Showing there exists a countable subfield $F\subseteq\Bbb{R}$ such that if $x\in F$, $\sin(x)\in F$

Question

I am asked to show that there exists a countable subfield $F\subseteq\Bbb{R}$ such that if $x\in F$, $\sin(x)\in F$.

I don't see a classic set-theory approach to solving this. I started out by looking at $\frac{\pi}2\in F$ trying to build such $F$, which requires that $-\frac{\pi}2, 1, -1, \pi, 0, -\pi \in F$ following the definition for a field, but this is where I get lost. Clearly, $\sin (1)$ isn't "pretty" and I can't figure out how to count such a set. Should I start with an element and see how the field forms, or rather use set theory arguments more generally?

Answer 1

Call $\Bbb Q(S)$ the smallest subfield of $\Bbb R$ that contains the set $S$. Proving that $\lvert \Bbb Q(S)\rvert=\max\{\lvert S\rvert,\aleph_0\}$ is an easy exercise. Call $H_0=\Bbb Q$ and define $H_{n+1}=\Bbb Q(H_n\cup\{\sin t\,:\, t\in H_n\})$. Then, $F=\bigcup_{n\in\Bbb N}H_n$ is a subfield of $\Bbb R$ such that $\sin(x)\in F$ for all $x\in F$ and, in point of fact, it's the smallest such subfield. It is countable because it's countable union of countable sets. The specific function $\sin$ is not relevant; in point of fact, the same construction works for any function $f$, and in point of fact even for closure under application of functions from some countable family $\mathcal G$.

Answer 2

Consider the set of all formal expressions formed from integers via the field operations and application of $\sin$. For example, $\frac{2 + \sin(3 + \sin(1))}{\sin(4 + 5)}$ could be one such formal expression.

Now, let us examine what happens when we try to evaluate such expressions. Some of them might be invalid because they include a division by 0, for example in $\frac{1}{\sin(0)}$. In other cases, we might get the same result from different expressions; for example, $\sin(4 + 5)$ and $\sin(9) + \sin(\sin(0))$ would give equal values.

However, what we do see is that if $E$ is the set of formal expressions, and $E'$ is the subset of expressions whose evaluation is valid, then we get a function $\operatorname{ev} : E' \to \mathbb{R}$. Now, $E$ is countable (for example because every such expression can be encoded as an ASCII string that could be evaluated by some programming language, and the set of ASCII strings is countable; alternatively you could use an argument similar to the argument in the answer by Saucy O'Path, in showing that the set of formal expressions whose abstract syntax tree has depth $n$ is countable for each $n$). Therefore, $E'$ is also countable, and so the image of $\operatorname{ev}$ is also a countable subset of $\mathbb{R}$.

The claim is now that $\operatorname{im}(\operatorname{ev})$ is a subfield of $\mathbb{R}$ which is closed under $\sin$. The proof of this is straightforward; for example, if $x \in \operatorname{im}(\operatorname{ev})$, then by definition there exists some expression $e \in E'$ such that $\operatorname{ev}(e) = x$. Then $\sin(e)$ forms another element of $E'$, and $\operatorname{ev}(\sin(e)) = \sin x$, so $\sin x \in \operatorname{im}(\operatorname{ev})$ also. This shows that $\operatorname{im}(\operatorname{ev})$ is closed under $\sin$. The proofs that $\operatorname{im}(\operatorname{ev})$ is also closed under addition, multiplication, negation, and under taking inverses of nonzero elements, will be similar. And likewise, the proofs that $0, 1 \in \operatorname{im}(\operatorname{ev})$ are immediate.