# Which step is wrong in the following simplification of Silver's forcing?

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 $$U\in M[G]$$.

• 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 Sep 1, 2021 at 7:10
• Indeed you seem to claim j(P) is $j(\kappa)$-closed in M[G], which is not true. Sep 1, 2021 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.

• Thank you. And how does the iterated forcing solved this problem? Sep 1, 2021 at 7:19
• 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". Sep 1, 2021 at 7:36
• With respect to the specific problem of requiring a specific closure, this is treated for instance in Jech by Lemma 21.9. Sep 1, 2021 at 7:38