# CH and wellfounded models of ZFC containing all reals

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

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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" 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 ${\mathbb R}$).

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Haha, 51 seconds! –  tci Apr 17 '11 at 23:54
Hehe. I had seen this happen to other people, but I think it is the first time I'm involved. –  Andres Caicedo Apr 17 '11 at 23:59
Well, I was pausing over the part I got wrong, so perhaps I should have taken longer. :-) –  tci Apr 18 '11 at 0:04
Thanks, Andres. –  Cole Leahy Apr 18 '11 at 3:19

Assuming the existence of suitable large cardinals, the Axiom of Determinacy holds in $L(\mathbb R)$, which implies or does not imply CH depending on what you take CH to be:

If you take CH to be the statement that no set of reals has cardinality between countable and the cardinality of the reals, then yes, CH does hold in this model.

However, if you take it to mean that $2^{\aleph_0}$=$\aleph_1$(in the sense of bijections), then no, CH does not hold. Assuming AD, there is no injection from $\omega_1$ to $2^{\aleph_0}$. If there were, you could build all sorts of pathological sets which AD does not allow(for example, an uncountable set which is not perfect).

I'm a little hazy about what follows:

Also, assuming $V=L$, then $L$ contains(?) all the reals, and CH also holds in it.

In any case, CH or not CH can be forced from any given model, so I would guess that from any model containing all the reals you can go to one where CH or not CH holds by doing the right sort of forcing. I also can't see why these forcing extensions would not have all the reals given that they are forcing extensions of models which did.

$\text{Edit}$: As Andres mentions in the comment, the above paragraph is not correct. Andres' answer explains how much of it can be fixed.

Anyhow, given the experts here, I'm sure you'll get a better, more detailed answer with fewer disclaimers.

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Tanmay: If CH holds, you cannot go by forcing to a model where it fails, unless you add reals. (Of course, if CH fails, there is no need to do anything.) And it is independent that CH holds and there is an inner model of choice where CH fails. (I address this issue in my answer.) –  Andres Caicedo Apr 17 '11 at 23:58
Deleted that comment. It didn't make any sense. I seem to be a little sleepy on top of every thing else! Again, nice answer. –  tci Apr 18 '11 at 0:32