Beginner logic question on the continuum hypothesis

Question

I am very new to logic and don't know very much about it.

One thing that I know is that there are models of ZFC in which no cardinal lies strictly between $|\mathbb{Z}|$ and $|\mathbb{R}|$. There are also models of ZFC in which the continuum hypothesis is true. Based on this, we know that ZFC is consistent with both situations.

However, maybe I think that this situation is much like how universal properties work: the "ring" axioms are consistent with the existence of an element $x$ with $xx = -1$, but the universal ring has no such element. If somebody told me that the ring axioms were consistent with both, I might be unconcerned if I only cared about the universal model of $\mathbb{Z}$, especially if I only wanted to add $5$ and $13$. Other situations could be similar, but what is a universal model?

I don't know what the "universal model of ZFC" really is supposed to be, but definitely it seems like if no symbol can be constructed for a cardinality strictly in between $|\mathbb{Z}|$ and $|\mathbb{R}|$, then the "initial" model of ZFC simply won't have the continuum hypothesis (edit: I meant will).

Let $T$ be ZFC + 1 Grothendieck universe + ZFC is consistent. Here is my question:

Is the continuum hypothesis for the universal model of ZFC true in $T$?

Answer 1

In my opinion, your question is fundamentally flawed (but for a really interesting reason!).

As Andreas Blass says, your question starts by assuming that "the universal[/initial/whatever] model of $\mathsf{ZFC}$" is a thing that makes sense. This is much more subtle than it may appear! In contrast with (say) the ring axioms, the axioms of $\mathsf{ZFC}$ are extremely complicated. I mean this in a precise sense: the $\mathsf{ZFC}$ axioms unavoidably include sentences of arbitrarily high quantifier complexity, while the class of rings can be equationally axiomatized.

That said, there is an important "minimal-model-ish" construction for $\mathsf{ZFC}$. Given a model $M$ of $\mathsf{ZFC}$, we can always form the "constructible universe relative to $M$" - this, denoted "$L^M$" (hence my tweaking of your notation!), will always be a model of $\mathsf{ZFC}$. Moreover, it is minimal in the sense that if $M\models\mathsf{ZFC}$ and $N$ is a definable inner model of $M$ with the same ordinals, then $L^M=L^N$ - in particular, there is no $M$-definable inner model with all the $M$-ordinals which is smaller than $L^M$.

As you may suspect, the (generalized) continuum hypothesis holds within $L^M$ regardless of what $M$ is. Moreover, "$L$-ness" is expressible in the sense that there is a sentence in the language of set theory - denoted "$\mathsf{V=L}$" - which holds in exactly those models of the form $L^M$ for some $M$ (incidentally the $L$-construction is idempotent in the sense that $L^{L^M}=L^M$). So we even get a nice $\mathsf{ZFC}$-theorem, namely $$\mathsf{ZFC}\vdash\mathsf{V=L}\rightarrow \mathsf{GCH}.$$

However, it would be very wrong to think of $\mathsf{V=L}$ as "cutting down $\mathsf{ZFC}$ to its essentials." This axiom also has lots of "positive" consequences, such as the existence of a $\Diamond$ sequence or the existence of a projective well-ordering of the continuum, which are precisely as dubious (from a "naive minimalism" perspective) as the existence of a set of reals of intermediate cardinality between $\aleph_0$ and $2^{\aleph_0}$. I would summarize things as follows:

$\mathsf{ZFC}$ is so rich a theory that no model of $\mathsf{ZFC}$ can be "free of unexpected objects."