If X is Hausdorff and K⊂X is compact then the contraction X/K is Hausdorff
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I found this claim as an answer to one problem. I tried to prove it in the following way: $X$ is Hausdorff so for every $x$ that does not lie in $K$ compact i can find two disjont open set A,B such that $x∈A$ and $K⊂B$.
[As the space is $T2$ both the point and the compact subspace are closed in $X$].
$A$ and $B$ are saturated so their images through the projection are the disjoint open sets that let $X/K$ be Hausdorff.
Is this proof correct? If not where does it fail?
Thanks in advance
general-topology
asked Dec 16 '18 at 12:27
SMCSMC
6010
$endgroup$
add a comment |
$begingroup$
I found this claim as an answer to one problem. I tried to prove it in the following way: $X$ is Hausdorff so for every $x$ that does not lie in $K$ compact i can find two disjont open set A,B such that $x∈A$ and $K⊂B$.
[As the space is $T2$ both the point and the compact subspace are closed in $X$].
$A$ and $B$ are saturated so their images through the projection are the disjoint open sets that let $X/K$ be Hausdorff.
Is this proof correct? If not where does it fail?
Thanks in advance
general-topology
asked Dec 16 '18 at 12:27
SMCSMC
6010
$endgroup$
1
$begingroup$
Seems correct to me. You do know the proof that for any point and any compact set in a Hausdorff space, they can be separated by open neighborhoods?
$endgroup$
– SmileyCraft
Dec 16 '18 at 12:56
$begingroup$
Also, you may have proved that any point is separable by open neighborhoods from the contracted point, but any two points must be separable by open neighborhoods. This should be an easy exercise, given what you have already shown.
$endgroup$
– SmileyCraft
Dec 16 '18 at 12:58
$begingroup$
Oh yes I know but I thought it wasn't key as I only need to consider the case in which the intersection between one of the neighborhoods and K is not empty.
$endgroup$
– SMC
Dec 16 '18 at 13:25
add a comment |
$begingroup$
I found this claim as an answer to one problem. I tried to prove it in the following way: $X$ is Hausdorff so for every $x$ that does not lie in $K$ compact i can find two disjont open set A,B such that $x∈A$ and $K⊂B$.
[As the space is $T2$ both the point and the compact subspace are closed in $X$].
$A$ and $B$ are saturated so their images through the projection are the disjoint open sets that let $X/K$ be Hausdorff.
Is this proof correct? If not where does it fail?
Thanks in advance
general-topology
asked Dec 16 '18 at 12:27
SMCSMC
6010
$endgroup$
I found this claim as an answer to one problem. I tried to prove it in the following way: $X$ is Hausdorff so for every $x$ that does not lie in $K$ compact i can find two disjont open set A,B such that $x∈A$ and $K⊂B$.
[As the space is $T2$ both the point and the compact subspace are closed in $X$].
$A$ and $B$ are saturated so their images through the projection are the disjoint open sets that let $X/K$ be Hausdorff.
Is this proof correct? If not where does it fail?
Thanks in advance
general-topology
general-topology
asked Dec 16 '18 at 12:27
SMCSMC
6010
asked Dec 16 '18 at 12:27
SMCSMC
6010
edited Dec 16 '18 at 16:29
SMC
asked Dec 16 '18 at 12:27
SMCSMC
6010
asked Dec 16 '18 at 12:27
SMCSMC
6010
asked Dec 16 '18 at 12:27
SMCSMC
6010
6010
1
$begingroup$
Seems correct to me. You do know the proof that for any point and any compact set in a Hausdorff space, they can be separated by open neighborhoods?
$endgroup$
– SmileyCraft
Dec 16 '18 at 12:56
$begingroup$
Also, you may have proved that any point is separable by open neighborhoods from the contracted point, but any two points must be separable by open neighborhoods. This should be an easy exercise, given what you have already shown.
$endgroup$
– SmileyCraft
Dec 16 '18 at 12:58
$begingroup$
Oh yes I know but I thought it wasn't key as I only need to consider the case in which the intersection between one of the neighborhoods and K is not empty.
$endgroup$
– SMC
Dec 16 '18 at 13:25
add a comment |
1
$begingroup$
Seems correct to me. You do know the proof that for any point and any compact set in a Hausdorff space, they can be separated by open neighborhoods?
$endgroup$
– SmileyCraft
Dec 16 '18 at 12:56
$begingroup$
Also, you may have proved that any point is separable by open neighborhoods from the contracted point, but any two points must be separable by open neighborhoods. This should be an easy exercise, given what you have already shown.
$endgroup$
– SmileyCraft
Dec 16 '18 at 12:58
$begingroup$
Oh yes I know but I thought it wasn't key as I only need to consider the case in which the intersection between one of the neighborhoods and K is not empty.
$endgroup$
– SMC
Dec 16 '18 at 13:25
1
1
$begingroup$
Seems correct to me. You do know the proof that for any point and any compact set in a Hausdorff space, they can be separated by open neighborhoods?
$endgroup$
– SmileyCraft
Dec 16 '18 at 12:56
$begingroup$
Seems correct to me. You do know the proof that for any point and any compact set in a Hausdorff space, they can be separated by open neighborhoods?
$endgroup$
– SmileyCraft
Dec 16 '18 at 12:56
$begingroup$
Also, you may have proved that any point is separable by open neighborhoods from the contracted point, but any two points must be separable by open neighborhoods. This should be an easy exercise, given what you have already shown.
$endgroup$
– SmileyCraft
Dec 16 '18 at 12:58
$begingroup$
Also, you may have proved that any point is separable by open neighborhoods from the contracted point, but any two points must be separable by open neighborhoods. This should be an easy exercise, given what you have already shown.
$endgroup$
– SmileyCraft
Dec 16 '18 at 12:58
$begingroup$
Oh yes I know but I thought it wasn't key as I only need to consider the case in which the intersection between one of the neighborhoods and K is not empty.
$endgroup$
– SMC
Dec 16 '18 at 13:25
$begingroup$
Oh yes I know but I thought it wasn't key as I only need to consider the case in which the intersection between one of the neighborhoods and K is not empty.
$endgroup$
– SMC
Dec 16 '18 at 13:25
add a comment |
1 Answer
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The proof is good to show that a point $xin Xsetminus K$ and $hat{K}$ (the contraction of $K$) are separated in $X/K$.
However this doesn't fully suffice. If $x,yin Xsetminus K$ and $xne y$ you have to prove that they can be separated by disjoint open sets.
Choose $A$ and $B$ as before for $x$, but also $A'$ and $B'$ for $y$ in the same way. Since $X$ is Hausdorff, you can conclude. Fill in the details.
How do you prove the key fact? For every $zin K$ choose $U_z$ and $V_z$ disjoint open sets containing $x$ and $z$ respectively. Then $(V_z)_{zin K}$ is an open cover of $K$ and so there are $z_1,dots,z_n$ such that $B=V_{z_1}cupdotscup V_{z_n}supseteq K$. Take $A=U_{z_1}capdotscap U_{z_n}$.
answered Dec 16 '18 at 14:27
egregegreg
182k1485203
$endgroup$
$begingroup$
Yes I thought that once you show it works for any point in outside K and K it follows easily for the other points. For the proof of the key fact i had already worked it out.
$endgroup$
– SMC
Dec 16 '18 at 15:27
add a comment |
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$begingroup$
The proof is good to show that a point $xin Xsetminus K$ and $hat{K}$ (the contraction of $K$) are separated in $X/K$.
However this doesn't fully suffice. If $x,yin Xsetminus K$ and $xne y$ you have to prove that they can be separated by disjoint open sets.
Choose $A$ and $B$ as before for $x$, but also $A'$ and $B'$ for $y$ in the same way. Since $X$ is Hausdorff, you can conclude. Fill in the details.
How do you prove the key fact? For every $zin K$ choose $U_z$ and $V_z$ disjoint open sets containing $x$ and $z$ respectively. Then $(V_z)_{zin K}$ is an open cover of $K$ and so there are $z_1,dots,z_n$ such that $B=V_{z_1}cupdotscup V_{z_n}supseteq K$. Take $A=U_{z_1}capdotscap U_{z_n}$.
answered Dec 16 '18 at 14:27
egregegreg
182k1485203
$endgroup$
$begingroup$
Yes I thought that once you show it works for any point in outside K and K it follows easily for the other points. For the proof of the key fact i had already worked it out.
$endgroup$
– SMC
Dec 16 '18 at 15:27
add a comment |
$begingroup$
The proof is good to show that a point $xin Xsetminus K$ and $hat{K}$ (the contraction of $K$) are separated in $X/K$.
However this doesn't fully suffice. If $x,yin Xsetminus K$ and $xne y$ you have to prove that they can be separated by disjoint open sets.
Choose $A$ and $B$ as before for $x$, but also $A'$ and $B'$ for $y$ in the same way. Since $X$ is Hausdorff, you can conclude. Fill in the details.
How do you prove the key fact? For every $zin K$ choose $U_z$ and $V_z$ disjoint open sets containing $x$ and $z$ respectively. Then $(V_z)_{zin K}$ is an open cover of $K$ and so there are $z_1,dots,z_n$ such that $B=V_{z_1}cupdotscup V_{z_n}supseteq K$. Take $A=U_{z_1}capdotscap U_{z_n}$.
answered Dec 16 '18 at 14:27
egregegreg
182k1485203
$endgroup$
$begingroup$
Yes I thought that once you show it works for any point in outside K and K it follows easily for the other points. For the proof of the key fact i had already worked it out.
$endgroup$
– SMC
Dec 16 '18 at 15:27
add a comment |
$begingroup$
The proof is good to show that a point $xin Xsetminus K$ and $hat{K}$ (the contraction of $K$) are separated in $X/K$.
However this doesn't fully suffice. If $x,yin Xsetminus K$ and $xne y$ you have to prove that they can be separated by disjoint open sets.
Choose $A$ and $B$ as before for $x$, but also $A'$ and $B'$ for $y$ in the same way. Since $X$ is Hausdorff, you can conclude. Fill in the details.
How do you prove the key fact? For every $zin K$ choose $U_z$ and $V_z$ disjoint open sets containing $x$ and $z$ respectively. Then $(V_z)_{zin K}$ is an open cover of $K$ and so there are $z_1,dots,z_n$ such that $B=V_{z_1}cupdotscup V_{z_n}supseteq K$. Take $A=U_{z_1}capdotscap U_{z_n}$.
answered Dec 16 '18 at 14:27
egregegreg
182k1485203
$endgroup$
The proof is good to show that a point $xin Xsetminus K$ and $hat{K}$ (the contraction of $K$) are separated in $X/K$.
However this doesn't fully suffice. If $x,yin Xsetminus K$ and $xne y$ you have to prove that they can be separated by disjoint open sets.
Choose $A$ and $B$ as before for $x$, but also $A'$ and $B'$ for $y$ in the same way. Since $X$ is Hausdorff, you can conclude. Fill in the details.
How do you prove the key fact? For every $zin K$ choose $U_z$ and $V_z$ disjoint open sets containing $x$ and $z$ respectively. Then $(V_z)_{zin K}$ is an open cover of $K$ and so there are $z_1,dots,z_n$ such that $B=V_{z_1}cupdotscup V_{z_n}supseteq K$. Take $A=U_{z_1}capdotscap U_{z_n}$.
answered Dec 16 '18 at 14:27
egregegreg
182k1485203
answered Dec 16 '18 at 14:27
egregegreg
182k1485203
answered Dec 16 '18 at 14:27
egregegreg
182k1485203
answered Dec 16 '18 at 14:27
egregegreg
182k1485203
182k1485203
$begingroup$
Yes I thought that once you show it works for any point in outside K and K it follows easily for the other points. For the proof of the key fact i had already worked it out.
$endgroup$
– SMC
Dec 16 '18 at 15:27
add a comment |
$begingroup$
Yes I thought that once you show it works for any point in outside K and K it follows easily for the other points. For the proof of the key fact i had already worked it out.
$endgroup$
– SMC
Dec 16 '18 at 15:27
$begingroup$
Yes I thought that once you show it works for any point in outside K and K it follows easily for the other points. For the proof of the key fact i had already worked it out.
$endgroup$
– SMC
Dec 16 '18 at 15:27
$begingroup$
Yes I thought that once you show it works for any point in outside K and K it follows easily for the other points. For the proof of the key fact i had already worked it out.
$endgroup$
– SMC
Dec 16 '18 at 15:27
add a comment |
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$begingroup$
Seems correct to me. You do know the proof that for any point and any compact set in a Hausdorff space, they can be separated by open neighborhoods?
$endgroup$
– SmileyCraft
Dec 16 '18 at 12:56
$begingroup$
Also, you may have proved that any point is separable by open neighborhoods from the contracted point, but any two points must be separable by open neighborhoods. This should be an easy exercise, given what you have already shown.
$endgroup$
– SmileyCraft
Dec 16 '18 at 12:58
$begingroup$
Oh yes I know but I thought it wasn't key as I only need to consider the case in which the intersection between one of the neighborhoods and K is not empty.
$endgroup$
– SMC
Dec 16 '18 at 13:25