Showing that the Zariski topology with polynomials in $Q[x_1,…,x_n]$ over $Bbb R^n$ is $T_0$
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I am considering the Zariski topology over $Q[x_1,...,x_n$] as a topology over $Bbb R^n$, so that a set is Zariski-closed if there exists a set of polynomials $I$ such that $V= {r in Bbb R^n | f(r)=0 $ for all $f in I }$
First off a topological space is $T_0$ if for distinct $x,y$ we have an open set that contains one and not the other, which I believe is equivalent to having a closed set that does not contain one and not the other.
Next I consider three cases
1) If $x,y$ have rational components, then I just take a polynomial $f(x)=t-x_i$ for some $i$ where $x_i,y_i$ are different.
2) If one of $x,y$ has all rational components and the other isn't , then again I take $f(x)=t-x_i$ for some $i$ where $x_i,y_i$ are different.
3) If both $x,y$ have irrational components, I am stuck. I would need to find a polynomial that is zero for $x$ but not $y$ (or vice versa) which is tripping me up. Any hints appreciated.
Edited to reflect that this is over $Bbb R_n$
general-topology polynomials
asked Nov 17 at 22:20
IntegrateThis
1,7081717
add a comment |
up vote
1
down vote
favorite
I am considering the Zariski topology over $Q[x_1,...,x_n$] as a topology over $Bbb R^n$, so that a set is Zariski-closed if there exists a set of polynomials $I$ such that $V= {r in Bbb R^n | f(r)=0 $ for all $f in I }$
First off a topological space is $T_0$ if for distinct $x,y$ we have an open set that contains one and not the other, which I believe is equivalent to having a closed set that does not contain one and not the other.
Next I consider three cases
1) If $x,y$ have rational components, then I just take a polynomial $f(x)=t-x_i$ for some $i$ where $x_i,y_i$ are different.
2) If one of $x,y$ has all rational components and the other isn't , then again I take $f(x)=t-x_i$ for some $i$ where $x_i,y_i$ are different.
3) If both $x,y$ have irrational components, I am stuck. I would need to find a polynomial that is zero for $x$ but not $y$ (or vice versa) which is tripping me up. Any hints appreciated.
Edited to reflect that this is over $Bbb R_n$
general-topology polynomials
asked Nov 17 at 22:20
IntegrateThis
1,7081717
@SaucyO'Path yes I know, I just phrased it badly, will edit.
– IntegrateThis
Nov 17 at 22:35
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am considering the Zariski topology over $Q[x_1,...,x_n$] as a topology over $Bbb R^n$, so that a set is Zariski-closed if there exists a set of polynomials $I$ such that $V= {r in Bbb R^n | f(r)=0 $ for all $f in I }$
First off a topological space is $T_0$ if for distinct $x,y$ we have an open set that contains one and not the other, which I believe is equivalent to having a closed set that does not contain one and not the other.
Next I consider three cases
1) If $x,y$ have rational components, then I just take a polynomial $f(x)=t-x_i$ for some $i$ where $x_i,y_i$ are different.
2) If one of $x,y$ has all rational components and the other isn't , then again I take $f(x)=t-x_i$ for some $i$ where $x_i,y_i$ are different.
3) If both $x,y$ have irrational components, I am stuck. I would need to find a polynomial that is zero for $x$ but not $y$ (or vice versa) which is tripping me up. Any hints appreciated.
Edited to reflect that this is over $Bbb R_n$
general-topology polynomials
asked Nov 17 at 22:20
IntegrateThis
1,7081717
I am considering the Zariski topology over $Q[x_1,...,x_n$] as a topology over $Bbb R^n$, so that a set is Zariski-closed if there exists a set of polynomials $I$ such that $V= {r in Bbb R^n | f(r)=0 $ for all $f in I }$
First off a topological space is $T_0$ if for distinct $x,y$ we have an open set that contains one and not the other, which I believe is equivalent to having a closed set that does not contain one and not the other.
Next I consider three cases
1) If $x,y$ have rational components, then I just take a polynomial $f(x)=t-x_i$ for some $i$ where $x_i,y_i$ are different.
2) If one of $x,y$ has all rational components and the other isn't , then again I take $f(x)=t-x_i$ for some $i$ where $x_i,y_i$ are different.
3) If both $x,y$ have irrational components, I am stuck. I would need to find a polynomial that is zero for $x$ but not $y$ (or vice versa) which is tripping me up. Any hints appreciated.
Edited to reflect that this is over $Bbb R_n$
general-topology polynomials
general-topology polynomials
asked Nov 17 at 22:20
IntegrateThis
1,7081717
asked Nov 17 at 22:20
IntegrateThis
1,7081717
edited Nov 17 at 22:43
asked Nov 17 at 22:20
IntegrateThis
1,7081717
asked Nov 17 at 22:20
IntegrateThis
1,7081717
asked Nov 17 at 22:20
IntegrateThis
1,7081717
1,7081717
@SaucyO'Path yes I know, I just phrased it badly, will edit.
– IntegrateThis
Nov 17 at 22:35
add a comment |
@SaucyO'Path yes I know, I just phrased it badly, will edit.
– IntegrateThis
Nov 17 at 22:35
@SaucyO'Path yes I know, I just phrased it badly, will edit.
– IntegrateThis
Nov 17 at 22:35
@SaucyO'Path yes I know, I just phrased it badly, will edit.
– IntegrateThis
Nov 17 at 22:35
add a comment |
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
This is not true. In fact, if $(a_1,cdots,a_n)$ is a point such that its coordinates are algebraically independent over $Bbb Q$, then by definition the only closed set that contains it is the whole $Bbb R^n$. So, any two such points are topologically indistinguishable.
answered Nov 17 at 22:45
Saucy O'Path
5,5811626
Right so in $n=1$ no polynomial (let alone with rational coefficients) would be zero for $pi$ or $e$, so there can't be an open set containing one but not the other?
– IntegrateThis
Nov 17 at 22:49
1
@IntegrateThis There is no polynomial with rational coefficients. With real coefficients there are plenty, namely $X-pi$ and such.
– Saucy O'Path
Nov 17 at 22:50
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
This is not true. In fact, if $(a_1,cdots,a_n)$ is a point such that its coordinates are algebraically independent over $Bbb Q$, then by definition the only closed set that contains it is the whole $Bbb R^n$. So, any two such points are topologically indistinguishable.
answered Nov 17 at 22:45
Saucy O'Path
5,5811626
Right so in $n=1$ no polynomial (let alone with rational coefficients) would be zero for $pi$ or $e$, so there can't be an open set containing one but not the other?
– IntegrateThis
Nov 17 at 22:49
1
@IntegrateThis There is no polynomial with rational coefficients. With real coefficients there are plenty, namely $X-pi$ and such.
– Saucy O'Path
Nov 17 at 22:50
add a comment |
up vote
3
down vote
accepted
This is not true. In fact, if $(a_1,cdots,a_n)$ is a point such that its coordinates are algebraically independent over $Bbb Q$, then by definition the only closed set that contains it is the whole $Bbb R^n$. So, any two such points are topologically indistinguishable.
answered Nov 17 at 22:45
Saucy O'Path
5,5811626
Right so in $n=1$ no polynomial (let alone with rational coefficients) would be zero for $pi$ or $e$, so there can't be an open set containing one but not the other?
– IntegrateThis
Nov 17 at 22:49
1
@IntegrateThis There is no polynomial with rational coefficients. With real coefficients there are plenty, namely $X-pi$ and such.
– Saucy O'Path
Nov 17 at 22:50
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
This is not true. In fact, if $(a_1,cdots,a_n)$ is a point such that its coordinates are algebraically independent over $Bbb Q$, then by definition the only closed set that contains it is the whole $Bbb R^n$. So, any two such points are topologically indistinguishable.
answered Nov 17 at 22:45
Saucy O'Path
5,5811626
This is not true. In fact, if $(a_1,cdots,a_n)$ is a point such that its coordinates are algebraically independent over $Bbb Q$, then by definition the only closed set that contains it is the whole $Bbb R^n$. So, any two such points are topologically indistinguishable.
answered Nov 17 at 22:45
Saucy O'Path
5,5811626
answered Nov 17 at 22:45
Saucy O'Path
5,5811626
answered Nov 17 at 22:45
Saucy O'Path
5,5811626
answered Nov 17 at 22:45
Saucy O'Path
5,5811626
5,5811626
Right so in $n=1$ no polynomial (let alone with rational coefficients) would be zero for $pi$ or $e$, so there can't be an open set containing one but not the other?
– IntegrateThis
Nov 17 at 22:49
1
@IntegrateThis There is no polynomial with rational coefficients. With real coefficients there are plenty, namely $X-pi$ and such.
– Saucy O'Path
Nov 17 at 22:50
add a comment |
Right so in $n=1$ no polynomial (let alone with rational coefficients) would be zero for $pi$ or $e$, so there can't be an open set containing one but not the other?
– IntegrateThis
Nov 17 at 22:49
1
@IntegrateThis There is no polynomial with rational coefficients. With real coefficients there are plenty, namely $X-pi$ and such.
– Saucy O'Path
Nov 17 at 22:50
Right so in $n=1$ no polynomial (let alone with rational coefficients) would be zero for $pi$ or $e$, so there can't be an open set containing one but not the other?
– IntegrateThis
Nov 17 at 22:49
Right so in $n=1$ no polynomial (let alone with rational coefficients) would be zero for $pi$ or $e$, so there can't be an open set containing one but not the other?
– IntegrateThis
Nov 17 at 22:49
1
1
@IntegrateThis There is no polynomial with rational coefficients. With real coefficients there are plenty, namely $X-pi$ and such.
– Saucy O'Path
Nov 17 at 22:50
@IntegrateThis There is no polynomial with rational coefficients. With real coefficients there are plenty, namely $X-pi$ and such.
– Saucy O'Path
Nov 17 at 22:50
add a comment |
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@SaucyO'Path yes I know, I just phrased it badly, will edit.
– IntegrateThis
Nov 17 at 22:35