Mixing
Is the closure of a compact set compact?
$begingroup$
While looking up information on compact operators I came across these two conflicting posts.
- If a set is compact then it is closed
- Topology: Example of a compact set but its closure not compact
So the first link says that if a set $U$ is compact then it is closed. $U$ closed means $U = overline{U}$ and hence $overline{U}$ is compact. This seems to be in direct contradiction with the second post?
general-topology functional-analysis compactness
edited Apr 13 '17 at 12:19
Community♦
1
asked Aug 25 '16 at 10:07
cssscsss
1,30221325
$endgroup$
add a comment |
$begingroup$
While looking up information on compact operators I came across these two conflicting posts.
- If a set is compact then it is closed
- Topology: Example of a compact set but its closure not compact
So the first link says that if a set $U$ is compact then it is closed. $U$ closed means $U = overline{U}$ and hence $overline{U}$ is compact. This seems to be in direct contradiction with the second post?
general-topology functional-analysis compactness
edited Apr 13 '17 at 12:19
Community♦
1
asked Aug 25 '16 at 10:07
cssscsss
1,30221325
$endgroup$
5
$begingroup$
A compact subspace of a Hausdorff space is closed.
$endgroup$
– Watson
Aug 25 '16 at 10:08
$begingroup$
@Watson That should be an answer, not just a comment :)
$endgroup$
– 5xum
Aug 25 '16 at 10:13
$begingroup$
@csss: see also math.stackexchange.com/questions/83355/…, math.stackexchange.com/questions/176458
$endgroup$
– Watson
Aug 25 '16 at 12:13
add a comment |
$begingroup$
While looking up information on compact operators I came across these two conflicting posts.
- If a set is compact then it is closed
- Topology: Example of a compact set but its closure not compact
So the first link says that if a set $U$ is compact then it is closed. $U$ closed means $U = overline{U}$ and hence $overline{U}$ is compact. This seems to be in direct contradiction with the second post?
general-topology functional-analysis compactness
edited Apr 13 '17 at 12:19
Community♦
1
asked Aug 25 '16 at 10:07
cssscsss
1,30221325
$endgroup$
While looking up information on compact operators I came across these two conflicting posts.
- If a set is compact then it is closed
- Topology: Example of a compact set but its closure not compact
So the first link says that if a set $U$ is compact then it is closed. $U$ closed means $U = overline{U}$ and hence $overline{U}$ is compact. This seems to be in direct contradiction with the second post?
general-topology functional-analysis compactness
general-topology functional-analysis compactness
edited Apr 13 '17 at 12:19
Community♦
1
asked Aug 25 '16 at 10:07
cssscsss
1,30221325
edited Apr 13 '17 at 12:19
Community♦
1
asked Aug 25 '16 at 10:07
cssscsss
1,30221325
edited Apr 13 '17 at 12:19
Community♦
1
edited Apr 13 '17 at 12:19
Community♦
1
edited Apr 13 '17 at 12:19
Community♦
1
1
asked Aug 25 '16 at 10:07
cssscsss
1,30221325
asked Aug 25 '16 at 10:07
cssscsss
1,30221325
asked Aug 25 '16 at 10:07
cssscsss
1,30221325
1,30221325
5
$begingroup$
A compact subspace of a Hausdorff space is closed.
$endgroup$
– Watson
Aug 25 '16 at 10:08
$begingroup$
@Watson That should be an answer, not just a comment :)
$endgroup$
– 5xum
Aug 25 '16 at 10:13
$begingroup$
@csss: see also math.stackexchange.com/questions/83355/…, math.stackexchange.com/questions/176458
$endgroup$
– Watson
Aug 25 '16 at 12:13
add a comment |
5
$begingroup$
A compact subspace of a Hausdorff space is closed.
$endgroup$
– Watson
Aug 25 '16 at 10:08
$begingroup$
@Watson That should be an answer, not just a comment :)
$endgroup$
– 5xum
Aug 25 '16 at 10:13
$begingroup$
@csss: see also math.stackexchange.com/questions/83355/…, math.stackexchange.com/questions/176458
$endgroup$
– Watson
Aug 25 '16 at 12:13
5
5
$begingroup$
A compact subspace of a Hausdorff space is closed.
$endgroup$
– Watson
Aug 25 '16 at 10:08
$begingroup$
A compact subspace of a Hausdorff space is closed.
$endgroup$
– Watson
Aug 25 '16 at 10:08
$begingroup$
@Watson That should be an answer, not just a comment :)
$endgroup$
– 5xum
Aug 25 '16 at 10:13
$begingroup$
@Watson That should be an answer, not just a comment :)
$endgroup$
– 5xum
Aug 25 '16 at 10:13
$begingroup$
@csss: see also math.stackexchange.com/questions/83355/…, math.stackexchange.com/questions/176458
$endgroup$
– Watson
Aug 25 '16 at 12:13
$begingroup$
@csss: see also math.stackexchange.com/questions/83355/…, math.stackexchange.com/questions/176458
$endgroup$
– Watson
Aug 25 '16 at 12:13
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
This is not a contradiction, because the main property is:
A compact subspace $K$ of a Hausdorff space $X$ is closed.
Indeed, we show that for every $x in X setminus K$, there is an open set $U$ such that $x in U subset X setminus K$. Fix such an $x$.
As $X$ is Hausdorff ($T2$), for every $y in K$, there are disjoint open sets $U_y,V_y$ such that $x in U_y$ and $y in V_y$. Now you can use the compactness of $K subset bigcup_{y in K} V_y$, so that
$$K subset bigcup_{y in E} V_y$$
for some finite subset $E subset K$.
Define $U = bigcap_{y in E} U_y$, which is an open set because $E$ is finite. You can finally show that $U$ satisfies the desired conditions. I leave it to!
answered Aug 25 '16 at 10:28
WatsonWatson
16k92970
$endgroup$
$begingroup$
And I take it a non-Hausdorff space is a very uncommon space in applications? That is, when dealing with spaces such as $mathbb{R}^n$ and Hilbert spaces such as $L^2(Omega)$ and $W^{1, 2}(Omega)$ then the closure of a compact set will be compact?
$endgroup$
– csss
Aug 25 '16 at 10:45
$begingroup$
@csss: Yes, in general, e.g. in (functional) analysis, we are dealing with metric spaces, which are Hausdorff, so that the above property always holds.
$endgroup$
– Watson
Aug 25 '16 at 10:46
$begingroup$
If $X$ is a topological space such that every compact subspace is closed, then it is clearly $T1$. This doesn't imply $T2$, see math.stackexchange.com/questions/518787.
$endgroup$
– Watson
Aug 25 '16 at 12:10
1
$begingroup$
@csss: In some branches of mathematics, non-Hausdorff spaces are more common. For example, in dynamics --- the study of group actions on topological spaces --- the orbit space or quotient of a group action is often non-Hausdorff.
$endgroup$
– Lee Mosher
Aug 25 '16 at 15:10
add a comment |
$begingroup$
Actually, the co-finite topology on R, the set of real numbers. Every subset of R is compact. However, it is not every subset of R is closed. There is a special spaces that assumed that every compact set is closed. Certainly, in Hausdorff spaces, every compact set is closed. For your the second question, Take X=R, the set of real numbers. Define topology on X by U is a subset of X is open if and only if U contains 1 or U is the empty set.Then K={1,2} is compact, but cl(K)=R which is not compact with this topology. Actually, every point x in not K, and any open subset U contain x then 1 is also in U. So, the intersection of U and K is not empty, then x is in the closure of K. Therefore, cl(K)=R. The collection U(x)={1,x} forms an open cover of X, obvious there is no finite sub-collection of it cover X.
answered Jan 28 at 4:59
Murtadha SarrayMurtadha Sarray
1
$endgroup$
add a comment |
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2 Answers
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$begingroup$
This is not a contradiction, because the main property is:
A compact subspace $K$ of a Hausdorff space $X$ is closed.
Indeed, we show that for every $x in X setminus K$, there is an open set $U$ such that $x in U subset X setminus K$. Fix such an $x$.
As $X$ is Hausdorff ($T2$), for every $y in K$, there are disjoint open sets $U_y,V_y$ such that $x in U_y$ and $y in V_y$. Now you can use the compactness of $K subset bigcup_{y in K} V_y$, so that
$$K subset bigcup_{y in E} V_y$$
for some finite subset $E subset K$.
Define $U = bigcap_{y in E} U_y$, which is an open set because $E$ is finite. You can finally show that $U$ satisfies the desired conditions. I leave it to!
answered Aug 25 '16 at 10:28
WatsonWatson
16k92970
$endgroup$
$begingroup$
And I take it a non-Hausdorff space is a very uncommon space in applications? That is, when dealing with spaces such as $mathbb{R}^n$ and Hilbert spaces such as $L^2(Omega)$ and $W^{1, 2}(Omega)$ then the closure of a compact set will be compact?
$endgroup$
– csss
Aug 25 '16 at 10:45
$begingroup$
@csss: Yes, in general, e.g. in (functional) analysis, we are dealing with metric spaces, which are Hausdorff, so that the above property always holds.
$endgroup$
– Watson
Aug 25 '16 at 10:46
$begingroup$
If $X$ is a topological space such that every compact subspace is closed, then it is clearly $T1$. This doesn't imply $T2$, see math.stackexchange.com/questions/518787.
$endgroup$
– Watson
Aug 25 '16 at 12:10
1
$begingroup$
@csss: In some branches of mathematics, non-Hausdorff spaces are more common. For example, in dynamics --- the study of group actions on topological spaces --- the orbit space or quotient of a group action is often non-Hausdorff.
$endgroup$
– Lee Mosher
Aug 25 '16 at 15:10
add a comment |
$begingroup$
This is not a contradiction, because the main property is:
A compact subspace $K$ of a Hausdorff space $X$ is closed.
Indeed, we show that for every $x in X setminus K$, there is an open set $U$ such that $x in U subset X setminus K$. Fix such an $x$.
As $X$ is Hausdorff ($T2$), for every $y in K$, there are disjoint open sets $U_y,V_y$ such that $x in U_y$ and $y in V_y$. Now you can use the compactness of $K subset bigcup_{y in K} V_y$, so that
$$K subset bigcup_{y in E} V_y$$
for some finite subset $E subset K$.
Define $U = bigcap_{y in E} U_y$, which is an open set because $E$ is finite. You can finally show that $U$ satisfies the desired conditions. I leave it to!
answered Aug 25 '16 at 10:28
WatsonWatson
16k92970
$endgroup$
$begingroup$
And I take it a non-Hausdorff space is a very uncommon space in applications? That is, when dealing with spaces such as $mathbb{R}^n$ and Hilbert spaces such as $L^2(Omega)$ and $W^{1, 2}(Omega)$ then the closure of a compact set will be compact?
$endgroup$
– csss
Aug 25 '16 at 10:45
$begingroup$
@csss: Yes, in general, e.g. in (functional) analysis, we are dealing with metric spaces, which are Hausdorff, so that the above property always holds.
$endgroup$
– Watson
Aug 25 '16 at 10:46
$begingroup$
If $X$ is a topological space such that every compact subspace is closed, then it is clearly $T1$. This doesn't imply $T2$, see math.stackexchange.com/questions/518787.
$endgroup$
– Watson
Aug 25 '16 at 12:10
1
$begingroup$
@csss: In some branches of mathematics, non-Hausdorff spaces are more common. For example, in dynamics --- the study of group actions on topological spaces --- the orbit space or quotient of a group action is often non-Hausdorff.
$endgroup$
– Lee Mosher
Aug 25 '16 at 15:10
add a comment |
$begingroup$
This is not a contradiction, because the main property is:
A compact subspace $K$ of a Hausdorff space $X$ is closed.
Indeed, we show that for every $x in X setminus K$, there is an open set $U$ such that $x in U subset X setminus K$. Fix such an $x$.
As $X$ is Hausdorff ($T2$), for every $y in K$, there are disjoint open sets $U_y,V_y$ such that $x in U_y$ and $y in V_y$. Now you can use the compactness of $K subset bigcup_{y in K} V_y$, so that
$$K subset bigcup_{y in E} V_y$$
for some finite subset $E subset K$.
Define $U = bigcap_{y in E} U_y$, which is an open set because $E$ is finite. You can finally show that $U$ satisfies the desired conditions. I leave it to!
answered Aug 25 '16 at 10:28
WatsonWatson
16k92970
$endgroup$
This is not a contradiction, because the main property is:
A compact subspace $K$ of a Hausdorff space $X$ is closed.
Indeed, we show that for every $x in X setminus K$, there is an open set $U$ such that $x in U subset X setminus K$. Fix such an $x$.
As $X$ is Hausdorff ($T2$), for every $y in K$, there are disjoint open sets $U_y,V_y$ such that $x in U_y$ and $y in V_y$. Now you can use the compactness of $K subset bigcup_{y in K} V_y$, so that
$$K subset bigcup_{y in E} V_y$$
for some finite subset $E subset K$.
Define $U = bigcap_{y in E} U_y$, which is an open set because $E$ is finite. You can finally show that $U$ satisfies the desired conditions. I leave it to!
answered Aug 25 '16 at 10:28
WatsonWatson
16k92970
answered Aug 25 '16 at 10:28
WatsonWatson
16k92970
answered Aug 25 '16 at 10:28
WatsonWatson
16k92970
answered Aug 25 '16 at 10:28
WatsonWatson
16k92970
16k92970
$begingroup$
And I take it a non-Hausdorff space is a very uncommon space in applications? That is, when dealing with spaces such as $mathbb{R}^n$ and Hilbert spaces such as $L^2(Omega)$ and $W^{1, 2}(Omega)$ then the closure of a compact set will be compact?
$endgroup$
– csss
Aug 25 '16 at 10:45
$begingroup$
@csss: Yes, in general, e.g. in (functional) analysis, we are dealing with metric spaces, which are Hausdorff, so that the above property always holds.
$endgroup$
– Watson
Aug 25 '16 at 10:46
$begingroup$
If $X$ is a topological space such that every compact subspace is closed, then it is clearly $T1$. This doesn't imply $T2$, see math.stackexchange.com/questions/518787.
$endgroup$
– Watson
Aug 25 '16 at 12:10
1
$begingroup$
@csss: In some branches of mathematics, non-Hausdorff spaces are more common. For example, in dynamics --- the study of group actions on topological spaces --- the orbit space or quotient of a group action is often non-Hausdorff.
$endgroup$
– Lee Mosher
Aug 25 '16 at 15:10
add a comment |
$begingroup$
And I take it a non-Hausdorff space is a very uncommon space in applications? That is, when dealing with spaces such as $mathbb{R}^n$ and Hilbert spaces such as $L^2(Omega)$ and $W^{1, 2}(Omega)$ then the closure of a compact set will be compact?
$endgroup$
– csss
Aug 25 '16 at 10:45
$begingroup$
@csss: Yes, in general, e.g. in (functional) analysis, we are dealing with metric spaces, which are Hausdorff, so that the above property always holds.
$endgroup$
– Watson
Aug 25 '16 at 10:46
$begingroup$
If $X$ is a topological space such that every compact subspace is closed, then it is clearly $T1$. This doesn't imply $T2$, see math.stackexchange.com/questions/518787.
$endgroup$
– Watson
Aug 25 '16 at 12:10
1
$begingroup$
@csss: In some branches of mathematics, non-Hausdorff spaces are more common. For example, in dynamics --- the study of group actions on topological spaces --- the orbit space or quotient of a group action is often non-Hausdorff.
$endgroup$
– Lee Mosher
Aug 25 '16 at 15:10
$begingroup$
And I take it a non-Hausdorff space is a very uncommon space in applications? That is, when dealing with spaces such as $mathbb{R}^n$ and Hilbert spaces such as $L^2(Omega)$ and $W^{1, 2}(Omega)$ then the closure of a compact set will be compact?
$endgroup$
– csss
Aug 25 '16 at 10:45
$begingroup$
And I take it a non-Hausdorff space is a very uncommon space in applications? That is, when dealing with spaces such as $mathbb{R}^n$ and Hilbert spaces such as $L^2(Omega)$ and $W^{1, 2}(Omega)$ then the closure of a compact set will be compact?
$endgroup$
– csss
Aug 25 '16 at 10:45
$begingroup$
@csss: Yes, in general, e.g. in (functional) analysis, we are dealing with metric spaces, which are Hausdorff, so that the above property always holds.
$endgroup$
– Watson
Aug 25 '16 at 10:46
$begingroup$
@csss: Yes, in general, e.g. in (functional) analysis, we are dealing with metric spaces, which are Hausdorff, so that the above property always holds.
$endgroup$
– Watson
Aug 25 '16 at 10:46
$begingroup$
If $X$ is a topological space such that every compact subspace is closed, then it is clearly $T1$. This doesn't imply $T2$, see math.stackexchange.com/questions/518787.
$endgroup$
– Watson
Aug 25 '16 at 12:10
$begingroup$
If $X$ is a topological space such that every compact subspace is closed, then it is clearly $T1$. This doesn't imply $T2$, see math.stackexchange.com/questions/518787.
$endgroup$
– Watson
Aug 25 '16 at 12:10
1
1
$begingroup$
@csss: In some branches of mathematics, non-Hausdorff spaces are more common. For example, in dynamics --- the study of group actions on topological spaces --- the orbit space or quotient of a group action is often non-Hausdorff.
$endgroup$
– Lee Mosher
Aug 25 '16 at 15:10
$begingroup$
@csss: In some branches of mathematics, non-Hausdorff spaces are more common. For example, in dynamics --- the study of group actions on topological spaces --- the orbit space or quotient of a group action is often non-Hausdorff.
$endgroup$
– Lee Mosher
Aug 25 '16 at 15:10
add a comment |
$begingroup$
Actually, the co-finite topology on R, the set of real numbers. Every subset of R is compact. However, it is not every subset of R is closed. There is a special spaces that assumed that every compact set is closed. Certainly, in Hausdorff spaces, every compact set is closed. For your the second question, Take X=R, the set of real numbers. Define topology on X by U is a subset of X is open if and only if U contains 1 or U is the empty set.Then K={1,2} is compact, but cl(K)=R which is not compact with this topology. Actually, every point x in not K, and any open subset U contain x then 1 is also in U. So, the intersection of U and K is not empty, then x is in the closure of K. Therefore, cl(K)=R. The collection U(x)={1,x} forms an open cover of X, obvious there is no finite sub-collection of it cover X.
answered Jan 28 at 4:59
Murtadha SarrayMurtadha Sarray
1
$endgroup$
add a comment |
$begingroup$
Actually, the co-finite topology on R, the set of real numbers. Every subset of R is compact. However, it is not every subset of R is closed. There is a special spaces that assumed that every compact set is closed. Certainly, in Hausdorff spaces, every compact set is closed. For your the second question, Take X=R, the set of real numbers. Define topology on X by U is a subset of X is open if and only if U contains 1 or U is the empty set.Then K={1,2} is compact, but cl(K)=R which is not compact with this topology. Actually, every point x in not K, and any open subset U contain x then 1 is also in U. So, the intersection of U and K is not empty, then x is in the closure of K. Therefore, cl(K)=R. The collection U(x)={1,x} forms an open cover of X, obvious there is no finite sub-collection of it cover X.
answered Jan 28 at 4:59
Murtadha SarrayMurtadha Sarray
1
$endgroup$
add a comment |
$begingroup$
Actually, the co-finite topology on R, the set of real numbers. Every subset of R is compact. However, it is not every subset of R is closed. There is a special spaces that assumed that every compact set is closed. Certainly, in Hausdorff spaces, every compact set is closed. For your the second question, Take X=R, the set of real numbers. Define topology on X by U is a subset of X is open if and only if U contains 1 or U is the empty set.Then K={1,2} is compact, but cl(K)=R which is not compact with this topology. Actually, every point x in not K, and any open subset U contain x then 1 is also in U. So, the intersection of U and K is not empty, then x is in the closure of K. Therefore, cl(K)=R. The collection U(x)={1,x} forms an open cover of X, obvious there is no finite sub-collection of it cover X.
answered Jan 28 at 4:59
Murtadha SarrayMurtadha Sarray
1
$endgroup$
Actually, the co-finite topology on R, the set of real numbers. Every subset of R is compact. However, it is not every subset of R is closed. There is a special spaces that assumed that every compact set is closed. Certainly, in Hausdorff spaces, every compact set is closed. For your the second question, Take X=R, the set of real numbers. Define topology on X by U is a subset of X is open if and only if U contains 1 or U is the empty set.Then K={1,2} is compact, but cl(K)=R which is not compact with this topology. Actually, every point x in not K, and any open subset U contain x then 1 is also in U. So, the intersection of U and K is not empty, then x is in the closure of K. Therefore, cl(K)=R. The collection U(x)={1,x} forms an open cover of X, obvious there is no finite sub-collection of it cover X.
answered Jan 28 at 4:59
Murtadha SarrayMurtadha Sarray
1
answered Jan 28 at 4:59
Murtadha SarrayMurtadha Sarray
1
answered Jan 28 at 4:59
Murtadha SarrayMurtadha Sarray
1
answered Jan 28 at 4:59
Murtadha SarrayMurtadha Sarray
1
1
add a comment |
add a comment |
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5
$begingroup$
A compact subspace of a Hausdorff space is closed.
$endgroup$
– Watson
Aug 25 '16 at 10:08
$begingroup$
@Watson That should be an answer, not just a comment :)
$endgroup$
– 5xum
Aug 25 '16 at 10:13
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@csss: see also math.stackexchange.com/questions/83355/…, math.stackexchange.com/questions/176458
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– Watson
Aug 25 '16 at 12:13