(f)$ will be a bijection between $j(\alpha) = \alpha$ and $j(x)$. So every element of $j(x)$ is of the form $j(f)(\beta)$ for some $\beta < \alpha$, but $j(f)(\beta) = j(f(\beta)) \in j''x$ since $\beta$ is also below the critical point.]

Also, $M$ will contain $h = j \upharpoonright \lambda$ by its closure under $\lambda$ sequences. Therefore, $M$ will have $j''P_{\kappa}\lambda = \{j(x)| x \in P_{\kappa}\lambda\} = \{j''x| x \in P_{\kappa}\lambda\} = \{h''x| x \in P_{\kappa}\lambda\}$. Now letting $g: P_{\kappa}\lambda \rightarrow \lambda^{{<}\kappa}$ be a bijection in $V$, we will have a bijection $j(g) \upharpoonright j''P_{\kappa}\lambda: j''P_{\kappa}\lambda \rightarrow j''\lambda^{{<}\kappa}$ in $M$. Therefore, $M$ will have the range of $j(g)$, which is exactly $j''\lambda^{{<}\kappa}$. Now, since $C$ has size at most $\lambda^{{<}\kappa}$ (in $V$), we may let $e: \lambda^{{<}\kappa} \rightarrow C$ be a surjection. Then $j(e) \upharpoonright j''\lambda^{{<}\kappa}: j''\lambda^{{<}\kappa} \rightarrow j''C$ is a surjection in $M$ so similarly, its range, $D = j''C$, will be in $M$. But $M$ will also know that $j''\lambda^{{<}\kappa}$ has size $\lambda^{{<}\kappa} < j(\kappa)$ because $M$ can construct $j \upharpoonright \lambda^{{<}\kappa}$ from $j''\lambda^{{<}\kappa}$ by virtue of $j$ being order-preserving. Therefore $\bigcup D \in j(C)$ by the ${<}j(\kappa)$-directed closure of $j(C)$ in $M$, as you mention.

Also, if $x \in C$, then $|x| < \kappa$ and $x \subseteq \lambda$ so $j(x) = j''x \subseteq j''\lambda$. Therefore, $\bigcup D = \bigcup j''C \subseteq j''\lambda$.

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  • $\begingroup$ Small omission: should say range of $j(g)$ restricted to $j''P_{\kappa}\lambda$, which is exactly $j''\lambda^{{<}\kappa}$. $\endgroup$
    – Jason
    Commented Apr 9, 2011 at 6:31
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    $\begingroup$ Thanks for the response! I think the argument in the second last paragraph can be simplified a little: If $g:P_{\kappa}(\lambda)\to C$ is a surjection in V, then $j(g)\upharpoonright j''P_{\kappa}(\lambda)$ belongs to $M$ by your observation that $j''P_{\kappa}(\lambda)$ does, and it surjects onto D. Also, by your observation that $P_{\kappa}^M(\lambda) = P_{\kappa}^V(\lambda)$, we can define a surjection $e : P_{\kappa}(\lambda) \to j''P_{\kappa}(\lambda)$ by $e(x) = h''x$. Thus in M, $|D| \leq |P_{\kappa}(\lambda)| = \lambda^{<\kappa} < j(\kappa)$ since $j(\kappa)$ is inaccessible in $M$. $\endgroup$
    – Amit Kumar Gupta
    Commented Apr 10, 2011 at 15:40

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