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 $T$?

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    $\begingroup$ The Continuum Hypothesis is that there is no cardinality between that of the integers and the reals. So, you need to flip one of your cases. en.m.wikipedia.org/wiki/Continuum_hypothesis $\endgroup$
    – badjohn
    Mar 19 at 15:52
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    $\begingroup$ The usual definition of "initial ..." is "having a unique morphism to any ...." So before asking what the initial model of ZFC looks like, you should specify what counts as a morphism between models of ZFC. And you should ask whether an initial model exists. $\endgroup$ Mar 19 at 17:39
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    $\begingroup$ I've changed your "$L$" to a "$T$." There's no reason for you to have known, but as it happens there is a very good reason for this! $\endgroup$ Mar 19 at 21:11
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    $\begingroup$ A minor note about the kind of heuristic used in your question. You argue that if no symbol can be constructed for a cardinality strictly in between |Z| and |R|, then CH should hold in <good> models of ZFC. This kind of argument rarely works, and two such arguments can often be played against each other. For example, you could equally well have argued that if no symbol can be constructed for a bijection between $\aleph_1$ and $2^{\aleph_0}$, then CH should fail in <good> models of ZFC. $\endgroup$
    – Z. A. K.
    Mar 20 at 7:34

1 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."

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    $\begingroup$ I always enjoy reading your answers. $\endgroup$
    – Alex
    Mar 19 at 22:11

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