Cardinality of subfields of real numbers

Question

I know that the set of real numbers is a field, and so is its subset, the set of rational numbers.

Are there any more fields which are subsets of the set of real numbers? If not, is there a proof?

Answer 1

Infinitely many. For instance, $\mathbf Q(\sqrt p)$ where $p$ is any prime number. If $p\ne p'$, $\;\mathbf Q(\sqrt p)\ne\mathbf Q(\sqrt p')$.

Answer 2

Examples are the set of rational numbers and the set of algebraic numbers. Another example is $\ \mathbb Q(\sqrt{2})=${$a+b\sqrt{2}\mid a,b\in \mathbb Q$}

Answer 3

Let $k\subset\Bbb{R}$ be any subfield and $r\in\Bbb{R}$ any element. Then $k(r)$ is again a subfield.

Starting from $\Bbb{Q}\subset\Bbb{R}$ you can keep adjoining elements to get (uncountably!) infinitely many subfields of $\Bbb{R}$.

Answer 4

The transcendence degree of $\Bbb R$ over $\Bbb Q$ is $2^{\aleph_0}$: there is a set $A$ of $2^{\aleph_0}$ algebraically independent transcendentals in $\Bbb R$. If $B\subseteq A$ we get a field $\Bbb Q(B)$ by adjoining the elements of $B$ to $\Bbb Q$. Different $B$ give different $\Bbb Q(B)$. This way we get $2^{2^{\aleph_0}}$ subfields of $\Bbb R$.