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This has now been cross posted to MO.

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.

Mason
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    Every set in the chain is countable, so you certainly don't hit $\mathbb{R}$ at any point. – Patrick Stevens Nov 26 '18 at 19:19
  • How do you propose to define $x^\alpha$ if $\alpha \notin \mathbb{Q}$? – Hans Engler Nov 26 '18 at 19:24
  • $2^{\sqrt2}$ is transcendental, but it satisfies $x^{\sqrt2}-4=0$, so it's in $A_2$. I'd conjecture that $2^{2^{\sqrt2}}$ is in $A_3$ and not $A_2$, that $2^{2^{2^{\sqrt2}}}$ is in $A_4$ and not $A_3$, etcetera. – Akiva Weinberger Dec 01 '18 at 21:34
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    I'm still somewhat concerned about the definition of $x^\alpha$ when $x<0$ and $\alpha$ is not an integer. But, there is a cool question in here (possibly a very difficult one). – Jyrki Lahtonen Dec 03 '18 at 04:31
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    Mason, I had some simple worries in my mind. Like is $S_F$ necessarily closed under negation? – Jyrki Lahtonen Dec 03 '18 at 14:06
  • @JyrkiLahtonen. Just to be clear about the concern raised: $ x^{\sqrt2}-2=0 \implies x= \sqrt{2}^{1/\sqrt{2}}$ What (generalized) polynomial is satisfied by $-x$? It could be that $ (-x)^{\sqrt2}-2=0$ is not a satisfactory answer. No. I would think this is indeed a satisfactory answer. The first polynomial has a non negative domain and the second has a non positive domain. What trouble do we get into if we take this naive approach? This route at least wins us closure under negation. Oh. The trouble we get into is when someone asks what the coefficient of the $x^{\sqrt 2}$ term is... – Mason Dec 03 '18 at 14:24
  • (continued) We should answer $e^{i \pi \sqrt{2}}$ but all of sudden it's start look like the question is better raised in the complex numbers. But maybe there is a different polynomial that satisfies $-x$? – Mason Dec 03 '18 at 14:29
  • Just a question: is $n$ (defined inside the definition of $S_F$ as a natural number) also arbitrary for any member of $S_F$? – Mostafa Ayaz Dec 03 '18 at 22:15
  • @JyrkiLahtonen, for this purpose I think we can define $(-|x|)^\alpha$ as $-(|x|^\alpha)$. –  Dec 06 '18 at 21:59
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    I don’t think we can rule out $e\in A_3$. My impression is that we know almost nothing about the transcendence of numbers like $3^\sqrt6-4^\sqrt3-1$. –  Dec 06 '18 at 22:35
  • How do we prove that $4^\sqrt3+1\in\cup A_n$? This seems like an argument for defining $S_F$ to include closure under field operations. –  Dec 07 '18 at 07:21
  • It seems like we might be able to achieve multiplicative inverses using a technique analogous to the one presented here. I haven't worked out all the details but it looks like it could be a good lead – Mason Dec 18 '18 at 19:41
  • Actually... $S_F$ is clearly closed under multiplicative inverses because $-e_i$ is in a field $F$ whenever $e_i$ is in $F$ – Mason Dec 18 '18 at 23:49
  • I am not so sure but maybe this piece by B Zilbner speaks to the question. – Mason Dec 21 '18 at 16:41
  • Why is $S_\mathbb Q$ equal to the set of algebraic real numbers? For instance, is any solution to $\sqrt x+\sqrt[3]x=0$ algebraic? – Jack M Mar 05 '19 at 18:59
  • @JackM, $x=0$ satisfies this and is algebraic. I am not sure I understand your meaning... But if you take $x=y^6$ then we can replace your equation with $y^3+y^2=0$ and we can always manage to do this type of change when the exponents are rational. – Mason Mar 05 '19 at 21:23

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