# Constructing an infinite chain of subsets of 'hyper' algebraic numbers?

This question is cross posted from MSE.

Let $$F$$ be a subset of $$\mathbb{R}$$ and let $$S_F$$ denote the set of values which satisfy some generalized polynomial whose exponents and coefficients are drawn from $$F$$. That is, we let $$S_F$$ denote $$\bigg \{x \in \mathbb{R}: 0=\sum_{i=1}^n{a_i x^{e_i}}: e_i \in F \text{ distinct}, a_i\in F \text{ non-zero}, n\in \mathbb{N} \bigg \}$$

Then $$S_{\mathbb{\mathbb{Q}}}$$ is the set of algebraic real numbers and we start to see the beginnings of a chain:

$$\mathbb{Q} \subsetneq S_\mathbb{Q} \subsetneq S_{S_\mathbb{Q}}$$

Main Question

Does this chain continue forever? That is, we let $$A_0= \mathbb{Q}$$ and let $$A_{n+1}=S_{A_{n}}$$. Is it the case that $$A_n \subsetneq A_{n+1}$$ for all $$n\in\mathbb{N}$$?

Other curiosities:

Is $$A_i$$ always a field? Perhaps, the argument is analogous to this. Or maybe this is just the case in a more general setting: Is it the case that $$F \subset \mathbb{R}$$, a field implies that $$S_F$$ is a field?

Is it possible to see that $$e\notin \cup A_i$$? Perhaps this is just a tweaking of LW Theorem.

• This is my first Q here and I am happy to modify anything. I know the tone of this website is slightly distinct from MSE. If you want to suggest edits feel free to ping me. Dec 20, 2018 at 21:07
• Welcome to Math Overflow. Dec 20, 2018 at 21:47
• AFAIK it could just well be that $e\in A_2$, and I wouldn't really hope for a proof one way or the other. Assuning Schanuel's conjecture might give you some results, though Dec 21, 2018 at 6:53
• @Wojowu. I think $e\notin A_2$ is precisely what LW gives us. You may mean $A_3$? Note that $\cup A_i$ is countable. So we should expect that an arbitrary real number is not a member of $\cup A_i$. However: Expectations are not mathematical proofs. Dec 21, 2018 at 12:20
• @Mason Right, for some reason I thought $A_1$ already contains numbers of the sort $2^{\sqrt{2}}$. Dec 21, 2018 at 12:38