Relative compact inside a subset with relative topology
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I am a little bit confused. Assume that $Y$ is a subset of $X$, with the inherited topology from $X$. What does it mean the $Ksubset Y$ is a relatively compact subset of $Y$?
One option, is that $K= Ccap Y$, where $C$ is a relatively compact subset in $X$.
Another option, is the $overlineKcap Y$ is compact in $Y$.
To be specific, I am looking on the following example:
Let $X$ be a topological (locally compact Hausdorff), $D_n_
ninmathbbZ$ be a sequence of open subsets inside $X$. For every $n$, let $T_n: D_nto D_-n$ be a homeomorphism.
Consider the set $Y:= xin D_n$ with the inherited topology from $Xtimes mathbbZtimes X$.
I want to say that if $Ksubset Y$ is relatively compact, then $K$ is contained in a set of the form $(T_n(x),n,x)in Xtimes mathbbZtimes X$ for some relatively compact set $C$ in $X$ and $NinmathbbN$. Is it true?
Thanks!
general-topology compactness
edited Sep 3 at 12:28
Arnaud D.
14.8k52142
asked Aug 25 at 11:01
Kiko
335111
add a comment |Â
up vote
1
down vote
favorite
I am a little bit confused. Assume that $Y$ is a subset of $X$, with the inherited topology from $X$. What does it mean the $Ksubset Y$ is a relatively compact subset of $Y$?
One option, is that $K= Ccap Y$, where $C$ is a relatively compact subset in $X$.
Another option, is the $overlineKcap Y$ is compact in $Y$.
To be specific, I am looking on the following example:
Let $X$ be a topological (locally compact Hausdorff), $D_n_
ninmathbbZ$ be a sequence of open subsets inside $X$. For every $n$, let $T_n: D_nto D_-n$ be a homeomorphism.
Consider the set $Y:= xin D_n$ with the inherited topology from $Xtimes mathbbZtimes X$.
I want to say that if $Ksubset Y$ is relatively compact, then $K$ is contained in a set of the form $(T_n(x),n,x)in Xtimes mathbbZtimes X$ for some relatively compact set $C$ in $X$ and $NinmathbbN$. Is it true?
Thanks!
general-topology compactness
edited Sep 3 at 12:28
Arnaud D.
14.8k52142
asked Aug 25 at 11:01
Kiko
335111
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am a little bit confused. Assume that $Y$ is a subset of $X$, with the inherited topology from $X$. What does it mean the $Ksubset Y$ is a relatively compact subset of $Y$?
One option, is that $K= Ccap Y$, where $C$ is a relatively compact subset in $X$.
Another option, is the $overlineKcap Y$ is compact in $Y$.
To be specific, I am looking on the following example:
Let $X$ be a topological (locally compact Hausdorff), $D_n_
ninmathbbZ$ be a sequence of open subsets inside $X$. For every $n$, let $T_n: D_nto D_-n$ be a homeomorphism.
Consider the set $Y:= xin D_n$ with the inherited topology from $Xtimes mathbbZtimes X$.
I want to say that if $Ksubset Y$ is relatively compact, then $K$ is contained in a set of the form $(T_n(x),n,x)in Xtimes mathbbZtimes X$ for some relatively compact set $C$ in $X$ and $NinmathbbN$. Is it true?
Thanks!
general-topology compactness
edited Sep 3 at 12:28
Arnaud D.
14.8k52142
asked Aug 25 at 11:01
Kiko
335111
I am a little bit confused. Assume that $Y$ is a subset of $X$, with the inherited topology from $X$. What does it mean the $Ksubset Y$ is a relatively compact subset of $Y$?
One option, is that $K= Ccap Y$, where $C$ is a relatively compact subset in $X$.
Another option, is the $overlineKcap Y$ is compact in $Y$.
To be specific, I am looking on the following example:
Let $X$ be a topological (locally compact Hausdorff), $D_n_
ninmathbbZ$ be a sequence of open subsets inside $X$. For every $n$, let $T_n: D_nto D_-n$ be a homeomorphism.
Consider the set $Y:= xin D_n$ with the inherited topology from $Xtimes mathbbZtimes X$.
I want to say that if $Ksubset Y$ is relatively compact, then $K$ is contained in a set of the form $(T_n(x),n,x)in Xtimes mathbbZtimes X$ for some relatively compact set $C$ in $X$ and $NinmathbbN$. Is it true?
Thanks!
general-topology compactness
edited Sep 3 at 12:28
Arnaud D.
14.8k52142
asked Aug 25 at 11:01
Kiko
335111
edited Sep 3 at 12:28
Arnaud D.
14.8k52142
edited Sep 3 at 12:28
Arnaud D.
14.8k52142
edited Sep 3 at 12:28
Arnaud D.
14.8k52142
14.8k52142
asked Aug 25 at 11:01
Kiko
335111
asked Aug 25 at 11:01
Kiko
335111
asked Aug 25 at 11:01
Kiko
335111
335111
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
According to definition a subset of a topological space is relatively compact if its closure (in that space) is compact.
So I suspect that $Ksubseteq Y$ is a relatively compact subset of $Y$ it is closure in $Y$ is compact.
Further if $overline K$ denotes the closure of $K$ in original space $X$ then the closure of $K$ as a subset of $Y$ equals $overline Kcap Y$.
This results in the second option that you mention:$$Ktext is a relatively compact subset of Ysubseteq Xtext iff overline Kcap Ytext is compact$$
I advice you not to accept this answer, but just think it over. Maybe a real topologist will pass by to adjust.
answered Aug 25 at 11:17
drhab
88.4k541120
Thank you very much, sounds good. It seems to me that in this case $overlineK$ is also a subset of $Y$: $overlineK=overlineKcap Ysubseteq overlineoverlineKcap Y=overlineKcap Y$, the last equality is since the intersection is compact and therefore closed in $X$. The inclusion implies that $overlineKsubseteq Y$. But then, the conclusion is that $K$ is a relatively compact subset of $Y$ if and only if $overlineK$ is contained in $Y$ and is compact(?)
– Kiko
Aug 26 at 7:29
1
"since the intersection is compact and therefore closed in $X$". Yes to this under the extra condition that $X$ is a Hausdorff space. In a Hausdorff space compact sets are indeed closed, and you end up with the conclusion you mention. But in general it is not true that compact sets are necessarily closed.
– drhab
Aug 26 at 11:23
Yes. Thank you.
– Kiko
Aug 26 at 14:39
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
According to definition a subset of a topological space is relatively compact if its closure (in that space) is compact.
So I suspect that $Ksubseteq Y$ is a relatively compact subset of $Y$ it is closure in $Y$ is compact.
Further if $overline K$ denotes the closure of $K$ in original space $X$ then the closure of $K$ as a subset of $Y$ equals $overline Kcap Y$.
This results in the second option that you mention:$$Ktext is a relatively compact subset of Ysubseteq Xtext iff overline Kcap Ytext is compact$$
I advice you not to accept this answer, but just think it over. Maybe a real topologist will pass by to adjust.
answered Aug 25 at 11:17
drhab
88.4k541120
Thank you very much, sounds good. It seems to me that in this case $overlineK$ is also a subset of $Y$: $overlineK=overlineKcap Ysubseteq overlineoverlineKcap Y=overlineKcap Y$, the last equality is since the intersection is compact and therefore closed in $X$. The inclusion implies that $overlineKsubseteq Y$. But then, the conclusion is that $K$ is a relatively compact subset of $Y$ if and only if $overlineK$ is contained in $Y$ and is compact(?)
– Kiko
Aug 26 at 7:29
1
"since the intersection is compact and therefore closed in $X$". Yes to this under the extra condition that $X$ is a Hausdorff space. In a Hausdorff space compact sets are indeed closed, and you end up with the conclusion you mention. But in general it is not true that compact sets are necessarily closed.
– drhab
Aug 26 at 11:23
Yes. Thank you.
– Kiko
Aug 26 at 14:39
add a comment |Â
up vote
1
down vote
accepted
According to definition a subset of a topological space is relatively compact if its closure (in that space) is compact.
So I suspect that $Ksubseteq Y$ is a relatively compact subset of $Y$ it is closure in $Y$ is compact.
Further if $overline K$ denotes the closure of $K$ in original space $X$ then the closure of $K$ as a subset of $Y$ equals $overline Kcap Y$.
This results in the second option that you mention:$$Ktext is a relatively compact subset of Ysubseteq Xtext iff overline Kcap Ytext is compact$$
I advice you not to accept this answer, but just think it over. Maybe a real topologist will pass by to adjust.
answered Aug 25 at 11:17
drhab
88.4k541120
Thank you very much, sounds good. It seems to me that in this case $overlineK$ is also a subset of $Y$: $overlineK=overlineKcap Ysubseteq overlineoverlineKcap Y=overlineKcap Y$, the last equality is since the intersection is compact and therefore closed in $X$. The inclusion implies that $overlineKsubseteq Y$. But then, the conclusion is that $K$ is a relatively compact subset of $Y$ if and only if $overlineK$ is contained in $Y$ and is compact(?)
– Kiko
Aug 26 at 7:29
1
"since the intersection is compact and therefore closed in $X$". Yes to this under the extra condition that $X$ is a Hausdorff space. In a Hausdorff space compact sets are indeed closed, and you end up with the conclusion you mention. But in general it is not true that compact sets are necessarily closed.
– drhab
Aug 26 at 11:23
Yes. Thank you.
– Kiko
Aug 26 at 14:39
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
According to definition a subset of a topological space is relatively compact if its closure (in that space) is compact.
So I suspect that $Ksubseteq Y$ is a relatively compact subset of $Y$ it is closure in $Y$ is compact.
Further if $overline K$ denotes the closure of $K$ in original space $X$ then the closure of $K$ as a subset of $Y$ equals $overline Kcap Y$.
This results in the second option that you mention:$$Ktext is a relatively compact subset of Ysubseteq Xtext iff overline Kcap Ytext is compact$$
I advice you not to accept this answer, but just think it over. Maybe a real topologist will pass by to adjust.
answered Aug 25 at 11:17
drhab
88.4k541120
According to definition a subset of a topological space is relatively compact if its closure (in that space) is compact.
So I suspect that $Ksubseteq Y$ is a relatively compact subset of $Y$ it is closure in $Y$ is compact.
Further if $overline K$ denotes the closure of $K$ in original space $X$ then the closure of $K$ as a subset of $Y$ equals $overline Kcap Y$.
This results in the second option that you mention:$$Ktext is a relatively compact subset of Ysubseteq Xtext iff overline Kcap Ytext is compact$$
I advice you not to accept this answer, but just think it over. Maybe a real topologist will pass by to adjust.
answered Aug 25 at 11:17
drhab
88.4k541120
answered Aug 25 at 11:17
drhab
88.4k541120
answered Aug 25 at 11:17
drhab
88.4k541120
answered Aug 25 at 11:17
drhab
88.4k541120
88.4k541120
Thank you very much, sounds good. It seems to me that in this case $overlineK$ is also a subset of $Y$: $overlineK=overlineKcap Ysubseteq overlineoverlineKcap Y=overlineKcap Y$, the last equality is since the intersection is compact and therefore closed in $X$. The inclusion implies that $overlineKsubseteq Y$. But then, the conclusion is that $K$ is a relatively compact subset of $Y$ if and only if $overlineK$ is contained in $Y$ and is compact(?)
– Kiko
Aug 26 at 7:29
1
"since the intersection is compact and therefore closed in $X$". Yes to this under the extra condition that $X$ is a Hausdorff space. In a Hausdorff space compact sets are indeed closed, and you end up with the conclusion you mention. But in general it is not true that compact sets are necessarily closed.
– drhab
Aug 26 at 11:23
Yes. Thank you.
– Kiko
Aug 26 at 14:39
add a comment |Â
Thank you very much, sounds good. It seems to me that in this case $overlineK$ is also a subset of $Y$: $overlineK=overlineKcap Ysubseteq overlineoverlineKcap Y=overlineKcap Y$, the last equality is since the intersection is compact and therefore closed in $X$. The inclusion implies that $overlineKsubseteq Y$. But then, the conclusion is that $K$ is a relatively compact subset of $Y$ if and only if $overlineK$ is contained in $Y$ and is compact(?)
– Kiko
Aug 26 at 7:29
1
"since the intersection is compact and therefore closed in $X$". Yes to this under the extra condition that $X$ is a Hausdorff space. In a Hausdorff space compact sets are indeed closed, and you end up with the conclusion you mention. But in general it is not true that compact sets are necessarily closed.
– drhab
Aug 26 at 11:23
Yes. Thank you.
– Kiko
Aug 26 at 14:39
Thank you very much, sounds good. It seems to me that in this case $overlineK$ is also a subset of $Y$: $overlineK=overlineKcap Ysubseteq overlineoverlineKcap Y=overlineKcap Y$, the last equality is since the intersection is compact and therefore closed in $X$. The inclusion implies that $overlineKsubseteq Y$. But then, the conclusion is that $K$ is a relatively compact subset of $Y$ if and only if $overlineK$ is contained in $Y$ and is compact(?)
– Kiko
Aug 26 at 7:29
Thank you very much, sounds good. It seems to me that in this case $overlineK$ is also a subset of $Y$: $overlineK=overlineKcap Ysubseteq overlineoverlineKcap Y=overlineKcap Y$, the last equality is since the intersection is compact and therefore closed in $X$. The inclusion implies that $overlineKsubseteq Y$. But then, the conclusion is that $K$ is a relatively compact subset of $Y$ if and only if $overlineK$ is contained in $Y$ and is compact(?)
– Kiko
Aug 26 at 7:29
1
1
"since the intersection is compact and therefore closed in $X$". Yes to this under the extra condition that $X$ is a Hausdorff space. In a Hausdorff space compact sets are indeed closed, and you end up with the conclusion you mention. But in general it is not true that compact sets are necessarily closed.
– drhab
Aug 26 at 11:23
"since the intersection is compact and therefore closed in $X$". Yes to this under the extra condition that $X$ is a Hausdorff space. In a Hausdorff space compact sets are indeed closed, and you end up with the conclusion you mention. But in general it is not true that compact sets are necessarily closed.
– drhab
Aug 26 at 11:23
Yes. Thank you.
– Kiko
Aug 26 at 14:39
Yes. Thank you.
– Kiko
Aug 26 at 14:39
add a comment |Â
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