Constructing an infinite chain of subfields of 'hyper' algebraic numbers?












24














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
$$bigg {x in mathbb{R}: 0=sum_{i=1}^n{a_i x^{e_i}}: e_i in F text{ distinct}, a_iin F text{ non-zero}, nin mathbb{N} bigg }$$



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 $ninmathbb{N}$?




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 $enotin cup A_i$? Perhaps this is just a tweaking of LW Theorem.






real-analysis field-theory transcendental-numbers







share|cite|improve this question




















  • 3




    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






  • 1




    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 3 '18 at 4:31






  • 1




    Mason, I had some simple worries in my mind. Like is $S_F$ necessarily closed under negation?
    – Jyrki Lahtonen
    Dec 3 '18 at 14:06






  • 2




    I don’t think we can rule out $ein A_3$. My impression is that we know almost nothing about the transcendence of numbers like $3^sqrt6-4^sqrt3-1$.
    – Matt F.
    Dec 6 '18 at 22:35


















24














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
$$bigg {x in mathbb{R}: 0=sum_{i=1}^n{a_i x^{e_i}}: e_i in F text{ distinct}, a_iin F text{ non-zero}, nin mathbb{N} bigg }$$



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 $ninmathbb{N}$?




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 $enotin cup A_i$? Perhaps this is just a tweaking of LW Theorem.






real-analysis field-theory transcendental-numbers







share|cite|improve this question




















  • 3




    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






  • 1




    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 3 '18 at 4:31






  • 1




    Mason, I had some simple worries in my mind. Like is $S_F$ necessarily closed under negation?
    – Jyrki Lahtonen
    Dec 3 '18 at 14:06






  • 2




    I don’t think we can rule out $ein A_3$. My impression is that we know almost nothing about the transcendence of numbers like $3^sqrt6-4^sqrt3-1$.
    – Matt F.
    Dec 6 '18 at 22:35
















24












24








24


2





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
$$bigg {x in mathbb{R}: 0=sum_{i=1}^n{a_i x^{e_i}}: e_i in F text{ distinct}, a_iin F text{ non-zero}, nin mathbb{N} bigg }$$



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 $ninmathbb{N}$?




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 $enotin cup A_i$? Perhaps this is just a tweaking of LW Theorem.






real-analysis field-theory transcendental-numbers







share|cite|improve this question















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
$$bigg {x in mathbb{R}: 0=sum_{i=1}^n{a_i x^{e_i}}: e_i in F text{ distinct}, a_iin F text{ non-zero}, nin mathbb{N} bigg }$$



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 $ninmathbb{N}$?




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 $enotin cup A_i$? Perhaps this is just a tweaking of LW Theorem.






real-analysis field-theory transcendental-numbers




real-analysis field-theory transcendental-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 20 '18 at 21:00

























asked Nov 26 '18 at 18:55









Mason

1,9641530




1,9641530








  • 3




    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






  • 1




    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 3 '18 at 4:31






  • 1




    Mason, I had some simple worries in my mind. Like is $S_F$ necessarily closed under negation?
    – Jyrki Lahtonen
    Dec 3 '18 at 14:06






  • 2




    I don’t think we can rule out $ein A_3$. My impression is that we know almost nothing about the transcendence of numbers like $3^sqrt6-4^sqrt3-1$.
    – Matt F.
    Dec 6 '18 at 22:35
















  • 3




    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






  • 1




    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 3 '18 at 4:31






  • 1




    Mason, I had some simple worries in my mind. Like is $S_F$ necessarily closed under negation?
    – Jyrki Lahtonen
    Dec 3 '18 at 14:06






  • 2




    I don’t think we can rule out $ein A_3$. My impression is that we know almost nothing about the transcendence of numbers like $3^sqrt6-4^sqrt3-1$.
    – Matt F.
    Dec 6 '18 at 22:35










3




3




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




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




How do you propose to define $x^alpha$ if $alpha notin mathbb{Q}$?
– Hans Engler
Nov 26 '18 at 19:24




1




1




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 3 '18 at 4:31




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 3 '18 at 4:31




1




1




Mason, I had some simple worries in my mind. Like is $S_F$ necessarily closed under negation?
– Jyrki Lahtonen
Dec 3 '18 at 14:06




Mason, I had some simple worries in my mind. Like is $S_F$ necessarily closed under negation?
– Jyrki Lahtonen
Dec 3 '18 at 14:06




2




2




I don’t think we can rule out $ein A_3$. My impression is that we know almost nothing about the transcendence of numbers like $3^sqrt6-4^sqrt3-1$.
– Matt F.
Dec 6 '18 at 22:35






I don’t think we can rule out $ein A_3$. My impression is that we know almost nothing about the transcendence of numbers like $3^sqrt6-4^sqrt3-1$.
– Matt F.
Dec 6 '18 at 22:35












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