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, 2021 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, 2021 at 7:17

1 Answer 1


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, 2021 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, 2021 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, 2021 at 7:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.