A contradictory result from a corollary with the fact that the quotient space of a Hausdorff space is not...
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It is known that a quotient space of a Hausdorff space is not necessarily Hausdorff; however, in the book of Topology by Munkres, at page 140, it is given that
But, we can always choose $Z = X$ and $g=i$ so that $g = i$ is a surjective continuous map. Hence, if $Z=X$ is Hausdorff, then by part $b$, $X^*$ must be Hausdorff, which is not true, as there are lots of counterexamples, by what is wrong with the above argument ?
general-topology
asked Dec 11 '18 at 5:51
onurcanbektasonurcanbektas
3,48911037
$endgroup$
add a comment |
$begingroup$
It is known that a quotient space of a Hausdorff space is not necessarily Hausdorff; however, in the book of Topology by Munkres, at page 140, it is given that
But, we can always choose $Z = X$ and $g=i$ so that $g = i$ is a surjective continuous map. Hence, if $Z=X$ is Hausdorff, then by part $b$, $X^*$ must be Hausdorff, which is not true, as there are lots of counterexamples, by what is wrong with the above argument ?
general-topology
asked Dec 11 '18 at 5:51
onurcanbektasonurcanbektas
3,48911037
$endgroup$
$begingroup$
In that case, $X^*$ is homemorphic to $X$ so it is Hausdorff
$endgroup$
– Dante Grevino
Dec 11 '18 at 5:55
$begingroup$
And by the existence of $f$, the topology in $X^*$ is finer than the initial topology with respecto to $f$, which is Hausdorff if $Z$ is Hausdorff. And we get $(b)$.
$endgroup$
– Dante Grevino
Dec 11 '18 at 6:00
add a comment |
$begingroup$
It is known that a quotient space of a Hausdorff space is not necessarily Hausdorff; however, in the book of Topology by Munkres, at page 140, it is given that
But, we can always choose $Z = X$ and $g=i$ so that $g = i$ is a surjective continuous map. Hence, if $Z=X$ is Hausdorff, then by part $b$, $X^*$ must be Hausdorff, which is not true, as there are lots of counterexamples, by what is wrong with the above argument ?
general-topology
asked Dec 11 '18 at 5:51
onurcanbektasonurcanbektas
3,48911037
$endgroup$
It is known that a quotient space of a Hausdorff space is not necessarily Hausdorff; however, in the book of Topology by Munkres, at page 140, it is given that
But, we can always choose $Z = X$ and $g=i$ so that $g = i$ is a surjective continuous map. Hence, if $Z=X$ is Hausdorff, then by part $b$, $X^*$ must be Hausdorff, which is not true, as there are lots of counterexamples, by what is wrong with the above argument ?
general-topology
general-topology
asked Dec 11 '18 at 5:51
onurcanbektasonurcanbektas
3,48911037
asked Dec 11 '18 at 5:51
onurcanbektasonurcanbektas
3,48911037
asked Dec 11 '18 at 5:51
onurcanbektasonurcanbektas
3,48911037
asked Dec 11 '18 at 5:51
onurcanbektasonurcanbektas
3,48911037
asked Dec 11 '18 at 5:51
onurcanbektasonurcanbektas
3,48911037
3,48911037
$begingroup$
In that case, $X^*$ is homemorphic to $X$ so it is Hausdorff
$endgroup$
– Dante Grevino
Dec 11 '18 at 5:55
$begingroup$
And by the existence of $f$, the topology in $X^*$ is finer than the initial topology with respecto to $f$, which is Hausdorff if $Z$ is Hausdorff. And we get $(b)$.
$endgroup$
– Dante Grevino
Dec 11 '18 at 6:00
add a comment |
$begingroup$
In that case, $X^*$ is homemorphic to $X$ so it is Hausdorff
$endgroup$
– Dante Grevino
Dec 11 '18 at 5:55
$begingroup$
And by the existence of $f$, the topology in $X^*$ is finer than the initial topology with respecto to $f$, which is Hausdorff if $Z$ is Hausdorff. And we get $(b)$.
$endgroup$
– Dante Grevino
Dec 11 '18 at 6:00
$begingroup$
In that case, $X^*$ is homemorphic to $X$ so it is Hausdorff
$endgroup$
– Dante Grevino
Dec 11 '18 at 5:55
$begingroup$
In that case, $X^*$ is homemorphic to $X$ so it is Hausdorff
$endgroup$
– Dante Grevino
Dec 11 '18 at 5:55
$begingroup$
And by the existence of $f$, the topology in $X^*$ is finer than the initial topology with respecto to $f$, which is Hausdorff if $Z$ is Hausdorff. And we get $(b)$.
$endgroup$
– Dante Grevino
Dec 11 '18 at 6:00
$begingroup$
And by the existence of $f$, the topology in $X^*$ is finer than the initial topology with respecto to $f$, which is Hausdorff if $Z$ is Hausdorff. And we get $(b)$.
$endgroup$
– Dante Grevino
Dec 11 '18 at 6:00
add a comment |
1 Answer
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If $Z=X$ and $g=id$ then
$$X^*={ g^{-1}(z) | z in Z }=X$$
I think that you are confusing $X^*$ with some quotient of $X$, which is not the case in this corollary.
answered Dec 11 '18 at 5:56
N. S.N. S.
105k7114210
$endgroup$
$begingroup$
Yeah, I assumed $X^*$ was some quotient space of $X$; didn't pay attention to the definition of $X^*$ there. Thanks for the answer.
$endgroup$
– onurcanbektas
Dec 11 '18 at 6:08
add a comment |
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1 Answer
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1 Answer
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$begingroup$
If $Z=X$ and $g=id$ then
$$X^*={ g^{-1}(z) | z in Z }=X$$
I think that you are confusing $X^*$ with some quotient of $X$, which is not the case in this corollary.
answered Dec 11 '18 at 5:56
N. S.N. S.
105k7114210
$endgroup$
$begingroup$
Yeah, I assumed $X^*$ was some quotient space of $X$; didn't pay attention to the definition of $X^*$ there. Thanks for the answer.
$endgroup$
– onurcanbektas
Dec 11 '18 at 6:08
add a comment |
$begingroup$
If $Z=X$ and $g=id$ then
$$X^*={ g^{-1}(z) | z in Z }=X$$
I think that you are confusing $X^*$ with some quotient of $X$, which is not the case in this corollary.
answered Dec 11 '18 at 5:56
N. S.N. S.
105k7114210
$endgroup$
$begingroup$
Yeah, I assumed $X^*$ was some quotient space of $X$; didn't pay attention to the definition of $X^*$ there. Thanks for the answer.
$endgroup$
– onurcanbektas
Dec 11 '18 at 6:08
add a comment |
$begingroup$
If $Z=X$ and $g=id$ then
$$X^*={ g^{-1}(z) | z in Z }=X$$
I think that you are confusing $X^*$ with some quotient of $X$, which is not the case in this corollary.
answered Dec 11 '18 at 5:56
N. S.N. S.
105k7114210
$endgroup$
If $Z=X$ and $g=id$ then
$$X^*={ g^{-1}(z) | z in Z }=X$$
I think that you are confusing $X^*$ with some quotient of $X$, which is not the case in this corollary.
answered Dec 11 '18 at 5:56
N. S.N. S.
105k7114210
answered Dec 11 '18 at 5:56
N. S.N. S.
105k7114210
answered Dec 11 '18 at 5:56
N. S.N. S.
105k7114210
answered Dec 11 '18 at 5:56
N. S.N. S.
105k7114210
105k7114210
$begingroup$
Yeah, I assumed $X^*$ was some quotient space of $X$; didn't pay attention to the definition of $X^*$ there. Thanks for the answer.
$endgroup$
– onurcanbektas
Dec 11 '18 at 6:08
add a comment |
$begingroup$
Yeah, I assumed $X^*$ was some quotient space of $X$; didn't pay attention to the definition of $X^*$ there. Thanks for the answer.
$endgroup$
– onurcanbektas
Dec 11 '18 at 6:08
$begingroup$
Yeah, I assumed $X^*$ was some quotient space of $X$; didn't pay attention to the definition of $X^*$ there. Thanks for the answer.
$endgroup$
– onurcanbektas
Dec 11 '18 at 6:08
$begingroup$
Yeah, I assumed $X^*$ was some quotient space of $X$; didn't pay attention to the definition of $X^*$ there. Thanks for the answer.
$endgroup$
– onurcanbektas
Dec 11 '18 at 6:08
add a comment |
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$begingroup$
In that case, $X^*$ is homemorphic to $X$ so it is Hausdorff
$endgroup$
– Dante Grevino
Dec 11 '18 at 5:55
$begingroup$
And by the existence of $f$, the topology in $X^*$ is finer than the initial topology with respecto to $f$, which is Hausdorff if $Z$ is Hausdorff. And we get $(b)$.
$endgroup$
– Dante Grevino
Dec 11 '18 at 6:00