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Is every compact set is closed in any topological space?

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3

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Is every compact set is closed in any topological space ?

My attempt: I can say in usual topologies that every compact set is closed but I'm confused in other topologies.

general-topology

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Is every compact set is closed in any topological space ?

My attempt: I can say in usual topologies that every compact set is closed but I'm confused in other topologies.

general-topology

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edited 7 hours ago

ZeroXLR

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asked 8 hours ago

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$begingroup$

Is every compact set is closed in any topological space ?

My attempt: I can say in usual topologies that every compact set is closed but I'm confused in other topologies.

general-topology

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edited 7 hours ago

ZeroXLR

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jasminejasmine

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share|cite|improve this question

edited 7 hours ago

ZeroXLR

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edited 7 hours ago

ZeroXLR

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3 Answers
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11

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No. The simplest counterexample is arguably this. Consider $X = 0, 1$ with the indiscrete topology i.e. the only two open sets are $emptyset$ and $X = 0, 1$ itself.

Then trivially any subset of $X$ is compact. After all, any open cover at most contains just those two open sets. But, for example, the singleton subset $0$ of $X$ is not closed because its complement $X setminus 0 = 1$ is not one of the two open sets listed above and therefore not open.

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answered 8 hours ago

ZeroXLRZeroXLR

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If the space is Hausdorff you can show any compact set is closed. In general it is not the case. For example, take any finite set with at least two points with the trivial topology then any singleton is compact but not closed.

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user504775user504775

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A different kind of example: The co-finite topology on an infinite set $X.$ That is, any $Asubset X$ is open iff (i) $A=emptyset$ or (ii) $Xsetminus A$ is finite. Equivalently any $Bsubset X$ is closed iff (i') $B=X$ or (ii') $B$ is finite. In this space every subset is compact.

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11

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No. The simplest counterexample is arguably this. Consider $X = 0, 1$ with the indiscrete topology i.e. the only two open sets are $emptyset$ and $X = 0, 1$ itself.

Then trivially any subset of $X$ is compact. After all, any open cover at most contains just those two open sets. But, for example, the singleton subset $0$ of $X$ is not closed because its complement $X setminus 0 = 1$ is not one of the two open sets listed above and therefore not open.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

ZeroXLRZeroXLR

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11

$begingroup$

No. The simplest counterexample is arguably this. Consider $X = 0, 1$ with the indiscrete topology i.e. the only two open sets are $emptyset$ and $X = 0, 1$ itself.

Then trivially any subset of $X$ is compact. After all, any open cover at most contains just those two open sets. But, for example, the singleton subset $0$ of $X$ is not closed because its complement $X setminus 0 = 1$ is not one of the two open sets listed above and therefore not open.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

ZeroXLRZeroXLR

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$endgroup$

add a comment | 

$begingroup$

No. The simplest counterexample is arguably this. Consider $X = 0, 1$ with the indiscrete topology i.e. the only two open sets are $emptyset$ and $X = 0, 1$ itself.

Then trivially any subset of $X$ is compact. After all, any open cover at most contains just those two open sets. But, for example, the singleton subset $0$ of $X$ is not closed because its complement $X setminus 0 = 1$ is not one of the two open sets listed above and therefore not open.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

ZeroXLRZeroXLR

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share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

ZeroXLRZeroXLR

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answered 8 hours ago

ZeroXLRZeroXLR

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answered 8 hours ago

ZeroXLRZeroXLR

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6

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If the space is Hausdorff you can show any compact set is closed. In general it is not the case. For example, take any finite set with at least two points with the trivial topology then any singleton is compact but not closed.

share|cite|improve this answer

answered 8 hours ago

user504775user504775

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6

$begingroup$

If the space is Hausdorff you can show any compact set is closed. In general it is not the case. For example, take any finite set with at least two points with the trivial topology then any singleton is compact but not closed.

share|cite|improve this answer

answered 8 hours ago

user504775user504775

15833 bronze badges

$endgroup$

add a comment | 

$begingroup$

If the space is Hausdorff you can show any compact set is closed. In general it is not the case. For example, take any finite set with at least two points with the trivial topology then any singleton is compact but not closed.

share|cite|improve this answer

answered 8 hours ago

user504775user504775

15833 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

user504775user504775

15833 bronze badges

answered 8 hours ago

user504775user504775

15833 bronze badges

answered 8 hours ago

user504775user504775

15833 bronze badges

1

$begingroup$

A different kind of example: The co-finite topology on an infinite set $X.$ That is, any $Asubset X$ is open iff (i) $A=emptyset$ or (ii) $Xsetminus A$ is finite. Equivalently any $Bsubset X$ is closed iff (i') $B=X$ or (ii') $B$ is finite. In this space every subset is compact.

share|cite|improve this answer

answered 5 hours ago

DanielWainfleetDanielWainfleet

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$endgroup$

add a comment | 

1

$begingroup$

A different kind of example: The co-finite topology on an infinite set $X.$ That is, any $Asubset X$ is open iff (i) $A=emptyset$ or (ii) $Xsetminus A$ is finite. Equivalently any $Bsubset X$ is closed iff (i') $B=X$ or (ii') $B$ is finite. In this space every subset is compact.

share|cite|improve this answer

answered 5 hours ago

DanielWainfleetDanielWainfleet

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$endgroup$

add a comment | 

$begingroup$

A different kind of example: The co-finite topology on an infinite set $X.$ That is, any $Asubset X$ is open iff (i) $A=emptyset$ or (ii) $Xsetminus A$ is finite. Equivalently any $Bsubset X$ is closed iff (i') $B=X$ or (ii') $B$ is finite. In this space every subset is compact.

share|cite|improve this answer

answered 5 hours ago

DanielWainfleetDanielWainfleet

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$endgroup$

share|cite|improve this answer

answered 5 hours ago

DanielWainfleetDanielWainfleet

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answered 5 hours ago

DanielWainfleetDanielWainfleet

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answered 5 hours ago

DanielWainfleetDanielWainfleet

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