Quotient of $R^2$
$begingroup$
Consider $X$=$R^2$ and let A $in X$ be A={$(-1,0),(1,0)$}. Consider now the quotient space ${X}/{A}$.
Is it Hausdorff? Connected? Compact?
My answer: the projection from $X$ to $A$ is surjective and continuous, $X$ is connected and so is ${X}/{A}$.
The projection of the open cover of $X$ made up of ball centered at $(0,0)$ is an open cover of $X/A$ as they are all both open and saturated. We can not extract any finite subcover and because of that ${X}/{A}$ is not compact.
Honestly I'm not sure about the proof about compactness and I'm left stuck with the Hausdorff question. If I could demonstrate that the projection is open I could work from there. Working with the intervals left me with a proof that I'm not 100% sure about.
Thanks in advance for any help.
general-topology
asked Dec 15 '18 at 15:21
SMCSMC
6010
$endgroup$
add a comment |
$begingroup$
Consider $X$=$R^2$ and let A $in X$ be A={$(-1,0),(1,0)$}. Consider now the quotient space ${X}/{A}$.
Is it Hausdorff? Connected? Compact?
My answer: the projection from $X$ to $A$ is surjective and continuous, $X$ is connected and so is ${X}/{A}$.
The projection of the open cover of $X$ made up of ball centered at $(0,0)$ is an open cover of $X/A$ as they are all both open and saturated. We can not extract any finite subcover and because of that ${X}/{A}$ is not compact.
Honestly I'm not sure about the proof about compactness and I'm left stuck with the Hausdorff question. If I could demonstrate that the projection is open I could work from there. Working with the intervals left me with a proof that I'm not 100% sure about.
Thanks in advance for any help.
general-topology
asked Dec 15 '18 at 15:21
SMCSMC
6010
$endgroup$
add a comment |
$begingroup$
Consider $X$=$R^2$ and let A $in X$ be A={$(-1,0),(1,0)$}. Consider now the quotient space ${X}/{A}$.
Is it Hausdorff? Connected? Compact?
My answer: the projection from $X$ to $A$ is surjective and continuous, $X$ is connected and so is ${X}/{A}$.
The projection of the open cover of $X$ made up of ball centered at $(0,0)$ is an open cover of $X/A$ as they are all both open and saturated. We can not extract any finite subcover and because of that ${X}/{A}$ is not compact.
Honestly I'm not sure about the proof about compactness and I'm left stuck with the Hausdorff question. If I could demonstrate that the projection is open I could work from there. Working with the intervals left me with a proof that I'm not 100% sure about.
Thanks in advance for any help.
general-topology
asked Dec 15 '18 at 15:21
SMCSMC
6010
$endgroup$
Consider $X$=$R^2$ and let A $in X$ be A={$(-1,0),(1,0)$}. Consider now the quotient space ${X}/{A}$.
Is it Hausdorff? Connected? Compact?
My answer: the projection from $X$ to $A$ is surjective and continuous, $X$ is connected and so is ${X}/{A}$.
The projection of the open cover of $X$ made up of ball centered at $(0,0)$ is an open cover of $X/A$ as they are all both open and saturated. We can not extract any finite subcover and because of that ${X}/{A}$ is not compact.
Honestly I'm not sure about the proof about compactness and I'm left stuck with the Hausdorff question. If I could demonstrate that the projection is open I could work from there. Working with the intervals left me with a proof that I'm not 100% sure about.
Thanks in advance for any help.
general-topology
general-topology
asked Dec 15 '18 at 15:21
SMCSMC
6010
asked Dec 15 '18 at 15:21
SMCSMC
6010
asked Dec 15 '18 at 15:21
SMCSMC
6010
asked Dec 15 '18 at 15:21
SMCSMC
6010
asked Dec 15 '18 at 15:21
SMCSMC
6010
6010
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
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Your proof of noncompactness is fine. For the Hausdorff condition, there's really only one unusual point in this space. For a direct proof, you just have to separate this point from points outside.
answered Dec 15 '18 at 15:38
Matt SamuelMatt Samuel
38.3k63768
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add a comment |
$begingroup$
Hint: Prove that, in general, if $X$ is Hausdorff and $K subset X$ is compact then the contraction $X/K$ is Hausdorff
answered Dec 15 '18 at 15:38
smantismanti
134
$endgroup$
$begingroup$
This is overkill on my opinion.
$endgroup$
– Matt Samuel
Dec 15 '18 at 15:39
$begingroup$
Maybe it is, but it's useful.
$endgroup$
– smanti
Dec 15 '18 at 15:42
$begingroup$
$X$ is Hausdorff so for every $x$ that does not lie in $K$ i can find two disjont open set $A,B$ such that $x in A$ and $K ⊂ B$ both A and B are saturated so their images through the projection are the disjoint open sets that let $X/K$ be Hausdorff.
$endgroup$
– SMC
Dec 15 '18 at 16:31
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your proof of noncompactness is fine. For the Hausdorff condition, there's really only one unusual point in this space. For a direct proof, you just have to separate this point from points outside.
answered Dec 15 '18 at 15:38
Matt SamuelMatt Samuel
38.3k63768
$endgroup$
add a comment |
$begingroup$
Your proof of noncompactness is fine. For the Hausdorff condition, there's really only one unusual point in this space. For a direct proof, you just have to separate this point from points outside.
answered Dec 15 '18 at 15:38
Matt SamuelMatt Samuel
38.3k63768
$endgroup$
add a comment |
$begingroup$
Your proof of noncompactness is fine. For the Hausdorff condition, there's really only one unusual point in this space. For a direct proof, you just have to separate this point from points outside.
answered Dec 15 '18 at 15:38
Matt SamuelMatt Samuel
38.3k63768
$endgroup$
Your proof of noncompactness is fine. For the Hausdorff condition, there's really only one unusual point in this space. For a direct proof, you just have to separate this point from points outside.
answered Dec 15 '18 at 15:38
Matt SamuelMatt Samuel
38.3k63768
answered Dec 15 '18 at 15:38
Matt SamuelMatt Samuel
38.3k63768
answered Dec 15 '18 at 15:38
Matt SamuelMatt Samuel
38.3k63768
answered Dec 15 '18 at 15:38
Matt SamuelMatt Samuel
38.3k63768
38.3k63768
add a comment |
add a comment |
$begingroup$
Hint: Prove that, in general, if $X$ is Hausdorff and $K subset X$ is compact then the contraction $X/K$ is Hausdorff
answered Dec 15 '18 at 15:38
smantismanti
134
$endgroup$
$begingroup$
This is overkill on my opinion.
$endgroup$
– Matt Samuel
Dec 15 '18 at 15:39
$begingroup$
Maybe it is, but it's useful.
$endgroup$
– smanti
Dec 15 '18 at 15:42
$begingroup$
$X$ is Hausdorff so for every $x$ that does not lie in $K$ i can find two disjont open set $A,B$ such that $x in A$ and $K ⊂ B$ both A and B are saturated so their images through the projection are the disjoint open sets that let $X/K$ be Hausdorff.
$endgroup$
– SMC
Dec 15 '18 at 16:31
add a comment |
$begingroup$
Hint: Prove that, in general, if $X$ is Hausdorff and $K subset X$ is compact then the contraction $X/K$ is Hausdorff
answered Dec 15 '18 at 15:38
smantismanti
134
$endgroup$
$begingroup$
This is overkill on my opinion.
$endgroup$
– Matt Samuel
Dec 15 '18 at 15:39
$begingroup$
Maybe it is, but it's useful.
$endgroup$
– smanti
Dec 15 '18 at 15:42
$begingroup$
$X$ is Hausdorff so for every $x$ that does not lie in $K$ i can find two disjont open set $A,B$ such that $x in A$ and $K ⊂ B$ both A and B are saturated so their images through the projection are the disjoint open sets that let $X/K$ be Hausdorff.
$endgroup$
– SMC
Dec 15 '18 at 16:31
add a comment |
$begingroup$
Hint: Prove that, in general, if $X$ is Hausdorff and $K subset X$ is compact then the contraction $X/K$ is Hausdorff
answered Dec 15 '18 at 15:38
smantismanti
134
$endgroup$
Hint: Prove that, in general, if $X$ is Hausdorff and $K subset X$ is compact then the contraction $X/K$ is Hausdorff
answered Dec 15 '18 at 15:38
smantismanti
134
edited Dec 15 '18 at 15:39
answered Dec 15 '18 at 15:38
smantismanti
134
answered Dec 15 '18 at 15:38
smantismanti
134
answered Dec 15 '18 at 15:38
smantismanti
134
134
$begingroup$
This is overkill on my opinion.
$endgroup$
– Matt Samuel
Dec 15 '18 at 15:39
$begingroup$
Maybe it is, but it's useful.
$endgroup$
– smanti
Dec 15 '18 at 15:42
$begingroup$
$X$ is Hausdorff so for every $x$ that does not lie in $K$ i can find two disjont open set $A,B$ such that $x in A$ and $K ⊂ B$ both A and B are saturated so their images through the projection are the disjoint open sets that let $X/K$ be Hausdorff.
$endgroup$
– SMC
Dec 15 '18 at 16:31
add a comment |
$begingroup$
This is overkill on my opinion.
$endgroup$
– Matt Samuel
Dec 15 '18 at 15:39
$begingroup$
Maybe it is, but it's useful.
$endgroup$
– smanti
Dec 15 '18 at 15:42
$begingroup$
$X$ is Hausdorff so for every $x$ that does not lie in $K$ i can find two disjont open set $A,B$ such that $x in A$ and $K ⊂ B$ both A and B are saturated so their images through the projection are the disjoint open sets that let $X/K$ be Hausdorff.
$endgroup$
– SMC
Dec 15 '18 at 16:31
$begingroup$
This is overkill on my opinion.
$endgroup$
– Matt Samuel
Dec 15 '18 at 15:39
$begingroup$
This is overkill on my opinion.
$endgroup$
– Matt Samuel
Dec 15 '18 at 15:39
$begingroup$
Maybe it is, but it's useful.
$endgroup$
– smanti
Dec 15 '18 at 15:42
$begingroup$
Maybe it is, but it's useful.
$endgroup$
– smanti
Dec 15 '18 at 15:42
$begingroup$
$X$ is Hausdorff so for every $x$ that does not lie in $K$ i can find two disjont open set $A,B$ such that $x in A$ and $K ⊂ B$ both A and B are saturated so their images through the projection are the disjoint open sets that let $X/K$ be Hausdorff.
$endgroup$
– SMC
Dec 15 '18 at 16:31
$begingroup$
$X$ is Hausdorff so for every $x$ that does not lie in $K$ i can find two disjont open set $A,B$ such that $x in A$ and $K ⊂ B$ both A and B are saturated so their images through the projection are the disjoint open sets that let $X/K$ be Hausdorff.
$endgroup$
– SMC
Dec 15 '18 at 16:31
add a comment |
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