Cardinality of collection of subfields of $\mathbb C$

Question

The question is just curiosity on my part. The title says it all. I can see that the cardinality is at least $\aleph_1$ (take simple extensions by an uncountable family of transcendental numbers). But is the cardinality the same as that of all subsets of $\mathbb C$? As a proof strategy, I'd wonder about finding an uncountable set such that any subset is "mutually transcendental", but this sounds like a lot to ask.

Answer 1

Obviously, the set $\{ \mbox{subfields of } \Bbb{C} \}$ has cardinality almost $2^{\mathfrak{c}}$.

If you fix a transcendence basis $B$ of the extension $\Bbb{C} / \Bbb{Q}$, necessarily $|B| = \mathfrak{c}$.

Now, $$\{ \Bbb{Q}(S): S \in \mathcal{P}(B)\}$$ is a set of distinct subfields of $\Bbb{C}$ whose cardinality is exactly $2^{\mathfrak{c}}$. It follows that the cardinality of the set $\{ \mbox{subfields of } \Bbb{C} \}$ is exactly $2^{\mathfrak{c}}$.