Characterize the compact subspaces in a topological space.
It’s known that every closed subset of a compact topological space is also compact. However, it’s not always true that compact subspaces are closed(by taking the cofinite topology on $mathbb{Z}$ the set of positive integers $mathbb{Z}+$ is compact but is not closed). The following is my attempt to replace “closed” with a weaker condition so that the converse actually holds.
Let $(X, mathcal{T})$ be a compact space. The idea for the following arguments is that in the proof of “closed subset $A$ of compact space is compact”, we show the complement $X-A$ can be removed so that the rest of the open sets also form a cover:
For a subset $A$ of $X$, define $$K(A)=bigcap{Uin mathcal{T}: Usubset A}.$$ If there exists a closed set $C$ in $X$ such that $Asubset Csubset K(A)$, we say $A$ is $K$-closed. Apparently every closed set is also $K$-closed.
Now if $A$ is $K$-closed, then by taking $X-C$ together with an open cover of $A$, we have an open cover, thus a finite subcover ${U_i}$of $X$. It can then be shown that if $X-C=U_i$ for some $U_i$, then it is removable so that the rest of the finite cover also covers $A$.
Now I wonder if the converse holds:
If $Asubset X$ is compact, is it true that $A$ must be $K$-closed?
Is there an alternate way to modify the result so that the converse holds?
——————
$mathbb{Z}+$ is compact in $mathbb{Z}$ with the cofinite topology, but $K(mathbb{Z}+)=mathbb{Z}+$ so it’s not $K$-closed. My attempt does not solve the problem. Does anyone have a better idea?
general-topology compactness
edited Dec 12 '18 at 23:18
Eric Wofsey
180k12208336
asked Dec 12 '18 at 22:53
William SunWilliam Sun
471111
add a comment |
It’s known that every closed subset of a compact topological space is also compact. However, it’s not always true that compact subspaces are closed(by taking the cofinite topology on $mathbb{Z}$ the set of positive integers $mathbb{Z}+$ is compact but is not closed). The following is my attempt to replace “closed” with a weaker condition so that the converse actually holds.
Let $(X, mathcal{T})$ be a compact space. The idea for the following arguments is that in the proof of “closed subset $A$ of compact space is compact”, we show the complement $X-A$ can be removed so that the rest of the open sets also form a cover:
For a subset $A$ of $X$, define $$K(A)=bigcap{Uin mathcal{T}: Usubset A}.$$ If there exists a closed set $C$ in $X$ such that $Asubset Csubset K(A)$, we say $A$ is $K$-closed. Apparently every closed set is also $K$-closed.
Now if $A$ is $K$-closed, then by taking $X-C$ together with an open cover of $A$, we have an open cover, thus a finite subcover ${U_i}$of $X$. It can then be shown that if $X-C=U_i$ for some $U_i$, then it is removable so that the rest of the finite cover also covers $A$.
Now I wonder if the converse holds:
If $Asubset X$ is compact, is it true that $A$ must be $K$-closed?
Is there an alternate way to modify the result so that the converse holds?
——————
$mathbb{Z}+$ is compact in $mathbb{Z}$ with the cofinite topology, but $K(mathbb{Z}+)=mathbb{Z}+$ so it’s not $K$-closed. My attempt does not solve the problem. Does anyone have a better idea?
general-topology compactness
edited Dec 12 '18 at 23:18
Eric Wofsey
180k12208336
asked Dec 12 '18 at 22:53
William SunWilliam Sun
471111
$K(A) = bigcap {O in mathcal{T}: A subseteq O}$ of course.
– Henno Brandsma
Dec 12 '18 at 23:03
Thank you for the comment. I have re-edited the question.
– William Sun
Dec 12 '18 at 23:11
add a comment |
It’s known that every closed subset of a compact topological space is also compact. However, it’s not always true that compact subspaces are closed(by taking the cofinite topology on $mathbb{Z}$ the set of positive integers $mathbb{Z}+$ is compact but is not closed). The following is my attempt to replace “closed” with a weaker condition so that the converse actually holds.
Let $(X, mathcal{T})$ be a compact space. The idea for the following arguments is that in the proof of “closed subset $A$ of compact space is compact”, we show the complement $X-A$ can be removed so that the rest of the open sets also form a cover:
For a subset $A$ of $X$, define $$K(A)=bigcap{Uin mathcal{T}: Usubset A}.$$ If there exists a closed set $C$ in $X$ such that $Asubset Csubset K(A)$, we say $A$ is $K$-closed. Apparently every closed set is also $K$-closed.
Now if $A$ is $K$-closed, then by taking $X-C$ together with an open cover of $A$, we have an open cover, thus a finite subcover ${U_i}$of $X$. It can then be shown that if $X-C=U_i$ for some $U_i$, then it is removable so that the rest of the finite cover also covers $A$.
Now I wonder if the converse holds:
If $Asubset X$ is compact, is it true that $A$ must be $K$-closed?
Is there an alternate way to modify the result so that the converse holds?
——————
$mathbb{Z}+$ is compact in $mathbb{Z}$ with the cofinite topology, but $K(mathbb{Z}+)=mathbb{Z}+$ so it’s not $K$-closed. My attempt does not solve the problem. Does anyone have a better idea?
general-topology compactness
edited Dec 12 '18 at 23:18
Eric Wofsey
180k12208336
asked Dec 12 '18 at 22:53
William SunWilliam Sun
471111
It’s known that every closed subset of a compact topological space is also compact. However, it’s not always true that compact subspaces are closed(by taking the cofinite topology on $mathbb{Z}$ the set of positive integers $mathbb{Z}+$ is compact but is not closed). The following is my attempt to replace “closed” with a weaker condition so that the converse actually holds.
Let $(X, mathcal{T})$ be a compact space. The idea for the following arguments is that in the proof of “closed subset $A$ of compact space is compact”, we show the complement $X-A$ can be removed so that the rest of the open sets also form a cover:
For a subset $A$ of $X$, define $$K(A)=bigcap{Uin mathcal{T}: Usubset A}.$$ If there exists a closed set $C$ in $X$ such that $Asubset Csubset K(A)$, we say $A$ is $K$-closed. Apparently every closed set is also $K$-closed.
Now if $A$ is $K$-closed, then by taking $X-C$ together with an open cover of $A$, we have an open cover, thus a finite subcover ${U_i}$of $X$. It can then be shown that if $X-C=U_i$ for some $U_i$, then it is removable so that the rest of the finite cover also covers $A$.
Now I wonder if the converse holds:
If $Asubset X$ is compact, is it true that $A$ must be $K$-closed?
Is there an alternate way to modify the result so that the converse holds?
——————
$mathbb{Z}+$ is compact in $mathbb{Z}$ with the cofinite topology, but $K(mathbb{Z}+)=mathbb{Z}+$ so it’s not $K$-closed. My attempt does not solve the problem. Does anyone have a better idea?
general-topology compactness
general-topology compactness
edited Dec 12 '18 at 23:18
Eric Wofsey
180k12208336
asked Dec 12 '18 at 22:53
William SunWilliam Sun
471111
edited Dec 12 '18 at 23:18
Eric Wofsey
180k12208336
asked Dec 12 '18 at 22:53
William SunWilliam Sun
471111
edited Dec 12 '18 at 23:18
Eric Wofsey
180k12208336
edited Dec 12 '18 at 23:18
Eric Wofsey
180k12208336
edited Dec 12 '18 at 23:18
Eric Wofsey
180k12208336
180k12208336
asked Dec 12 '18 at 22:53
William SunWilliam Sun
471111
asked Dec 12 '18 at 22:53
William SunWilliam Sun
471111
asked Dec 12 '18 at 22:53
William SunWilliam Sun
471111
471111
$K(A) = bigcap {O in mathcal{T}: A subseteq O}$ of course.
– Henno Brandsma
Dec 12 '18 at 23:03
Thank you for the comment. I have re-edited the question.
– William Sun
Dec 12 '18 at 23:11
add a comment |
$K(A) = bigcap {O in mathcal{T}: A subseteq O}$ of course.
– Henno Brandsma
Dec 12 '18 at 23:03
Thank you for the comment. I have re-edited the question.
– William Sun
Dec 12 '18 at 23:11
$K(A) = bigcap {O in mathcal{T}: A subseteq O}$ of course.
– Henno Brandsma
Dec 12 '18 at 23:03
$K(A) = bigcap {O in mathcal{T}: A subseteq O}$ of course.
– Henno Brandsma
Dec 12 '18 at 23:03
Thank you for the comment. I have re-edited the question.
– William Sun
Dec 12 '18 at 23:11
Thank you for the comment. I have re-edited the question.
– William Sun
Dec 12 '18 at 23:11
add a comment |
1 Answer
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No. For instance, let $X={0,1}$ with ${0}$ open but ${1}$ not open. Then $A={0}$ is compact but $K(A)=A={0}$ is not closed so $A$ is not $K$-closed.
Ultimately, the issue is that the "reason" $A$ is compact could be totally unrelated to the "reason" $X$ is compact. In particular, $X$ might have a point $x$ (like $1$ in the example above) whose only neighborhood is $X$ itself, which automatically guarantees that $X$ is compact. If $xnotin A$, there's never going to be any nice way to relate open covers of $A$ to open covers of $X$ to learn anything interesting using the compactness of $A$, since open covers of $X$ are all trivial ($X$ itself must always be one of the sets).
answered Dec 12 '18 at 23:18
Eric WofseyEric Wofsey
180k12208336
add a comment |
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No. For instance, let $X={0,1}$ with ${0}$ open but ${1}$ not open. Then $A={0}$ is compact but $K(A)=A={0}$ is not closed so $A$ is not $K$-closed.
Ultimately, the issue is that the "reason" $A$ is compact could be totally unrelated to the "reason" $X$ is compact. In particular, $X$ might have a point $x$ (like $1$ in the example above) whose only neighborhood is $X$ itself, which automatically guarantees that $X$ is compact. If $xnotin A$, there's never going to be any nice way to relate open covers of $A$ to open covers of $X$ to learn anything interesting using the compactness of $A$, since open covers of $X$ are all trivial ($X$ itself must always be one of the sets).
answered Dec 12 '18 at 23:18
Eric WofseyEric Wofsey
180k12208336
add a comment |
No. For instance, let $X={0,1}$ with ${0}$ open but ${1}$ not open. Then $A={0}$ is compact but $K(A)=A={0}$ is not closed so $A$ is not $K$-closed.
Ultimately, the issue is that the "reason" $A$ is compact could be totally unrelated to the "reason" $X$ is compact. In particular, $X$ might have a point $x$ (like $1$ in the example above) whose only neighborhood is $X$ itself, which automatically guarantees that $X$ is compact. If $xnotin A$, there's never going to be any nice way to relate open covers of $A$ to open covers of $X$ to learn anything interesting using the compactness of $A$, since open covers of $X$ are all trivial ($X$ itself must always be one of the sets).
answered Dec 12 '18 at 23:18
Eric WofseyEric Wofsey
180k12208336
add a comment |
No. For instance, let $X={0,1}$ with ${0}$ open but ${1}$ not open. Then $A={0}$ is compact but $K(A)=A={0}$ is not closed so $A$ is not $K$-closed.
Ultimately, the issue is that the "reason" $A$ is compact could be totally unrelated to the "reason" $X$ is compact. In particular, $X$ might have a point $x$ (like $1$ in the example above) whose only neighborhood is $X$ itself, which automatically guarantees that $X$ is compact. If $xnotin A$, there's never going to be any nice way to relate open covers of $A$ to open covers of $X$ to learn anything interesting using the compactness of $A$, since open covers of $X$ are all trivial ($X$ itself must always be one of the sets).
answered Dec 12 '18 at 23:18
Eric WofseyEric Wofsey
180k12208336
No. For instance, let $X={0,1}$ with ${0}$ open but ${1}$ not open. Then $A={0}$ is compact but $K(A)=A={0}$ is not closed so $A$ is not $K$-closed.
Ultimately, the issue is that the "reason" $A$ is compact could be totally unrelated to the "reason" $X$ is compact. In particular, $X$ might have a point $x$ (like $1$ in the example above) whose only neighborhood is $X$ itself, which automatically guarantees that $X$ is compact. If $xnotin A$, there's never going to be any nice way to relate open covers of $A$ to open covers of $X$ to learn anything interesting using the compactness of $A$, since open covers of $X$ are all trivial ($X$ itself must always be one of the sets).
answered Dec 12 '18 at 23:18
Eric WofseyEric Wofsey
180k12208336
answered Dec 12 '18 at 23:18
Eric WofseyEric Wofsey
180k12208336
answered Dec 12 '18 at 23:18
Eric WofseyEric Wofsey
180k12208336
answered Dec 12 '18 at 23:18
Eric WofseyEric Wofsey
180k12208336
180k12208336
add a comment |
add a comment |
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$K(A) = bigcap {O in mathcal{T}: A subseteq O}$ of course.
– Henno Brandsma
Dec 12 '18 at 23:03
Thank you for the comment. I have re-edited the question.
– William Sun
Dec 12 '18 at 23:11