# Con ZF implies Con ZFC using set sized models

Can we use forcing to construct models of ZFC and ZFC + GCH starting from c.t.m s of ZF? The usual way to obtain the associated relative consistency results (Con ZF implies Con ZFC and Con ZF implies Con ZF + GCH)is to look at the class sized model L (of ZF) and I was interested in seeing if there was an alternate way to do this. Is there any reference material I could look up?

(I do realize that a model of ZF + GCH will model AC so the the question was phrased this way in case it was possible to do the first construction but not the second)

## 1 Answer

Not always.

There are models of $\sf ZF$ which cannot be extended to models of $\sf ZFC$ without adding ordinals. Since forcing is a technique which does not add ordinals, this means that this is impossible. Note that when I say that, I include class-forcing as well.

Moreover, if you only limit yourself to set forcing, then the answer is an even simpler no. One can easily construct (using class forcing) models of $\sf ZF$ such that no set forcing extension can satisfy the axiom of choice. Similarly one can produce models of $\sf ZFC$ that have no set forcing extension which satisfy $\sf GCH$.

Do note, though, that given a set model of $\sf ZF$ one can produce a set model of $\sf ZFC+GCH$ by simply taking the $L$ of that model, which will be a model of $\sf ZFC$ satisfying $V=L$ as well.

• Thank you. That helps a lot. Are there any reference sources for this that you know of? – UserB1234 Oct 6 '13 at 19:19
• Which part of "this"? – Asaf Karagila Oct 6 '13 at 19:19
• For the "models of ZF which cannot be extended to models of ZFC without adding ordinals" and "one can produce models of ZFC that have no set forcing extension which satisfy GCH". – UserB1234 Oct 6 '13 at 19:26
• The former is an obvious observation on Gitik's model in which every $\aleph$ number has a countable cofinality; but you can find a more difficult example due to D. Morris in Jech's The Axiom of Choice (problem 14, or so, in chapter 5). The latter is by using Easton's theorem by having $2^\kappa=\kappa^{++}$ for every regular $\kappa$; then noting that $\sf GCH$ can only be achieved by collapsing a class of cardinals, and no set forcing can collapse a class of cardinals. – Asaf Karagila Oct 6 '13 at 19:30
• It is not a more difficult example, Asaf. – Andrés E. Caicedo Oct 6 '13 at 19:39