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Compact subset in two different topologies
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Let $(X,tau_1)$ and $(Y,tau_2)$ be two Hausdorff topological spaces such that $Xsubset Y$. Let $Ksubset X$ be compact. Is $K$ also compact in $Y$? If not, what if both $X$ and $Y$ are compact spaces?
general-topology
asked Jul 22 at 10:25
Ronald
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Let $(X,tau_1)$ and $(Y,tau_2)$ be two Hausdorff topological spaces such that $Xsubset Y$. Let $Ksubset X$ be compact. Is $K$ also compact in $Y$? If not, what if both $X$ and $Y$ are compact spaces?
general-topology
asked Jul 22 at 10:25
Ronald
1,5841821
2
A counterexample for the first question: $X=mathbb R$ with Euclidean topology, $Y=mathbb R$ with discrete topology. Both are metrizable, thus Hausdorff. $K=[0,1]$ is compact in $X$, but not compact in $Y$.
– suhogrozdje
Jul 22 at 10:51
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Let $(X,tau_1)$ and $(Y,tau_2)$ be two Hausdorff topological spaces such that $Xsubset Y$. Let $Ksubset X$ be compact. Is $K$ also compact in $Y$? If not, what if both $X$ and $Y$ are compact spaces?
general-topology
asked Jul 22 at 10:25
Ronald
1,5841821
Let $(X,tau_1)$ and $(Y,tau_2)$ be two Hausdorff topological spaces such that $Xsubset Y$. Let $Ksubset X$ be compact. Is $K$ also compact in $Y$? If not, what if both $X$ and $Y$ are compact spaces?
general-topology
asked Jul 22 at 10:25
Ronald
1,5841821
asked Jul 22 at 10:25
Ronald
1,5841821
asked Jul 22 at 10:25
Ronald
1,5841821
asked Jul 22 at 10:25
Ronald
1,5841821
1,5841821
2
A counterexample for the first question: $X=mathbb R$ with Euclidean topology, $Y=mathbb R$ with discrete topology. Both are metrizable, thus Hausdorff. $K=[0,1]$ is compact in $X$, but not compact in $Y$.
– suhogrozdje
Jul 22 at 10:51
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2
A counterexample for the first question: $X=mathbb R$ with Euclidean topology, $Y=mathbb R$ with discrete topology. Both are metrizable, thus Hausdorff. $K=[0,1]$ is compact in $X$, but not compact in $Y$.
– suhogrozdje
Jul 22 at 10:51
2
2
A counterexample for the first question: $X=mathbb R$ with Euclidean topology, $Y=mathbb R$ with discrete topology. Both are metrizable, thus Hausdorff. $K=[0,1]$ is compact in $X$, but not compact in $Y$.
– suhogrozdje
Jul 22 at 10:51
A counterexample for the first question: $X=mathbb R$ with Euclidean topology, $Y=mathbb R$ with discrete topology. Both are metrizable, thus Hausdorff. $K=[0,1]$ is compact in $X$, but not compact in $Y$.
– suhogrozdje
Jul 22 at 10:51
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2 Answers
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2
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Compactness depends on the topology, so if $Xsubset Y$ as a set, but not as a subspace, you cannot guarantee that $K$ will be compact with both topologies. For example, the one suhogrozdje gave in the comments, take $X=mathbbR$ with Euclidean topology and $Y=mathbbR$ with discrete topology. Then, $K=[0,1]$ is compact in $X$, but not compact in $Y$ (can you see why?).
Conversely, if $X$ is a subspace of $Y$, then $K$ is compact in $Y$, since for any open cover in $Y$, you can restrict it to $X$, where you have a finite cover by sets of the form $Xcap U$ with $Uintau_2$. Then just use the sets $U$ to finitely cover $K$ in $Y$.
Compactness is independent of the ambient space, i.e., requiring $X$ and $Y$ to be compact doesn't actually add anything, unless you demand some extra properties on $K$ like being closed in $Y$ (a closed subset of a compact space is always compact).
answered Jul 22 at 10:40
Javi
2,1631725
So in case subspace topology everything follows?
– Ronald
Jul 22 at 10:47
Yes, since in that case you can always construct a finite cover using open subsets of $Y$ from a finite cover in $X$. I've added it to the answer
– Javi
Jul 22 at 10:50
add a comment |Â
up vote
1
down vote
If $Ksubseteq X$ then $K$ is compact in $X$ if and only if $K$ is a compact topological space when it is equipped with the subspace topology inherited from $X$.
So if also $Ksubseteq Y$ then it will be a compact subset of $Y$ if it inherits from $Y$ the same subspace topology as it inherits from $X$. This will be the case if $X$ is a subspace of $Y$ (which is a stronger condition than only $Xsubseteq Y$).
Moreover it concerns a sufficient condition (not a necessary condition) for $K$ being compact.
answered Jul 22 at 10:50
drhab
86.4k541118
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Compactness depends on the topology, so if $Xsubset Y$ as a set, but not as a subspace, you cannot guarantee that $K$ will be compact with both topologies. For example, the one suhogrozdje gave in the comments, take $X=mathbbR$ with Euclidean topology and $Y=mathbbR$ with discrete topology. Then, $K=[0,1]$ is compact in $X$, but not compact in $Y$ (can you see why?).
Conversely, if $X$ is a subspace of $Y$, then $K$ is compact in $Y$, since for any open cover in $Y$, you can restrict it to $X$, where you have a finite cover by sets of the form $Xcap U$ with $Uintau_2$. Then just use the sets $U$ to finitely cover $K$ in $Y$.
Compactness is independent of the ambient space, i.e., requiring $X$ and $Y$ to be compact doesn't actually add anything, unless you demand some extra properties on $K$ like being closed in $Y$ (a closed subset of a compact space is always compact).
answered Jul 22 at 10:40
Javi
2,1631725
So in case subspace topology everything follows?
– Ronald
Jul 22 at 10:47
Yes, since in that case you can always construct a finite cover using open subsets of $Y$ from a finite cover in $X$. I've added it to the answer
– Javi
Jul 22 at 10:50
add a comment |Â
up vote
2
down vote
accepted
Compactness depends on the topology, so if $Xsubset Y$ as a set, but not as a subspace, you cannot guarantee that $K$ will be compact with both topologies. For example, the one suhogrozdje gave in the comments, take $X=mathbbR$ with Euclidean topology and $Y=mathbbR$ with discrete topology. Then, $K=[0,1]$ is compact in $X$, but not compact in $Y$ (can you see why?).
Conversely, if $X$ is a subspace of $Y$, then $K$ is compact in $Y$, since for any open cover in $Y$, you can restrict it to $X$, where you have a finite cover by sets of the form $Xcap U$ with $Uintau_2$. Then just use the sets $U$ to finitely cover $K$ in $Y$.
Compactness is independent of the ambient space, i.e., requiring $X$ and $Y$ to be compact doesn't actually add anything, unless you demand some extra properties on $K$ like being closed in $Y$ (a closed subset of a compact space is always compact).
answered Jul 22 at 10:40
Javi
2,1631725
So in case subspace topology everything follows?
– Ronald
Jul 22 at 10:47
Yes, since in that case you can always construct a finite cover using open subsets of $Y$ from a finite cover in $X$. I've added it to the answer
– Javi
Jul 22 at 10:50
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Compactness depends on the topology, so if $Xsubset Y$ as a set, but not as a subspace, you cannot guarantee that $K$ will be compact with both topologies. For example, the one suhogrozdje gave in the comments, take $X=mathbbR$ with Euclidean topology and $Y=mathbbR$ with discrete topology. Then, $K=[0,1]$ is compact in $X$, but not compact in $Y$ (can you see why?).
Conversely, if $X$ is a subspace of $Y$, then $K$ is compact in $Y$, since for any open cover in $Y$, you can restrict it to $X$, where you have a finite cover by sets of the form $Xcap U$ with $Uintau_2$. Then just use the sets $U$ to finitely cover $K$ in $Y$.
Compactness is independent of the ambient space, i.e., requiring $X$ and $Y$ to be compact doesn't actually add anything, unless you demand some extra properties on $K$ like being closed in $Y$ (a closed subset of a compact space is always compact).
answered Jul 22 at 10:40
Javi
2,1631725
Compactness depends on the topology, so if $Xsubset Y$ as a set, but not as a subspace, you cannot guarantee that $K$ will be compact with both topologies. For example, the one suhogrozdje gave in the comments, take $X=mathbbR$ with Euclidean topology and $Y=mathbbR$ with discrete topology. Then, $K=[0,1]$ is compact in $X$, but not compact in $Y$ (can you see why?).
Conversely, if $X$ is a subspace of $Y$, then $K$ is compact in $Y$, since for any open cover in $Y$, you can restrict it to $X$, where you have a finite cover by sets of the form $Xcap U$ with $Uintau_2$. Then just use the sets $U$ to finitely cover $K$ in $Y$.
Compactness is independent of the ambient space, i.e., requiring $X$ and $Y$ to be compact doesn't actually add anything, unless you demand some extra properties on $K$ like being closed in $Y$ (a closed subset of a compact space is always compact).
answered Jul 22 at 10:40
Javi
2,1631725
edited Jul 22 at 11:03
answered Jul 22 at 10:40
Javi
2,1631725
answered Jul 22 at 10:40
Javi
2,1631725
answered Jul 22 at 10:40
Javi
2,1631725
2,1631725
So in case subspace topology everything follows?
– Ronald
Jul 22 at 10:47
Yes, since in that case you can always construct a finite cover using open subsets of $Y$ from a finite cover in $X$. I've added it to the answer
– Javi
Jul 22 at 10:50
add a comment |Â
So in case subspace topology everything follows?
– Ronald
Jul 22 at 10:47
Yes, since in that case you can always construct a finite cover using open subsets of $Y$ from a finite cover in $X$. I've added it to the answer
– Javi
Jul 22 at 10:50
So in case subspace topology everything follows?
– Ronald
Jul 22 at 10:47
So in case subspace topology everything follows?
– Ronald
Jul 22 at 10:47
Yes, since in that case you can always construct a finite cover using open subsets of $Y$ from a finite cover in $X$. I've added it to the answer
– Javi
Jul 22 at 10:50
Yes, since in that case you can always construct a finite cover using open subsets of $Y$ from a finite cover in $X$. I've added it to the answer
– Javi
Jul 22 at 10:50
add a comment |Â
up vote
1
down vote
If $Ksubseteq X$ then $K$ is compact in $X$ if and only if $K$ is a compact topological space when it is equipped with the subspace topology inherited from $X$.
So if also $Ksubseteq Y$ then it will be a compact subset of $Y$ if it inherits from $Y$ the same subspace topology as it inherits from $X$. This will be the case if $X$ is a subspace of $Y$ (which is a stronger condition than only $Xsubseteq Y$).
Moreover it concerns a sufficient condition (not a necessary condition) for $K$ being compact.
answered Jul 22 at 10:50
drhab
86.4k541118
add a comment |Â
up vote
1
down vote
If $Ksubseteq X$ then $K$ is compact in $X$ if and only if $K$ is a compact topological space when it is equipped with the subspace topology inherited from $X$.
So if also $Ksubseteq Y$ then it will be a compact subset of $Y$ if it inherits from $Y$ the same subspace topology as it inherits from $X$. This will be the case if $X$ is a subspace of $Y$ (which is a stronger condition than only $Xsubseteq Y$).
Moreover it concerns a sufficient condition (not a necessary condition) for $K$ being compact.
answered Jul 22 at 10:50
drhab
86.4k541118
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If $Ksubseteq X$ then $K$ is compact in $X$ if and only if $K$ is a compact topological space when it is equipped with the subspace topology inherited from $X$.
So if also $Ksubseteq Y$ then it will be a compact subset of $Y$ if it inherits from $Y$ the same subspace topology as it inherits from $X$. This will be the case if $X$ is a subspace of $Y$ (which is a stronger condition than only $Xsubseteq Y$).
Moreover it concerns a sufficient condition (not a necessary condition) for $K$ being compact.
answered Jul 22 at 10:50
drhab
86.4k541118
If $Ksubseteq X$ then $K$ is compact in $X$ if and only if $K$ is a compact topological space when it is equipped with the subspace topology inherited from $X$.
So if also $Ksubseteq Y$ then it will be a compact subset of $Y$ if it inherits from $Y$ the same subspace topology as it inherits from $X$. This will be the case if $X$ is a subspace of $Y$ (which is a stronger condition than only $Xsubseteq Y$).
Moreover it concerns a sufficient condition (not a necessary condition) for $K$ being compact.
answered Jul 22 at 10:50
drhab
86.4k541118
edited Jul 22 at 10:56
answered Jul 22 at 10:50
drhab
86.4k541118
answered Jul 22 at 10:50
drhab
86.4k541118
answered Jul 22 at 10:50
drhab
86.4k541118
86.4k541118
add a comment |Â
add a comment |Â
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2
A counterexample for the first question: $X=mathbb R$ with Euclidean topology, $Y=mathbb R$ with discrete topology. Both are metrizable, thus Hausdorff. $K=[0,1]$ is compact in $X$, but not compact in $Y$.
– suhogrozdje
Jul 22 at 10:51