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i.e. what is cardinality of $\{A \mid \ A\subset \mathbb R, A \text{ is a field} \}$?

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why not $2^c$, $c$ is continuum ? –  Leitingok Aug 16 '11 at 10:52
    
@ Leitingok: why do you answer a question with another question? [The standard answer is "why not?" etc., etc.] –  Georges Elencwajg Aug 16 '11 at 10:59
    
@Georges: note that Leitingok is the OP. I believe they're responding to a now-deleted comment asserting the answer to be $\mathfrak{c}$. –  Chris Eagle Aug 16 '11 at 11:02
    
Yes, sorry, that now-deleted comment was by me. It should have read "At least $\mathfrak{c}$" (you get a continuum by extending your field by a single element, but Georges Elencwajg takes this to its proper conclusion, below). –  user1729 Aug 16 '11 at 11:15
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Dear Chris, I didn't know that, thanks for your explanation. But actually it doesn't matter: I just wanted to recycle an old joke (probably not very funny) about recursion and about committing the very sin I was pretending to criticize. –  Georges Elencwajg Aug 16 '11 at 11:17

2 Answers 2

up vote 11 down vote accepted

The number $x$ of subfields of $\mathbb R$ is $x=2^{\mathfrak c}$ where ${\mathfrak c}=2^{\aleph_0}$. Here is a proof.

Choose a transcendence basis $(x_i)_{i \in I}$ of $\mathbb R$ over $\mathbb Q$, where $I$ necessarily has cardinality $\mathfrak c$ .
a) For each subset $J\subset I$ you get a subfield $\mathbb Q ((x_j)_{j \in J})\subset \mathbb R$ and so you already get as many subfields as there are subsets $J\subset I$ , namely $2^{\mathfrak c}$.
Hence $x \geq 2^{\mathfrak c}$

b) Of course you can't have more subfields than subsets of $\mathbb R$.
Hence $x \leq 2^{\mathfrak c}$

c) From a) , b) and Cantor-Schroeder-Bernstein, deduce $x = 2^{\mathfrak c}$

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Without AC, Can this hold ? –  Leitingok Aug 18 '11 at 6:35

$\mathbb{R}$ has $2^{\mathfrak{c}}=2^{2^{\aleph_0}}$ subfields.

Let $\mathcal{B}$ be a transcendence basis for $\mathbb{R}$ over $\mathbb{Q}$. Then $|\mathcal{B}|=|\mathbb{R}|=\mathfrak{c}$. Every subset of $\mathcal{B}$ generates a different subfield of $\mathbb{R}$, so this gives us at least $|\mathcal{P}(\mathcal{B})|=2^{\mathfrak{c}}$ subfields. Since there are only $2^{\mathfrak{c}}$ subsets of $\mathbb{R}$ in total, there are clearly at most $2^{\mathfrak{c}}$ subfields. Thus by Schroeder–Bernstein there are exactly $2^{\mathfrak{c}}$.

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