Is there a ZFC-definable subset of the real numbers whose cardinality is provably $\aleph_1$, without assuming the Continuum Hypothesis?

  • $\begingroup$ I know of a subset of $P(P(\Bbb Q))$ which is provably $\aleph_1$. But that's one too many power set operations. $\endgroup$ – Arthur Mar 29 at 14:43
  • $\begingroup$ @Arthur I haven't heard of this set. Is it easily describable? $\endgroup$ – Don Thousand Mar 29 at 14:51
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    $\begingroup$ What do you mean by "definable"? $\endgroup$ – Asaf Karagila Mar 29 at 15:16
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    $\begingroup$ @DonThousand Any countable ordinal can be embedded as a subset of $\Bbb Q$ with the standard order. In other words, for each countable ordinal, there are elements of $P(\Bbb Q)$ which are order isomorphic to it. So each countable ordinal has a corresponding subset of $P(\Bbb Q)$, which is to say element of $P(P(\Bbb Q))$, of all sets of rationals order isomorphic to it. Now take the set of all of those, and you have a subset of $P(P(\Bbb Q))$ which is bijective to the first uncountable ordinal. $\endgroup$ – Arthur Mar 29 at 15:20
  • $\begingroup$ If I recall correctly, we have a very strong negative answer: there is a model of ZFC with no parameter-freely definable set of reals of (internal) cardinality $\aleph_1$. I don't remember how to get this, though. The most obvious approach is to start with $W$ being the $L(\mathbb{R})$ of some ZFC-model with large cardinals, and force over $W$ with (say) countable partial functions from $\omega_2$ to $\mathbb{R}$. The result satisfies ZFC (once we well-order $\mathbb{R}$, everything else well-orders itself) and since the forcing is homogeneous and $W$ had the perfect set property, (cont'd) $\endgroup$ – Noah Schweber Mar 29 at 16:53

The situation is complicated.

There is a surjective map from $\Bbb R$ onto $\omega_1$, this much is provably in $\sf ZF$, and so using choice this map has an injective inverse which is a subset of $\Bbb R$ that has size $\aleph_1$. The function, by the way, is the following:

Fix a bijection between $\Bbb R$ and $\Bbb{R^N}$, then map $x$ to $\alpha$ if its corresponding sequence in $\Bbb{R^N}$ is an enumeration of a set which is well-ordered in the standard $<$ relation of $\Bbb R$. Otherwise, map $x$ to $0$. Since every countable ordinal embeds into the rationals, this is certainly a surjective function.

But is this set definable? In what sense? In any reasonable sense, you can claim that it is not definable, unless the injective inverse is definable (which could be the case, depending on additional axioms such as $V=L$). Let me point out that here by definable we mean "definable in the set theoretic universe without parameters".

We can interpret definable as Borel, which would be a sensible way to interpret this outside of the context of set theory, and then the answer is certainly negative. All uncountable Borel sets have a copy of the Cantor set inside them, which makes them the size of the real numbers. In other words, Borel sets are not counterexamples to the Continuum Hypothesis, and cannot provably have size $\aleph_1$.

Okay. So going back to the original question. The naive approach to a counterexample would be the collapse the continuum and add many Cohen reals. This, in principle, should work. But the direct proof seems to hit a snag whenever you remember that you're talking about sets of reals, and not sets of ordinals. David Asperó suggested a different solution: $\Bbb P_{\max}$ forcing over $L(\Bbb R)$ as a model of $\sf AD$.

Since $\Bbb P_\max$ is homogeneous, adds no reals, and forces $2^{\aleph_0}=\aleph_2$, we can show that any definable set of reals was already in $L(\Bbb R)$ and by $\sf AD$ has a perfect subset. Therefore has size $\aleph_2$.

But this requires quite a strong consistency strength. And it would very interesting to see if the large cardinals can be removed.


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