$F$ be the smallest subfield of the real numbers which contains irrational $a$. Prove that $F$ is countable.

Question

Let $a$ be an irrational number and let $F$ be the smallest subfield of the real numbers which contains $a$. Prove that $F$ is countable.

Answer 1

Define $F_0 = \mathbb{Q} \cup \{a\}$, which is clearly countable, as the union of a countable set and a finite set.

Having defined $F_n$, define $F_{n+1}$ by $$F_{n+1} = \{x+y: x,y \in F_n\} \cup \{-x: x \in F_n\} \cup \{x\cdot y: x,y \in F_n\} \cup \{\frac{1}{x}: x \in F_n, x \neq 0\}$$

This defines a sequence of subsets of $\mathbb{R}$, and they are all countable, as shown by induction: $F_0$ is countable, as stated, and if $F_n$ is, so is $F_{n+1}$: we only need to show that each of the four sets of the union is countable (a countable or finite union of countable sets is countable). For the sets of inverses and negatives this is clear (we have a bijection with $F_n$ or a subset thereof), and for the sums and products it's easy to see as $+$ and $\cdot$ form surjections from $F_n \times F_n$ onto these sets resp. (and a product of two countable sets is countable).

Also, handy to observe: $F_n \subset F_{n+1}$, as $x = x + 0$, and $0$ is in all $F_n$ (we could have defined $F_n$ to be the fifth of the sets in the defining union of $F_{n+1}$ as well, if we're lazy).

Now, $F = \cup_n F_n$ is countable (as a countable union of countable sets again) and is a field: e.g. let $x, y \in F$, so $x \in F_n, y \in F_m$ for some $n,m \in \mathbb{N}$. Then by the increasingness, setting $k = \max(n,m) \in \mathbb{N}$, we have that $x,y \in F_k$, so by definition, $x + y \in F_{k+1} \subset F$ as well. Similarly for multiplication, inversion and negation, plus we already have $0$ and $1$.

And it's the smallest field as well, because if $G$ is any field containg the rationals and $a$, then $F_0 \subset G$, and then by induction $F_n \subset G$ for all $G$, as $G$ is closed under all field operations already. So $F \subset G$. This last step is not strictly necessary to see that the smallest such field is countable, as we already see there exists at least one countable such field, so the smallest one would be a subset so also countable... But it's nice to know, anyway. We added no more than needed to make $F$ a field.

This idea of "exhausting", i.e. adding points to make a set closed under some operations(s), is a very common technique, and is one of the motivating examples to start using ordinal numbers later, e.g.