Equivalent form of CH in ZF

Question

I know that in $ZFC$ the continuum hypothesis (CH) is equivalent to the statement: "a subset $A\subset\mathbb R$ is either finite, infinite countable or $A$ has the same cardinality of $\mathbb R$, i.e. $|A|=|\mathbb R|$". We can call this statement $CH(\mathbb R)$, but what's going on if we work in $ZF$? Clearly $CH\Rightarrow CH(\mathbb R)$, but does the other implication hold? What I want to know is if there is a model of $ZF$ in which $CH(\mathbb R)$ is true but $CH$ not.

Answer 1

There are, as you point out, two versions of CH which one can consider:

• CH$(\mathbb{R})$: "Every set of reals is either countable or has size continuum." (To me, "countable" includes "finite," and I believe this is the general usage.)

• CH: "$2^{\aleph_0}=\aleph_1$." Note that both of these are perfectly well-defined: $2^{\aleph_0}$ is the cardinality of the set of reals, and $\aleph_1$ is the cardinality of the set of countable ordinals (in fact, it is the set of countable ordinals if you want to go there).

It is consistent with ZF that CH$(\mathbb{R})$ holds but CH fails. The nicest way this can happen is if the axiom of determinacy holds. In ZF+AD we have:

• By the determinacy of perfect set games, every uncountable set of reals contains a perfect subset and hence has cardinality $2^{\aleph_0}$. So CH$(\mathbb{R})$ holds.

• On the other hand, it can be shown that there is no injection of $\omega_1$ into $\mathbb{R}$ or of $\mathbb{R}$ into $\omega_1$, so $\aleph_1$ and $2^{\aleph_0}$ are incomparable.

You may find these notes by Sherwood Hachtman useful.

(Indeed, the arrangement of cardinals under ZF+AD is quite interesting; for example, $\mathbb{R}/\mathbb{Q}$ (the set of equivalence classes of reals modulo rational difference) has greater cardinality than $\mathbb{R}$! Remember that we measure cardinality in terms of injections: $\vert A\vert\le\vert B\vert$ iff there is an injection from $A$ to $B$. In the absence of choice, a surjection from $B$ to $A$ does not imply $\vert A\vert\le\vert B\vert$ in general ...)

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The axiom of determinacy has the slight drawback of high consistency strength - it is possible that ZFC is consistent but ZF+AD is not (by contrast, ZFC is consistent iff ZF is consistent - adding choice to ZF won't result in a contradiction unless one was there already). We can get the same picture without a jump in consistency strength via forcing and symmetric submodels, but that gets a bit technical.

In case you're interested, there are a few different tools to calibrate the consistency strength of a theory. On the proof-theoretic side we can associate ordinals to theories: a bigger ordinal means greater consistency strength. This ordinal analysis, however, quickly becomes extremely difficult to carry out; the current state-of-the-art is somewhere around $\Pi^1_2$-CA, which is a tiny tiny fragment of ZF. If you're interested in ordinal analysis, I recommend this paper by Rathjen.

A much more useful approach in set theory is given by the large cardinal hierarchy. Large cardinal axioms express the existence of certain "extremely infinite" sets in reasonably well-understood ways, and they provide a useful yardstick for measuring the consistency strength of extensions of ZF (for example, ZF+AD is consistent iff the theory ZFC+"There are infinitely many Woodin cardinals" is consistent).