Constructing an infinite chain of subfields of 'hyper' algebraic numbers?
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
asked Nov 26 '18 at 18:55
Mason
1,9641530
|
show 18 more comments
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
asked Nov 26 '18 at 18:55
Mason
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
|
show 18 more comments
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
asked Nov 26 '18 at 18:55
Mason
1,9641530
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
asked Nov 26 '18 at 18:55
Mason
1,9641530
asked Nov 26 '18 at 18:55
Mason
1,9641530
edited Dec 20 '18 at 21:00
asked Nov 26 '18 at 18:55
Mason
1,9641530
asked Nov 26 '18 at 18:55
Mason
1,9641530
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
|
show 18 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 '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
|
show 18 more comments
0
active
oldest
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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 '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