Theorem: If M is a countable transitive model of ZFC, and $\kappa$ is a supercompact cardinal in M, and $2^\kappa=\kappa^+$. Then there exists a forcing extension M[G] such that $\kappa$ becomes a measurable cardinal and $2^\kappa=\kappa^{++}$.

I want to know why we need to use iterated forcing in this theorem. What if we make $2^\kappa=\kappa^{++}$ in the simplest way?

Proof: Let $j:M\rightarrow N$ be the elementary embedding such that $crit(j)=\kappa$, $j(\kappa)>\kappa^{++}$, $N^{\kappa^{++}}\subset N$.

Take P be subsets of $\kappa\times \kappa^{++}$ with cardinal smaller than $\kappa$. The order on P is inclusion.

Let G be a generic filter of P. We prove that $\kappa$ is measurable in M[G].

Because $j''G$ is pairwise compatible, and $j(P)$ is a $j(\kappa)$-closed forcing condition. So there is a generic filter K on $j(P)$ such that $j''G\subset K$.

For every $M^P$ name $\dot{x}$, $j(\dot{x})$ is an $N^{j(P)}$-name. We extend j to $M[G]\rightarrow N[K]$ in the following way: $j(x):=j(\dot{x})^K$.

If $p\in G$, $p\Vdash \dot{x}=\dot{y}$, then $j(p)\in K$, and $j(p)\Vdash j(\dot{x})=j(\dot{y})$, so j is well defined. Next we prove j is an elementary ebmedding.

It's not hard to check $\left \| \varphi(j(\dot{x_1}),...,j(\dot{x_n})) \right \|$ in B(j(P)) is equal to $j(\left \| \varphi(\dot{x_1},...,\dot{x_n}) \right \|)$ , so

$N[K]\models \varphi(j(\dot{x_1}),...,j(\dot{x_n}))\Leftrightarrow \left \| \varphi(j(\dot{x_1}),...,j(\dot{x_n})) \right \|\in K \Leftrightarrow \left \| \varphi(\dot{x_1},...,\dot{x_n}) \right \|\in G\Leftrightarrow M[G]\models \varphi(x_1,...,x_n)$.

Now j can be defined in $M[G\times K]$, and it induces an ultrafilter U on $\kappa$, but j(P) is $j(\kappa)$-closed and $|U|=2^\kappa$, so $U\in M[G]$.

  • 1
    $\begingroup$ j(P) is $j(\kappa)$-closed in N, while what you are claiming is that going from M[G] to M[$G \times K$] can be done by a very closed forcing $\endgroup$ Sep 1 at 7:10
  • $\begingroup$ Indeed you seem to claim j(P) is $j(\kappa)$-closed in M[G], which is not true. $\endgroup$ Sep 1 at 7:17

The simplest reason we have to use iterations to violate GCH below $\kappa$ is because it's required. More specifically, we have:

If $\kappa$ is a measurable cardinal and $2^\kappa>\kappa^+$, then there is a normal measure $U$ and some $A\in U$ such that $2^\gamma>\gamma^+$ for all $\gamma\in U$. (In fact, every normal measure will satisfy this).

This fact is proven via a quick analysis of ultrapowers coming from normal measures. For example, see Lemma 17.11 in Jech. So to violate the GCH at $\kappa$, it is necessary to violate it very often below $\kappa$.

In particular, this fact allows one to see that (assuming GCH, for instance) $\mathrm{Add}(\kappa,\kappa^{++})$ can destroy the measurability of $\kappa$, since it makes GCH fail at it and nowhere below.

As to where the proof fails: we only have $N\vDash j(P)\text{ is } j(\kappa)\text{-closed}$ by elementarity. At the same time, $M\vDash [\mathrm{Add}(j(\kappa),j(\kappa^{++}))]^M \text{ is }j(\kappa)\text{-closed}$, but $[\mathrm{Add}(j(\kappa),j(\kappa^{++}))]^M\neq j(P)=[\mathrm{Add}(j(\kappa),j(\kappa^{++}))]^N$. So the required filter $K$ is not guaranteed to exist.

  • $\begingroup$ Thank you. And how does the iterated forcing solved this problem? $\endgroup$ Sep 1 at 7:19
  • $\begingroup$ Since we'll need to disturb the continuum function very often below $\kappa$, and we won't know beforehand what the last bit of the forcing will look like (i.e. the poset named by the final factor in the iteration, to add $\kappa^{++}$ subsets to $\kappa$), so we let iteration take care of that by letting the final factor just be "that poset to add this many subsets to $\kappa$ when we get there". $\endgroup$ Sep 1 at 7:36
  • $\begingroup$ With respect to the specific problem of requiring a specific closure, this is treated for instance in Jech by Lemma 21.9. $\endgroup$ Sep 1 at 7:38

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