Mixing
Show the set is closed
$begingroup$
Let $X$ be a Hausdorff locally compact topological space, $s:Xtimes Xto mathbb{R}$ be a continuous map such that $s(x,x)=0$ for all $xin X$. Set $triangle={(x,y): x=y, x,yin X}$ and $A={(x,y): s(x,y)=0}$, so $trianglesubset A$. If for each fixed $xin X$, $s(x,cdot)$ is a local homeomorphism near $y=x$ in the second variable, show that the set $Y=Abackslash triangle$ is closed in $Xtimes X$.
(I'm not certain the condition in the problem is sufficient, the real problem I met is that $X$ is a complex manifold and $mathbb{R}$ being replaced by another complex manifold, $s$ is a holomorphic map and it's a local biholomorphism near the diagonal in the second variable. But I believe it's really a topological problem..)
general-topology complex-geometry several-complex-variables
asked Jan 29 at 15:29
Chun GanChun Gan
291110
$endgroup$
add a comment |
$begingroup$
Let $X$ be a Hausdorff locally compact topological space, $s:Xtimes Xto mathbb{R}$ be a continuous map such that $s(x,x)=0$ for all $xin X$. Set $triangle={(x,y): x=y, x,yin X}$ and $A={(x,y): s(x,y)=0}$, so $trianglesubset A$. If for each fixed $xin X$, $s(x,cdot)$ is a local homeomorphism near $y=x$ in the second variable, show that the set $Y=Abackslash triangle$ is closed in $Xtimes X$.
(I'm not certain the condition in the problem is sufficient, the real problem I met is that $X$ is a complex manifold and $mathbb{R}$ being replaced by another complex manifold, $s$ is a holomorphic map and it's a local biholomorphism near the diagonal in the second variable. But I believe it's really a topological problem..)
general-topology complex-geometry several-complex-variables
asked Jan 29 at 15:29
Chun GanChun Gan
291110
$endgroup$
$begingroup$
How is "$s(x,cdot)$ is a local homeomorphism near $y=x$" defined precisely?
$endgroup$
– Henno Brandsma
Jan 29 at 17:28
$begingroup$
Fix $xin X$, then there exists an open neighborhood $Usubset X$ near $y=x$, and $s(x,cdot):Uto Vsubset mathbb{R}$ is a homeomorphism.
$endgroup$
– Chun Gan
Jan 29 at 17:30
add a comment |
$begingroup$
Let $X$ be a Hausdorff locally compact topological space, $s:Xtimes Xto mathbb{R}$ be a continuous map such that $s(x,x)=0$ for all $xin X$. Set $triangle={(x,y): x=y, x,yin X}$ and $A={(x,y): s(x,y)=0}$, so $trianglesubset A$. If for each fixed $xin X$, $s(x,cdot)$ is a local homeomorphism near $y=x$ in the second variable, show that the set $Y=Abackslash triangle$ is closed in $Xtimes X$.
(I'm not certain the condition in the problem is sufficient, the real problem I met is that $X$ is a complex manifold and $mathbb{R}$ being replaced by another complex manifold, $s$ is a holomorphic map and it's a local biholomorphism near the diagonal in the second variable. But I believe it's really a topological problem..)
general-topology complex-geometry several-complex-variables
asked Jan 29 at 15:29
Chun GanChun Gan
291110
$endgroup$
Let $X$ be a Hausdorff locally compact topological space, $s:Xtimes Xto mathbb{R}$ be a continuous map such that $s(x,x)=0$ for all $xin X$. Set $triangle={(x,y): x=y, x,yin X}$ and $A={(x,y): s(x,y)=0}$, so $trianglesubset A$. If for each fixed $xin X$, $s(x,cdot)$ is a local homeomorphism near $y=x$ in the second variable, show that the set $Y=Abackslash triangle$ is closed in $Xtimes X$.
(I'm not certain the condition in the problem is sufficient, the real problem I met is that $X$ is a complex manifold and $mathbb{R}$ being replaced by another complex manifold, $s$ is a holomorphic map and it's a local biholomorphism near the diagonal in the second variable. But I believe it's really a topological problem..)
general-topology complex-geometry several-complex-variables
general-topology complex-geometry several-complex-variables
asked Jan 29 at 15:29
Chun GanChun Gan
291110
asked Jan 29 at 15:29
Chun GanChun Gan
291110
edited Jan 29 at 15:58
Chun Gan
asked Jan 29 at 15:29
Chun GanChun Gan
291110
asked Jan 29 at 15:29
Chun GanChun Gan
291110
asked Jan 29 at 15:29
Chun GanChun Gan
291110
291110
$begingroup$
How is "$s(x,cdot)$ is a local homeomorphism near $y=x$" defined precisely?
$endgroup$
– Henno Brandsma
Jan 29 at 17:28
$begingroup$
Fix $xin X$, then there exists an open neighborhood $Usubset X$ near $y=x$, and $s(x,cdot):Uto Vsubset mathbb{R}$ is a homeomorphism.
$endgroup$
– Chun Gan
Jan 29 at 17:30
add a comment |
$begingroup$
How is "$s(x,cdot)$ is a local homeomorphism near $y=x$" defined precisely?
$endgroup$
– Henno Brandsma
Jan 29 at 17:28
$begingroup$
Fix $xin X$, then there exists an open neighborhood $Usubset X$ near $y=x$, and $s(x,cdot):Uto Vsubset mathbb{R}$ is a homeomorphism.
$endgroup$
– Chun Gan
Jan 29 at 17:30
$begingroup$
How is "$s(x,cdot)$ is a local homeomorphism near $y=x$" defined precisely?
$endgroup$
– Henno Brandsma
Jan 29 at 17:28
$begingroup$
How is "$s(x,cdot)$ is a local homeomorphism near $y=x$" defined precisely?
$endgroup$
– Henno Brandsma
Jan 29 at 17:28
$begingroup$
Fix $xin X$, then there exists an open neighborhood $Usubset X$ near $y=x$, and $s(x,cdot):Uto Vsubset mathbb{R}$ is a homeomorphism.
$endgroup$
– Chun Gan
Jan 29 at 17:30
$begingroup$
Fix $xin X$, then there exists an open neighborhood $Usubset X$ near $y=x$, and $s(x,cdot):Uto Vsubset mathbb{R}$ is a homeomorphism.
$endgroup$
– Chun Gan
Jan 29 at 17:30
add a comment |
1 Answer
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$begingroup$
The statement is not valid. Let $X=mathbb{R}$ and let $s(x,y)=y(x-y)(x+y)=x^2y-y^3$ for all $x,yinmathbb{R}$. Clearly, $s$ is continuous. For a fixed $xinmathbb{R}$, the derivative of $s(x,y)$ with respect to $y$ is $-3y^2+x^2$. For $xneq 0$, this derivative is nonzero at $y=x$, so $s(x,cdot)$ is a local homeomorphism at $y=x$. For $x=0$, $s(0,y)=-y^3$, so again $s(0,cdot)$ is a homeomorphism. However, the set $AsetminusDelta={(x,y)inmathbb{R}^2!:y=-x wedge xneq 0}$ is not closed in $mathbb{R}^2$.
answered Jan 29 at 18:34
Peter EliasPeter Elias
938415
$endgroup$
$begingroup$
Hi Peter, thanks for the example! I think this did answer my question in the topological category. Now I have to try to use some complex analysis to attack the problem...
$endgroup$
– Chun Gan
Jan 29 at 20:17
add a comment |
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1 Answer
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$begingroup$
The statement is not valid. Let $X=mathbb{R}$ and let $s(x,y)=y(x-y)(x+y)=x^2y-y^3$ for all $x,yinmathbb{R}$. Clearly, $s$ is continuous. For a fixed $xinmathbb{R}$, the derivative of $s(x,y)$ with respect to $y$ is $-3y^2+x^2$. For $xneq 0$, this derivative is nonzero at $y=x$, so $s(x,cdot)$ is a local homeomorphism at $y=x$. For $x=0$, $s(0,y)=-y^3$, so again $s(0,cdot)$ is a homeomorphism. However, the set $AsetminusDelta={(x,y)inmathbb{R}^2!:y=-x wedge xneq 0}$ is not closed in $mathbb{R}^2$.
answered Jan 29 at 18:34
Peter EliasPeter Elias
938415
$endgroup$
$begingroup$
Hi Peter, thanks for the example! I think this did answer my question in the topological category. Now I have to try to use some complex analysis to attack the problem...
$endgroup$
– Chun Gan
Jan 29 at 20:17
add a comment |
$begingroup$
The statement is not valid. Let $X=mathbb{R}$ and let $s(x,y)=y(x-y)(x+y)=x^2y-y^3$ for all $x,yinmathbb{R}$. Clearly, $s$ is continuous. For a fixed $xinmathbb{R}$, the derivative of $s(x,y)$ with respect to $y$ is $-3y^2+x^2$. For $xneq 0$, this derivative is nonzero at $y=x$, so $s(x,cdot)$ is a local homeomorphism at $y=x$. For $x=0$, $s(0,y)=-y^3$, so again $s(0,cdot)$ is a homeomorphism. However, the set $AsetminusDelta={(x,y)inmathbb{R}^2!:y=-x wedge xneq 0}$ is not closed in $mathbb{R}^2$.
answered Jan 29 at 18:34
Peter EliasPeter Elias
938415
$endgroup$
$begingroup$
Hi Peter, thanks for the example! I think this did answer my question in the topological category. Now I have to try to use some complex analysis to attack the problem...
$endgroup$
– Chun Gan
Jan 29 at 20:17
add a comment |
$begingroup$
The statement is not valid. Let $X=mathbb{R}$ and let $s(x,y)=y(x-y)(x+y)=x^2y-y^3$ for all $x,yinmathbb{R}$. Clearly, $s$ is continuous. For a fixed $xinmathbb{R}$, the derivative of $s(x,y)$ with respect to $y$ is $-3y^2+x^2$. For $xneq 0$, this derivative is nonzero at $y=x$, so $s(x,cdot)$ is a local homeomorphism at $y=x$. For $x=0$, $s(0,y)=-y^3$, so again $s(0,cdot)$ is a homeomorphism. However, the set $AsetminusDelta={(x,y)inmathbb{R}^2!:y=-x wedge xneq 0}$ is not closed in $mathbb{R}^2$.
answered Jan 29 at 18:34
Peter EliasPeter Elias
938415
$endgroup$
The statement is not valid. Let $X=mathbb{R}$ and let $s(x,y)=y(x-y)(x+y)=x^2y-y^3$ for all $x,yinmathbb{R}$. Clearly, $s$ is continuous. For a fixed $xinmathbb{R}$, the derivative of $s(x,y)$ with respect to $y$ is $-3y^2+x^2$. For $xneq 0$, this derivative is nonzero at $y=x$, so $s(x,cdot)$ is a local homeomorphism at $y=x$. For $x=0$, $s(0,y)=-y^3$, so again $s(0,cdot)$ is a homeomorphism. However, the set $AsetminusDelta={(x,y)inmathbb{R}^2!:y=-x wedge xneq 0}$ is not closed in $mathbb{R}^2$.
answered Jan 29 at 18:34
Peter EliasPeter Elias
938415
edited Jan 29 at 18:53
answered Jan 29 at 18:34
Peter EliasPeter Elias
938415
answered Jan 29 at 18:34
Peter EliasPeter Elias
938415
answered Jan 29 at 18:34
Peter EliasPeter Elias
938415
938415
$begingroup$
Hi Peter, thanks for the example! I think this did answer my question in the topological category. Now I have to try to use some complex analysis to attack the problem...
$endgroup$
– Chun Gan
Jan 29 at 20:17
add a comment |
$begingroup$
Hi Peter, thanks for the example! I think this did answer my question in the topological category. Now I have to try to use some complex analysis to attack the problem...
$endgroup$
– Chun Gan
Jan 29 at 20:17
$begingroup$
Hi Peter, thanks for the example! I think this did answer my question in the topological category. Now I have to try to use some complex analysis to attack the problem...
$endgroup$
– Chun Gan
Jan 29 at 20:17
$begingroup$
Hi Peter, thanks for the example! I think this did answer my question in the topological category. Now I have to try to use some complex analysis to attack the problem...
$endgroup$
– Chun Gan
Jan 29 at 20:17
add a comment |
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$begingroup$
How is "$s(x,cdot)$ is a local homeomorphism near $y=x$" defined precisely?
$endgroup$
– Henno Brandsma
Jan 29 at 17:28
$begingroup$
Fix $xin X$, then there exists an open neighborhood $Usubset X$ near $y=x$, and $s(x,cdot):Uto Vsubset mathbb{R}$ is a homeomorphism.
$endgroup$
– Chun Gan
Jan 29 at 17:30