Can it be proven in ZFC that there exists a field with $\aleph_1$ subfields? I am not asking whether such a field is definable in ZFC, I am asking whether the statement "there exists a field $K$ with $\aleph_1$ subfields", after a suitable translation into first-order logic, can be proven in ZFC without using the continuum hypothesis.
Can it be proven in ZFC that there exists a field with $\aleph_1$ subfields
field-theoryset-theory
Related Solutions
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.
The basic strategy of an earlier answer of Asaf Karagila's can be used here too to show that there is no formula $\varphi$ which $\mathsf{ZFC}$ proves defines a set of reals of size $\aleph_1$. (Really, one proves: if $\mathsf{ZFC}\vdash\{x\in\mathbb{R}:\varphi(x)\}$ is uncountable then there is a forcing extension in which that set has size continuum.) The details are a bit technical, but I suspect this is folklore.
This, combined with the usual descriptive set theoretic obstacles (assuming $\neg\mathsf{CH}$ there is no Borel set of size $\aleph_1$, nor a projective set of size $\aleph_1$ if there are also large cardinals), gives a pretty strong negative answer.
Beyond this, you will need to make more precise your notion of "explicit construction."
Best Answer
No; indeed, any field $K$ with uncountably many subfields must have at least $2^{\aleph_0}$ subfields, so your statement is equivalent to CH. If $K$ has infinite transcendence degree over the prime field, you can get $2^{\aleph_0}$ subfields by taking subfields generated by arbitrary subsets of an infinite algebraically independent set. If $K$ has finite transcendence degree of the prime field, then in particular it is countable. Now observe that the set of (characteristic functions of) subfields of $K$ is a closed subset of $\{0,1\}^K$ (in the product topology). Since $K$ is countable, $\{0,1\}^K$ is just a Cantor set, so by the Cantor-Bendixson theorem, every uncountably closed subset of it has cardinality $2^{\aleph_0}$.