Constructing an infinite chain of subfields of 'hyper' algebraic numbers?
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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 numbers and we start to see the beginnings of a chain:
$S_{S_mathbb{Q}} supsetneq S_mathbb{Q} supsetneq mathbb{Q}$.
Main Question
Does this chain continue forever? That is, we let $A_0= mathbb{Q}$ and let $A_n =S_{A_{n-1}}$. Is it the case that $A_n supsetneq A_{n-1}$ 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
asked Nov 26 at 18:55
Mason
1,8641527
This question has an open bounty worth +150
reputation from Mason ending in 3 days.
The question is widely applicable to a large audience. A detailed canonical answer is required to address all the concerns.
|
show 21 more comments
up vote
13
down vote
favorite
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 numbers and we start to see the beginnings of a chain:
$S_{S_mathbb{Q}} supsetneq S_mathbb{Q} supsetneq mathbb{Q}$.
Main Question
Does this chain continue forever? That is, we let $A_0= mathbb{Q}$ and let $A_n =S_{A_{n-1}}$. Is it the case that $A_n supsetneq A_{n-1}$ 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
asked Nov 26 at 18:55
Mason
1,8641527
This question has an open bounty worth +150
reputation from Mason ending in 3 days.
The question is widely applicable to a large audience. A detailed canonical answer is required to address all the concerns.
3
Every set in the chain is countable, so you certainly don't hit $mathbb{R}$ at any point.
– Patrick Stevens
Nov 26 at 19:19
1
@Mason - are you doing this over $mathbb{R}$ or $mathbb{C}$?
– Hans Engler
Nov 26 at 21:25
4
An aged mathematician asks you never to use both “a” and “alpha” in similar contexts. It’s late in the day, my eyes are smarting, and I honestly can not tell the one from the other.
– Lubin
Nov 27 at 2:01
1
This is just not well defined over the complex numbers since also the limit does not exist. (There are many values of $i^i$ for example.) I think over the reals, the question is meaningful and interesting.
– Ben
Nov 29 at 16:33
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 at 4:31
|
show 21 more comments
up vote
13
down vote
favorite
up vote
13
down vote
favorite
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 numbers and we start to see the beginnings of a chain:
$S_{S_mathbb{Q}} supsetneq S_mathbb{Q} supsetneq mathbb{Q}$.
Main Question
Does this chain continue forever? That is, we let $A_0= mathbb{Q}$ and let $A_n =S_{A_{n-1}}$. Is it the case that $A_n supsetneq A_{n-1}$ 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
asked Nov 26 at 18:55
Mason
1,8641527
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 numbers and we start to see the beginnings of a chain:
$S_{S_mathbb{Q}} supsetneq S_mathbb{Q} supsetneq mathbb{Q}$.
Main Question
Does this chain continue forever? That is, we let $A_0= mathbb{Q}$ and let $A_n =S_{A_{n-1}}$. Is it the case that $A_n supsetneq A_{n-1}$ 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
asked Nov 26 at 18:55
Mason
1,8641527
asked Nov 26 at 18:55
Mason
1,8641527
edited Dec 2 at 21:12
asked Nov 26 at 18:55
Mason
1,8641527
asked Nov 26 at 18:55
Mason
1,8641527
asked Nov 26 at 18:55
Mason
1,8641527
1,8641527
This question has an open bounty worth +150
reputation from Mason ending in 3 days.
The question is widely applicable to a large audience. A detailed canonical answer is required to address all the concerns.
This question has an open bounty worth +150
reputation from Mason ending in 3 days.
The question is widely applicable to a large audience. A detailed canonical answer is required to address all the concerns.
3
Every set in the chain is countable, so you certainly don't hit $mathbb{R}$ at any point.
– Patrick Stevens
Nov 26 at 19:19
1
@Mason - are you doing this over $mathbb{R}$ or $mathbb{C}$?
– Hans Engler
Nov 26 at 21:25
4
An aged mathematician asks you never to use both “a” and “alpha” in similar contexts. It’s late in the day, my eyes are smarting, and I honestly can not tell the one from the other.
– Lubin
Nov 27 at 2:01
1
This is just not well defined over the complex numbers since also the limit does not exist. (There are many values of $i^i$ for example.) I think over the reals, the question is meaningful and interesting.
– Ben
Nov 29 at 16:33
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 at 4:31
|
show 21 more comments
3
Every set in the chain is countable, so you certainly don't hit $mathbb{R}$ at any point.
– Patrick Stevens
Nov 26 at 19:19
1
@Mason - are you doing this over $mathbb{R}$ or $mathbb{C}$?
– Hans Engler
Nov 26 at 21:25
4
An aged mathematician asks you never to use both “a” and “alpha” in similar contexts. It’s late in the day, my eyes are smarting, and I honestly can not tell the one from the other.
– Lubin
Nov 27 at 2:01
1
This is just not well defined over the complex numbers since also the limit does not exist. (There are many values of $i^i$ for example.) I think over the reals, the question is meaningful and interesting.
– Ben
Nov 29 at 16:33
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 at 4:31
3
3
Every set in the chain is countable, so you certainly don't hit $mathbb{R}$ at any point.
– Patrick Stevens
Nov 26 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 at 19:19
1
1
@Mason - are you doing this over $mathbb{R}$ or $mathbb{C}$?
– Hans Engler
Nov 26 at 21:25
@Mason - are you doing this over $mathbb{R}$ or $mathbb{C}$?
– Hans Engler
Nov 26 at 21:25
4
4
An aged mathematician asks you never to use both “a” and “alpha” in similar contexts. It’s late in the day, my eyes are smarting, and I honestly can not tell the one from the other.
– Lubin
Nov 27 at 2:01
An aged mathematician asks you never to use both “a” and “alpha” in similar contexts. It’s late in the day, my eyes are smarting, and I honestly can not tell the one from the other.
– Lubin
Nov 27 at 2:01
1
1
This is just not well defined over the complex numbers since also the limit does not exist. (There are many values of $i^i$ for example.) I think over the reals, the question is meaningful and interesting.
– Ben
Nov 29 at 16:33
This is just not well defined over the complex numbers since also the limit does not exist. (There are many values of $i^i$ for example.) I think over the reals, the question is meaningful and interesting.
– Ben
Nov 29 at 16:33
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 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 at 4:31
|
show 21 more comments
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3
Every set in the chain is countable, so you certainly don't hit $mathbb{R}$ at any point.
– Patrick Stevens
Nov 26 at 19:19
1
@Mason - are you doing this over $mathbb{R}$ or $mathbb{C}$?
– Hans Engler
Nov 26 at 21:25
4
An aged mathematician asks you never to use both “a” and “alpha” in similar contexts. It’s late in the day, my eyes are smarting, and I honestly can not tell the one from the other.
– Lubin
Nov 27 at 2:01
1
This is just not well defined over the complex numbers since also the limit does not exist. (There are many values of $i^i$ for example.) I think over the reals, the question is meaningful and interesting.
– Ben
Nov 29 at 16:33
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 at 4:31