Are subspaces of paracompact spaces normal?
$begingroup$
Are all subspaces of a paracompact space normal?
This is what I think about this question...
First a paracompact Hausdorff space turns out to be Normal, second the paracompact property is not hereditary, meaning any subspace need to be closed+Hausdorff to be normal. So I think the answer is no, but I have to find a counter-example.
Am I right about this and what counter-example do we have?
general-topology examples-counterexamples separation-axioms paracompactness
edited Dec 9 '18 at 3:22
Eric Wofsey
184k13212338
asked Dec 4 '14 at 3:25
BirkaryBirkary
397
$endgroup$
add a comment |
$begingroup$
Are all subspaces of a paracompact space normal?
This is what I think about this question...
First a paracompact Hausdorff space turns out to be Normal, second the paracompact property is not hereditary, meaning any subspace need to be closed+Hausdorff to be normal. So I think the answer is no, but I have to find a counter-example.
Am I right about this and what counter-example do we have?
general-topology examples-counterexamples separation-axioms paracompactness
edited Dec 9 '18 at 3:22
Eric Wofsey
184k13212338
asked Dec 4 '14 at 3:25
BirkaryBirkary
397
$endgroup$
add a comment |
$begingroup$
Are all subspaces of a paracompact space normal?
This is what I think about this question...
First a paracompact Hausdorff space turns out to be Normal, second the paracompact property is not hereditary, meaning any subspace need to be closed+Hausdorff to be normal. So I think the answer is no, but I have to find a counter-example.
Am I right about this and what counter-example do we have?
general-topology examples-counterexamples separation-axioms paracompactness
edited Dec 9 '18 at 3:22
Eric Wofsey
184k13212338
asked Dec 4 '14 at 3:25
BirkaryBirkary
397
$endgroup$
Are all subspaces of a paracompact space normal?
This is what I think about this question...
First a paracompact Hausdorff space turns out to be Normal, second the paracompact property is not hereditary, meaning any subspace need to be closed+Hausdorff to be normal. So I think the answer is no, but I have to find a counter-example.
Am I right about this and what counter-example do we have?
general-topology examples-counterexamples separation-axioms paracompactness
general-topology examples-counterexamples separation-axioms paracompactness
edited Dec 9 '18 at 3:22
Eric Wofsey
184k13212338
asked Dec 4 '14 at 3:25
BirkaryBirkary
397
edited Dec 9 '18 at 3:22
Eric Wofsey
184k13212338
asked Dec 4 '14 at 3:25
BirkaryBirkary
397
edited Dec 9 '18 at 3:22
Eric Wofsey
184k13212338
edited Dec 9 '18 at 3:22
Eric Wofsey
184k13212338
edited Dec 9 '18 at 3:22
Eric Wofsey
184k13212338
184k13212338
asked Dec 4 '14 at 3:25
BirkaryBirkary
397
asked Dec 4 '14 at 3:25
BirkaryBirkary
397
asked Dec 4 '14 at 3:25
BirkaryBirkary
397
397
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Every compact space is paracompact. Let $X$ be the one-point compactification of a countably infinite discrete space $X_0$, and let $Y$ be the one-point compactification of an uncountable discrete space $Y_0$. Let $p$ be the non-isolated point of $X$ and $q$ the non-isolated point of $Y$. Then $Xtimes Y$ is compact, so it’s paracompact. However, the subspace
$$Z=(Xtimes Y)setminus{langle p,qrangle}$$
is not even normal, so it’s certainly not paracompact. Specifically, the closed sets $H=X_0times{q}$ and $K={p}times Y_0$ cannot be separated by disjoint open sets.
To see this, suppose that $U$ is an open nbhd of $H$. For each $xin X_0$ there is a finite $F_xsubseteq Y_0$ such that $langle x,qranglein{x}times(Ysetminus F_x)subseteq U$. Let $F=bigcup_{xin X_0}F_x$; then $F$ is countable, so there is a $yin Y_0setminus F$. But then $Xtimes{y}subseteq U$, so no open nbhd of $langle p,yrangle$ is disjoint from $U$, and therefore no open nbhd of $K$ is disjoint from $U$.
answered Dec 4 '14 at 3:36
Brian M. ScottBrian M. Scott
456k38509909
$endgroup$
$begingroup$
your answer was clear,just wonder if we add the closeness to our subspace in this case we will get the paracompact for free, but are still we need Hausdorff to be normal ( the subspace ).
$endgroup$
– Birkary
Dec 4 '14 at 3:58
$begingroup$
@Birkary: Yes, if you restrict yourself to closed subspaces you get paracompactness for free, but you don’t necessarily get normality unless the subspace is Hausdorff.
$endgroup$
– Brian M. Scott
Dec 4 '14 at 4:06
$begingroup$
thanks,this was very helpful. I will make sure to understand the answer you give.
$endgroup$
– Birkary
Dec 4 '14 at 4:08
$begingroup$
@Birkary: You’re welcome.
$endgroup$
– Brian M. Scott
Dec 4 '14 at 4:09
$begingroup$
More generally, any non-normal space has a compactification $id_X:Xto cX.$ So $X$ is a non-paracompact sub-space of the compact Hausdorff space $cX.$....We can also say this if $X$ is normal but not paracompact. For example if $X$ is $omega_1$ with the $in$-order topology, and $cX=omega_1+1$.
$endgroup$
– DanielWainfleet
Dec 9 '18 at 3:36
add a comment |
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1 Answer
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$begingroup$
Every compact space is paracompact. Let $X$ be the one-point compactification of a countably infinite discrete space $X_0$, and let $Y$ be the one-point compactification of an uncountable discrete space $Y_0$. Let $p$ be the non-isolated point of $X$ and $q$ the non-isolated point of $Y$. Then $Xtimes Y$ is compact, so it’s paracompact. However, the subspace
$$Z=(Xtimes Y)setminus{langle p,qrangle}$$
is not even normal, so it’s certainly not paracompact. Specifically, the closed sets $H=X_0times{q}$ and $K={p}times Y_0$ cannot be separated by disjoint open sets.
To see this, suppose that $U$ is an open nbhd of $H$. For each $xin X_0$ there is a finite $F_xsubseteq Y_0$ such that $langle x,qranglein{x}times(Ysetminus F_x)subseteq U$. Let $F=bigcup_{xin X_0}F_x$; then $F$ is countable, so there is a $yin Y_0setminus F$. But then $Xtimes{y}subseteq U$, so no open nbhd of $langle p,yrangle$ is disjoint from $U$, and therefore no open nbhd of $K$ is disjoint from $U$.
answered Dec 4 '14 at 3:36
Brian M. ScottBrian M. Scott
456k38509909
$endgroup$
$begingroup$
your answer was clear,just wonder if we add the closeness to our subspace in this case we will get the paracompact for free, but are still we need Hausdorff to be normal ( the subspace ).
$endgroup$
– Birkary
Dec 4 '14 at 3:58
$begingroup$
@Birkary: Yes, if you restrict yourself to closed subspaces you get paracompactness for free, but you don’t necessarily get normality unless the subspace is Hausdorff.
$endgroup$
– Brian M. Scott
Dec 4 '14 at 4:06
$begingroup$
thanks,this was very helpful. I will make sure to understand the answer you give.
$endgroup$
– Birkary
Dec 4 '14 at 4:08
$begingroup$
@Birkary: You’re welcome.
$endgroup$
– Brian M. Scott
Dec 4 '14 at 4:09
$begingroup$
More generally, any non-normal space has a compactification $id_X:Xto cX.$ So $X$ is a non-paracompact sub-space of the compact Hausdorff space $cX.$....We can also say this if $X$ is normal but not paracompact. For example if $X$ is $omega_1$ with the $in$-order topology, and $cX=omega_1+1$.
$endgroup$
– DanielWainfleet
Dec 9 '18 at 3:36
add a comment |
$begingroup$
Every compact space is paracompact. Let $X$ be the one-point compactification of a countably infinite discrete space $X_0$, and let $Y$ be the one-point compactification of an uncountable discrete space $Y_0$. Let $p$ be the non-isolated point of $X$ and $q$ the non-isolated point of $Y$. Then $Xtimes Y$ is compact, so it’s paracompact. However, the subspace
$$Z=(Xtimes Y)setminus{langle p,qrangle}$$
is not even normal, so it’s certainly not paracompact. Specifically, the closed sets $H=X_0times{q}$ and $K={p}times Y_0$ cannot be separated by disjoint open sets.
To see this, suppose that $U$ is an open nbhd of $H$. For each $xin X_0$ there is a finite $F_xsubseteq Y_0$ such that $langle x,qranglein{x}times(Ysetminus F_x)subseteq U$. Let $F=bigcup_{xin X_0}F_x$; then $F$ is countable, so there is a $yin Y_0setminus F$. But then $Xtimes{y}subseteq U$, so no open nbhd of $langle p,yrangle$ is disjoint from $U$, and therefore no open nbhd of $K$ is disjoint from $U$.
answered Dec 4 '14 at 3:36
Brian M. ScottBrian M. Scott
456k38509909
$endgroup$
$begingroup$
your answer was clear,just wonder if we add the closeness to our subspace in this case we will get the paracompact for free, but are still we need Hausdorff to be normal ( the subspace ).
$endgroup$
– Birkary
Dec 4 '14 at 3:58
$begingroup$
@Birkary: Yes, if you restrict yourself to closed subspaces you get paracompactness for free, but you don’t necessarily get normality unless the subspace is Hausdorff.
$endgroup$
– Brian M. Scott
Dec 4 '14 at 4:06
$begingroup$
thanks,this was very helpful. I will make sure to understand the answer you give.
$endgroup$
– Birkary
Dec 4 '14 at 4:08
$begingroup$
@Birkary: You’re welcome.
$endgroup$
– Brian M. Scott
Dec 4 '14 at 4:09
$begingroup$
More generally, any non-normal space has a compactification $id_X:Xto cX.$ So $X$ is a non-paracompact sub-space of the compact Hausdorff space $cX.$....We can also say this if $X$ is normal but not paracompact. For example if $X$ is $omega_1$ with the $in$-order topology, and $cX=omega_1+1$.
$endgroup$
– DanielWainfleet
Dec 9 '18 at 3:36
add a comment |
$begingroup$
Every compact space is paracompact. Let $X$ be the one-point compactification of a countably infinite discrete space $X_0$, and let $Y$ be the one-point compactification of an uncountable discrete space $Y_0$. Let $p$ be the non-isolated point of $X$ and $q$ the non-isolated point of $Y$. Then $Xtimes Y$ is compact, so it’s paracompact. However, the subspace
$$Z=(Xtimes Y)setminus{langle p,qrangle}$$
is not even normal, so it’s certainly not paracompact. Specifically, the closed sets $H=X_0times{q}$ and $K={p}times Y_0$ cannot be separated by disjoint open sets.
To see this, suppose that $U$ is an open nbhd of $H$. For each $xin X_0$ there is a finite $F_xsubseteq Y_0$ such that $langle x,qranglein{x}times(Ysetminus F_x)subseteq U$. Let $F=bigcup_{xin X_0}F_x$; then $F$ is countable, so there is a $yin Y_0setminus F$. But then $Xtimes{y}subseteq U$, so no open nbhd of $langle p,yrangle$ is disjoint from $U$, and therefore no open nbhd of $K$ is disjoint from $U$.
answered Dec 4 '14 at 3:36
Brian M. ScottBrian M. Scott
456k38509909
$endgroup$
Every compact space is paracompact. Let $X$ be the one-point compactification of a countably infinite discrete space $X_0$, and let $Y$ be the one-point compactification of an uncountable discrete space $Y_0$. Let $p$ be the non-isolated point of $X$ and $q$ the non-isolated point of $Y$. Then $Xtimes Y$ is compact, so it’s paracompact. However, the subspace
$$Z=(Xtimes Y)setminus{langle p,qrangle}$$
is not even normal, so it’s certainly not paracompact. Specifically, the closed sets $H=X_0times{q}$ and $K={p}times Y_0$ cannot be separated by disjoint open sets.
To see this, suppose that $U$ is an open nbhd of $H$. For each $xin X_0$ there is a finite $F_xsubseteq Y_0$ such that $langle x,qranglein{x}times(Ysetminus F_x)subseteq U$. Let $F=bigcup_{xin X_0}F_x$; then $F$ is countable, so there is a $yin Y_0setminus F$. But then $Xtimes{y}subseteq U$, so no open nbhd of $langle p,yrangle$ is disjoint from $U$, and therefore no open nbhd of $K$ is disjoint from $U$.
answered Dec 4 '14 at 3:36
Brian M. ScottBrian M. Scott
456k38509909
edited Dec 4 '14 at 3:41
answered Dec 4 '14 at 3:36
Brian M. ScottBrian M. Scott
456k38509909
answered Dec 4 '14 at 3:36
Brian M. ScottBrian M. Scott
456k38509909
answered Dec 4 '14 at 3:36
Brian M. ScottBrian M. Scott
456k38509909
456k38509909
$begingroup$
your answer was clear,just wonder if we add the closeness to our subspace in this case we will get the paracompact for free, but are still we need Hausdorff to be normal ( the subspace ).
$endgroup$
– Birkary
Dec 4 '14 at 3:58
$begingroup$
@Birkary: Yes, if you restrict yourself to closed subspaces you get paracompactness for free, but you don’t necessarily get normality unless the subspace is Hausdorff.
$endgroup$
– Brian M. Scott
Dec 4 '14 at 4:06
$begingroup$
thanks,this was very helpful. I will make sure to understand the answer you give.
$endgroup$
– Birkary
Dec 4 '14 at 4:08
$begingroup$
@Birkary: You’re welcome.
$endgroup$
– Brian M. Scott
Dec 4 '14 at 4:09
$begingroup$
More generally, any non-normal space has a compactification $id_X:Xto cX.$ So $X$ is a non-paracompact sub-space of the compact Hausdorff space $cX.$....We can also say this if $X$ is normal but not paracompact. For example if $X$ is $omega_1$ with the $in$-order topology, and $cX=omega_1+1$.
$endgroup$
– DanielWainfleet
Dec 9 '18 at 3:36
add a comment |
$begingroup$
your answer was clear,just wonder if we add the closeness to our subspace in this case we will get the paracompact for free, but are still we need Hausdorff to be normal ( the subspace ).
$endgroup$
– Birkary
Dec 4 '14 at 3:58
$begingroup$
@Birkary: Yes, if you restrict yourself to closed subspaces you get paracompactness for free, but you don’t necessarily get normality unless the subspace is Hausdorff.
$endgroup$
– Brian M. Scott
Dec 4 '14 at 4:06
$begingroup$
thanks,this was very helpful. I will make sure to understand the answer you give.
$endgroup$
– Birkary
Dec 4 '14 at 4:08
$begingroup$
@Birkary: You’re welcome.
$endgroup$
– Brian M. Scott
Dec 4 '14 at 4:09
$begingroup$
More generally, any non-normal space has a compactification $id_X:Xto cX.$ So $X$ is a non-paracompact sub-space of the compact Hausdorff space $cX.$....We can also say this if $X$ is normal but not paracompact. For example if $X$ is $omega_1$ with the $in$-order topology, and $cX=omega_1+1$.
$endgroup$
– DanielWainfleet
Dec 9 '18 at 3:36
$begingroup$
your answer was clear,just wonder if we add the closeness to our subspace in this case we will get the paracompact for free, but are still we need Hausdorff to be normal ( the subspace ).
$endgroup$
– Birkary
Dec 4 '14 at 3:58
$begingroup$
your answer was clear,just wonder if we add the closeness to our subspace in this case we will get the paracompact for free, but are still we need Hausdorff to be normal ( the subspace ).
$endgroup$
– Birkary
Dec 4 '14 at 3:58
$begingroup$
@Birkary: Yes, if you restrict yourself to closed subspaces you get paracompactness for free, but you don’t necessarily get normality unless the subspace is Hausdorff.
$endgroup$
– Brian M. Scott
Dec 4 '14 at 4:06
$begingroup$
@Birkary: Yes, if you restrict yourself to closed subspaces you get paracompactness for free, but you don’t necessarily get normality unless the subspace is Hausdorff.
$endgroup$
– Brian M. Scott
Dec 4 '14 at 4:06
$begingroup$
thanks,this was very helpful. I will make sure to understand the answer you give.
$endgroup$
– Birkary
Dec 4 '14 at 4:08
$begingroup$
thanks,this was very helpful. I will make sure to understand the answer you give.
$endgroup$
– Birkary
Dec 4 '14 at 4:08
$begingroup$
@Birkary: You’re welcome.
$endgroup$
– Brian M. Scott
Dec 4 '14 at 4:09
$begingroup$
@Birkary: You’re welcome.
$endgroup$
– Brian M. Scott
Dec 4 '14 at 4:09
$begingroup$
More generally, any non-normal space has a compactification $id_X:Xto cX.$ So $X$ is a non-paracompact sub-space of the compact Hausdorff space $cX.$....We can also say this if $X$ is normal but not paracompact. For example if $X$ is $omega_1$ with the $in$-order topology, and $cX=omega_1+1$.
$endgroup$
– DanielWainfleet
Dec 9 '18 at 3:36
$begingroup$
More generally, any non-normal space has a compactification $id_X:Xto cX.$ So $X$ is a non-paracompact sub-space of the compact Hausdorff space $cX.$....We can also say this if $X$ is normal but not paracompact. For example if $X$ is $omega_1$ with the $in$-order topology, and $cX=omega_1+1$.
$endgroup$
– DanielWainfleet
Dec 9 '18 at 3:36
add a comment |
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