Mixing
Show that the set of real numbers is not definable in the field of complex number
$begingroup$
Like what the title suggest, I wish to show that $mathbb{R}$ cannot be define in $mathbb{C}$. I want to make use of the following proposition ;
(David Marker) Fix a structure $M$, if $X subset M$ is definable over a set of parameters $A$, then every automorphisms of $M$ that fixes $A$ pointwise must fixes $X$ setwise.
So if we assume the contrary that $mathbb{R}$ is definable in $mathbb{C}$, then it is definable over a finite set of parameters, $A$, from $mathbb{C}$. Now all that remains is to find an automorphism $sigma$ that fixes everypoint in $A$ but maps some real numbeers into complex numbers.
For whatever reason, I'm having some difficulties coming up with such an explicit $sigma$. Any help or insights is deeply appreciated.
logic model-theory
asked Feb 1 at 5:48
some1fromhellsome1fromhell
935412
$endgroup$
add a comment |
$begingroup$
Like what the title suggest, I wish to show that $mathbb{R}$ cannot be define in $mathbb{C}$. I want to make use of the following proposition ;
(David Marker) Fix a structure $M$, if $X subset M$ is definable over a set of parameters $A$, then every automorphisms of $M$ that fixes $A$ pointwise must fixes $X$ setwise.
So if we assume the contrary that $mathbb{R}$ is definable in $mathbb{C}$, then it is definable over a finite set of parameters, $A$, from $mathbb{C}$. Now all that remains is to find an automorphism $sigma$ that fixes everypoint in $A$ but maps some real numbeers into complex numbers.
For whatever reason, I'm having some difficulties coming up with such an explicit $sigma$. Any help or insights is deeply appreciated.
logic model-theory
asked Feb 1 at 5:48
some1fromhellsome1fromhell
935412
$endgroup$
2
$begingroup$
You can't write one down explicitly, you'll have to use something like transfinite induction. See math.stackexchange.com/a/2092817/86856 for an example of the sort of argument involved.
$endgroup$
– Eric Wofsey
Feb 1 at 6:14
$begingroup$
See this similar question for a stronger result: $mathbb{R}$ is not even interpretable in $mathbb{C}$. The question gives an argument using the notion of stability, and my answer gives a more elementary argument using automorphisms.
$endgroup$
– Alex Kruckman
Feb 1 at 15:12
add a comment |
$begingroup$
Like what the title suggest, I wish to show that $mathbb{R}$ cannot be define in $mathbb{C}$. I want to make use of the following proposition ;
(David Marker) Fix a structure $M$, if $X subset M$ is definable over a set of parameters $A$, then every automorphisms of $M$ that fixes $A$ pointwise must fixes $X$ setwise.
So if we assume the contrary that $mathbb{R}$ is definable in $mathbb{C}$, then it is definable over a finite set of parameters, $A$, from $mathbb{C}$. Now all that remains is to find an automorphism $sigma$ that fixes everypoint in $A$ but maps some real numbeers into complex numbers.
For whatever reason, I'm having some difficulties coming up with such an explicit $sigma$. Any help or insights is deeply appreciated.
logic model-theory
asked Feb 1 at 5:48
some1fromhellsome1fromhell
935412
$endgroup$
Like what the title suggest, I wish to show that $mathbb{R}$ cannot be define in $mathbb{C}$. I want to make use of the following proposition ;
(David Marker) Fix a structure $M$, if $X subset M$ is definable over a set of parameters $A$, then every automorphisms of $M$ that fixes $A$ pointwise must fixes $X$ setwise.
So if we assume the contrary that $mathbb{R}$ is definable in $mathbb{C}$, then it is definable over a finite set of parameters, $A$, from $mathbb{C}$. Now all that remains is to find an automorphism $sigma$ that fixes everypoint in $A$ but maps some real numbeers into complex numbers.
For whatever reason, I'm having some difficulties coming up with such an explicit $sigma$. Any help or insights is deeply appreciated.
logic model-theory
logic model-theory
asked Feb 1 at 5:48
some1fromhellsome1fromhell
935412
asked Feb 1 at 5:48
some1fromhellsome1fromhell
935412
asked Feb 1 at 5:48
some1fromhellsome1fromhell
935412
asked Feb 1 at 5:48
some1fromhellsome1fromhell
935412
asked Feb 1 at 5:48
some1fromhellsome1fromhell
935412
935412
2
$begingroup$
You can't write one down explicitly, you'll have to use something like transfinite induction. See math.stackexchange.com/a/2092817/86856 for an example of the sort of argument involved.
$endgroup$
– Eric Wofsey
Feb 1 at 6:14
$begingroup$
See this similar question for a stronger result: $mathbb{R}$ is not even interpretable in $mathbb{C}$. The question gives an argument using the notion of stability, and my answer gives a more elementary argument using automorphisms.
$endgroup$
– Alex Kruckman
Feb 1 at 15:12
add a comment |
2
$begingroup$
You can't write one down explicitly, you'll have to use something like transfinite induction. See math.stackexchange.com/a/2092817/86856 for an example of the sort of argument involved.
$endgroup$
– Eric Wofsey
Feb 1 at 6:14
$begingroup$
See this similar question for a stronger result: $mathbb{R}$ is not even interpretable in $mathbb{C}$. The question gives an argument using the notion of stability, and my answer gives a more elementary argument using automorphisms.
$endgroup$
– Alex Kruckman
Feb 1 at 15:12
2
2
$begingroup$
You can't write one down explicitly, you'll have to use something like transfinite induction. See math.stackexchange.com/a/2092817/86856 for an example of the sort of argument involved.
$endgroup$
– Eric Wofsey
Feb 1 at 6:14
$begingroup$
You can't write one down explicitly, you'll have to use something like transfinite induction. See math.stackexchange.com/a/2092817/86856 for an example of the sort of argument involved.
$endgroup$
– Eric Wofsey
Feb 1 at 6:14
$begingroup$
See this similar question for a stronger result: $mathbb{R}$ is not even interpretable in $mathbb{C}$. The question gives an argument using the notion of stability, and my answer gives a more elementary argument using automorphisms.
$endgroup$
– Alex Kruckman
Feb 1 at 15:12
$begingroup$
See this similar question for a stronger result: $mathbb{R}$ is not even interpretable in $mathbb{C}$. The question gives an argument using the notion of stability, and my answer gives a more elementary argument using automorphisms.
$endgroup$
– Alex Kruckman
Feb 1 at 15:12
add a comment |
1 Answer
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oldest
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$begingroup$
As Eric Wofsey said, actually building an automorphism of $mathbb{C}$ which moves a real is difficult: while we can prove that for every transcendental real $r$ there is a field automorphism of $mathbb{C}$ which moves $r$, this proof is $(i)$ a transfinite recursion argument and $(ii)$ relies on the axiom of choice.
There is, however, an automorphism argument which does work:
We'll show a stronger result - that for any transcendental real number $r$ and any formula $varphi(x)$ such that $mathbb{C}modelsvarphi(r)$, there is a non-real complex number $s$ such that $mathbb{C}modelsvarphi(s)$. The argument uses automorphisms and elementary submodels, and goes as follows (outlined only - it's a good exercise to fill it all in):
Fix a transcendental real $r$, a transcendental non-real complex number $s$, and a formula $varphi(x)$ such that $mathbb{C}modelsvarphi(r)$. Let $M$ be a countable elementary submodel of $mathbb{C}$ which contains $r$ and $s$ - we know such an $M$ exists by the downward Lowenheim-Skolem theorem.
- Note that we've replacing the not-obviously-well-orderable field $mathbb{C}$ by a countable, and hence well-orderable, field $M$. Intuitively, this should (and does) make the axiom of choice irrelevant here.
Now show that $M$ is a countable algebraically closed field of characteristic zero and infinite transcendence degree.
By a standard back-and-forth argument, there is an automorphism of $M$ swapping $r$ and $s$, so $Mmodelsvarphi(s)$. Back-and-forth arguments are still recursive constructions, but they don't use transfinite recursion; also, this is probably a result you've already proved.
Remembering how $M$ and $mathbb{C}$ are related, conclude that $mathbb{C}modelsvarphi(s)$ as desired.
Remark $1$. We can avoid elementary submodels entirely by working with Ehrenfeucht-Fraisse games instead; however, elementary submodel juggling is an important technique to learn.
Remark $2$. An even stronger fact is true: $mathbb{C}$ is strongly minimal, meaning that for any model $N$ of $Th(mathbb{C})$, every definable-with-parameters subset of $N$ is either finite or cofinite. (First, prove that $mathbb{C}$ has quantifier elimination, and then show that every quantifier-free formula with parameters defines either a finite or a cofinite set; then lift this to every model of $Th(mathbb{C})$.)
answered Feb 1 at 14:20
Noah SchweberNoah Schweber
128k10152294
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add a comment |
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1 Answer
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$begingroup$
As Eric Wofsey said, actually building an automorphism of $mathbb{C}$ which moves a real is difficult: while we can prove that for every transcendental real $r$ there is a field automorphism of $mathbb{C}$ which moves $r$, this proof is $(i)$ a transfinite recursion argument and $(ii)$ relies on the axiom of choice.
There is, however, an automorphism argument which does work:
We'll show a stronger result - that for any transcendental real number $r$ and any formula $varphi(x)$ such that $mathbb{C}modelsvarphi(r)$, there is a non-real complex number $s$ such that $mathbb{C}modelsvarphi(s)$. The argument uses automorphisms and elementary submodels, and goes as follows (outlined only - it's a good exercise to fill it all in):
Fix a transcendental real $r$, a transcendental non-real complex number $s$, and a formula $varphi(x)$ such that $mathbb{C}modelsvarphi(r)$. Let $M$ be a countable elementary submodel of $mathbb{C}$ which contains $r$ and $s$ - we know such an $M$ exists by the downward Lowenheim-Skolem theorem.
- Note that we've replacing the not-obviously-well-orderable field $mathbb{C}$ by a countable, and hence well-orderable, field $M$. Intuitively, this should (and does) make the axiom of choice irrelevant here.
Now show that $M$ is a countable algebraically closed field of characteristic zero and infinite transcendence degree.
By a standard back-and-forth argument, there is an automorphism of $M$ swapping $r$ and $s$, so $Mmodelsvarphi(s)$. Back-and-forth arguments are still recursive constructions, but they don't use transfinite recursion; also, this is probably a result you've already proved.
Remembering how $M$ and $mathbb{C}$ are related, conclude that $mathbb{C}modelsvarphi(s)$ as desired.
Remark $1$. We can avoid elementary submodels entirely by working with Ehrenfeucht-Fraisse games instead; however, elementary submodel juggling is an important technique to learn.
Remark $2$. An even stronger fact is true: $mathbb{C}$ is strongly minimal, meaning that for any model $N$ of $Th(mathbb{C})$, every definable-with-parameters subset of $N$ is either finite or cofinite. (First, prove that $mathbb{C}$ has quantifier elimination, and then show that every quantifier-free formula with parameters defines either a finite or a cofinite set; then lift this to every model of $Th(mathbb{C})$.)
answered Feb 1 at 14:20
Noah SchweberNoah Schweber
128k10152294
$endgroup$
add a comment |
$begingroup$
As Eric Wofsey said, actually building an automorphism of $mathbb{C}$ which moves a real is difficult: while we can prove that for every transcendental real $r$ there is a field automorphism of $mathbb{C}$ which moves $r$, this proof is $(i)$ a transfinite recursion argument and $(ii)$ relies on the axiom of choice.
There is, however, an automorphism argument which does work:
We'll show a stronger result - that for any transcendental real number $r$ and any formula $varphi(x)$ such that $mathbb{C}modelsvarphi(r)$, there is a non-real complex number $s$ such that $mathbb{C}modelsvarphi(s)$. The argument uses automorphisms and elementary submodels, and goes as follows (outlined only - it's a good exercise to fill it all in):
Fix a transcendental real $r$, a transcendental non-real complex number $s$, and a formula $varphi(x)$ such that $mathbb{C}modelsvarphi(r)$. Let $M$ be a countable elementary submodel of $mathbb{C}$ which contains $r$ and $s$ - we know such an $M$ exists by the downward Lowenheim-Skolem theorem.
- Note that we've replacing the not-obviously-well-orderable field $mathbb{C}$ by a countable, and hence well-orderable, field $M$. Intuitively, this should (and does) make the axiom of choice irrelevant here.
Now show that $M$ is a countable algebraically closed field of characteristic zero and infinite transcendence degree.
By a standard back-and-forth argument, there is an automorphism of $M$ swapping $r$ and $s$, so $Mmodelsvarphi(s)$. Back-and-forth arguments are still recursive constructions, but they don't use transfinite recursion; also, this is probably a result you've already proved.
Remembering how $M$ and $mathbb{C}$ are related, conclude that $mathbb{C}modelsvarphi(s)$ as desired.
Remark $1$. We can avoid elementary submodels entirely by working with Ehrenfeucht-Fraisse games instead; however, elementary submodel juggling is an important technique to learn.
Remark $2$. An even stronger fact is true: $mathbb{C}$ is strongly minimal, meaning that for any model $N$ of $Th(mathbb{C})$, every definable-with-parameters subset of $N$ is either finite or cofinite. (First, prove that $mathbb{C}$ has quantifier elimination, and then show that every quantifier-free formula with parameters defines either a finite or a cofinite set; then lift this to every model of $Th(mathbb{C})$.)
answered Feb 1 at 14:20
Noah SchweberNoah Schweber
128k10152294
$endgroup$
add a comment |
$begingroup$
As Eric Wofsey said, actually building an automorphism of $mathbb{C}$ which moves a real is difficult: while we can prove that for every transcendental real $r$ there is a field automorphism of $mathbb{C}$ which moves $r$, this proof is $(i)$ a transfinite recursion argument and $(ii)$ relies on the axiom of choice.
There is, however, an automorphism argument which does work:
We'll show a stronger result - that for any transcendental real number $r$ and any formula $varphi(x)$ such that $mathbb{C}modelsvarphi(r)$, there is a non-real complex number $s$ such that $mathbb{C}modelsvarphi(s)$. The argument uses automorphisms and elementary submodels, and goes as follows (outlined only - it's a good exercise to fill it all in):
Fix a transcendental real $r$, a transcendental non-real complex number $s$, and a formula $varphi(x)$ such that $mathbb{C}modelsvarphi(r)$. Let $M$ be a countable elementary submodel of $mathbb{C}$ which contains $r$ and $s$ - we know such an $M$ exists by the downward Lowenheim-Skolem theorem.
- Note that we've replacing the not-obviously-well-orderable field $mathbb{C}$ by a countable, and hence well-orderable, field $M$. Intuitively, this should (and does) make the axiom of choice irrelevant here.
Now show that $M$ is a countable algebraically closed field of characteristic zero and infinite transcendence degree.
By a standard back-and-forth argument, there is an automorphism of $M$ swapping $r$ and $s$, so $Mmodelsvarphi(s)$. Back-and-forth arguments are still recursive constructions, but they don't use transfinite recursion; also, this is probably a result you've already proved.
Remembering how $M$ and $mathbb{C}$ are related, conclude that $mathbb{C}modelsvarphi(s)$ as desired.
Remark $1$. We can avoid elementary submodels entirely by working with Ehrenfeucht-Fraisse games instead; however, elementary submodel juggling is an important technique to learn.
Remark $2$. An even stronger fact is true: $mathbb{C}$ is strongly minimal, meaning that for any model $N$ of $Th(mathbb{C})$, every definable-with-parameters subset of $N$ is either finite or cofinite. (First, prove that $mathbb{C}$ has quantifier elimination, and then show that every quantifier-free formula with parameters defines either a finite or a cofinite set; then lift this to every model of $Th(mathbb{C})$.)
answered Feb 1 at 14:20
Noah SchweberNoah Schweber
128k10152294
$endgroup$
As Eric Wofsey said, actually building an automorphism of $mathbb{C}$ which moves a real is difficult: while we can prove that for every transcendental real $r$ there is a field automorphism of $mathbb{C}$ which moves $r$, this proof is $(i)$ a transfinite recursion argument and $(ii)$ relies on the axiom of choice.
There is, however, an automorphism argument which does work:
We'll show a stronger result - that for any transcendental real number $r$ and any formula $varphi(x)$ such that $mathbb{C}modelsvarphi(r)$, there is a non-real complex number $s$ such that $mathbb{C}modelsvarphi(s)$. The argument uses automorphisms and elementary submodels, and goes as follows (outlined only - it's a good exercise to fill it all in):
Fix a transcendental real $r$, a transcendental non-real complex number $s$, and a formula $varphi(x)$ such that $mathbb{C}modelsvarphi(r)$. Let $M$ be a countable elementary submodel of $mathbb{C}$ which contains $r$ and $s$ - we know such an $M$ exists by the downward Lowenheim-Skolem theorem.
- Note that we've replacing the not-obviously-well-orderable field $mathbb{C}$ by a countable, and hence well-orderable, field $M$. Intuitively, this should (and does) make the axiom of choice irrelevant here.
Now show that $M$ is a countable algebraically closed field of characteristic zero and infinite transcendence degree.
By a standard back-and-forth argument, there is an automorphism of $M$ swapping $r$ and $s$, so $Mmodelsvarphi(s)$. Back-and-forth arguments are still recursive constructions, but they don't use transfinite recursion; also, this is probably a result you've already proved.
Remembering how $M$ and $mathbb{C}$ are related, conclude that $mathbb{C}modelsvarphi(s)$ as desired.
Remark $1$. We can avoid elementary submodels entirely by working with Ehrenfeucht-Fraisse games instead; however, elementary submodel juggling is an important technique to learn.
Remark $2$. An even stronger fact is true: $mathbb{C}$ is strongly minimal, meaning that for any model $N$ of $Th(mathbb{C})$, every definable-with-parameters subset of $N$ is either finite or cofinite. (First, prove that $mathbb{C}$ has quantifier elimination, and then show that every quantifier-free formula with parameters defines either a finite or a cofinite set; then lift this to every model of $Th(mathbb{C})$.)
answered Feb 1 at 14:20
Noah SchweberNoah Schweber
128k10152294
answered Feb 1 at 14:20
Noah SchweberNoah Schweber
128k10152294
answered Feb 1 at 14:20
Noah SchweberNoah Schweber
128k10152294
answered Feb 1 at 14:20
Noah SchweberNoah Schweber
128k10152294
128k10152294
add a comment |
add a comment |
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$begingroup$
You can't write one down explicitly, you'll have to use something like transfinite induction. See math.stackexchange.com/a/2092817/86856 for an example of the sort of argument involved.
$endgroup$
– Eric Wofsey
Feb 1 at 6:14
$begingroup$
See this similar question for a stronger result: $mathbb{R}$ is not even interpretable in $mathbb{C}$. The question gives an argument using the notion of stability, and my answer gives a more elementary argument using automorphisms.
$endgroup$
– Alex Kruckman
Feb 1 at 15:12