Is it known whether there is a wellfounded model of ZFC, containing all reals, in which CH fails? In which it obtains?



CH and wellfounded models of ZFC containing all reals
Cole: 


Suppose first that $V=L$. Then certainly there is no well founded model as you want, because any well-founded model $M$ is correct about $L$, meaning $L^M=L\_{ORD^M}$, and if $M$ contains all the reals, then this model must contain $L\_{\omega\_1}$ and must therefore satisfy CH. 


(The assumption of $V=L$ is certainly an overkill here, one can easily produce by forcing many other examples of models where CH holds and no well-founded model with all the reals will satisfy not-CH. In fact, much more is true. For example, Jensen's remarkable "[coding the universe](http://books.google.com/books?id=WiA4AAAAIAAJ&printsec=frontcover&dq=coding+the+universe&hl=en&src=bmrr&ei=qIqrTbm0EdHUiALDztHvDA&sa=X&oi=book_result&ct=result&resnum=1&ved=0CDAQ6AEwAA#v=onepage&q&f=false)" theorem shows that using class forcing techniques, there is an extension of the universe that has the form $L[r]$ for some real $r$; but in $L[r]$ we have CH for the same reason that we have it in $L$. Now, note that no inner model with all the reals can satisfy not-CH, since of course any such model contains $r$ and so is correct about $L[r]$, so the argument from the previous paragraph applies.)


However, the technique of forcing gives us that it is also consistent that there is a model as you ask. For CH could fail, but one can always force to add CH without adding any reals (simply add generically an enumeration of the reals in type $\omega\_1$ using as conditions initial countable initial segments of the enumeration). This is actually a useful thing to know. It gives us that, as long as we are only interested in first-order statements about the reals, assuming CH is harmless: Any first order statement about the reals that holds under CH is in fact true. This is used sometimes in descriptive set theory, since amenability of groups is particularly nice in the presence of CH.


This shows that the answer to your question is independent of the usual axioms of set theory. (Of course, the answer is yes if we assume not-CH to begin with.)


Let me close with the comment that, if you are willing to forgo the axiom of choice, there are interesting ways in which you can have such models. For example, assuming enough large cardinals, then the inner model $L({\mathbb R})$ (the smallest transitive model of ZF that contains all the ordinals and all the reals) *is* an example of a model where CH fails in one sense (there is a surjection from ${\mathbb R}$ onto $\omega\_2$, for example), independently of whether CH fails or holds in the universe. Note this is now a theorem, not just a consistency result. The situation is in fact a bit stranger, because in $L({\mathbb R})$, CH holds in another sense (every subset of ${\mathbb R}$ is either countable or has the same size as ${\mathbb R}$).


