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.

  • $\begingroup$ 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. $\endgroup$
    – Mason
    Dec 20, 2018 at 21:07
  • 3
    $\begingroup$ Welcome to Math Overflow. $\endgroup$ Dec 20, 2018 at 21:47
  • $\begingroup$ 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 $\endgroup$
    – Wojowu
    Dec 21, 2018 at 6:53
  • $\begingroup$ @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. $\endgroup$
    – Mason
    Dec 21, 2018 at 12:20
  • $\begingroup$ @Mason Right, for some reason I thought $A_1$ already contains numbers of the sort $2^{\sqrt{2}}$. $\endgroup$
    – Wojowu
    Dec 21, 2018 at 12:38


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