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Counterexample finite intersection property

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Counterexample finite intersection property

Countably Compact vs Compact vs Finite Intersection PropertyUnions and intersections of compact subsetsSome compactness related propertyDoes this characterize compactness?Kelley's topology: Compact spaces and finite intersection propertyProve $X$ is compact if every family of closed sets with the finite intersection property have non-empty intersectionEvery closed subspace of a compact space is compact using finite intersection propertyIntersection of a family of compact sets having finite intersection property in a Hausdorff spaceFinite intersection property for a subsetCheck My Proof: Finite Intersection Property

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}

4

$begingroup$

Let $(X,mathcal T)$ be a (not necessarily Hausdorff) topological space. It is well-known that $X$ is compact (in the sense of every open cover has a finite subcover) if and only if for every family of closed subsets $(C_i)_{iin I}$ of $X$ satisfying the finite intersection property, $bigcap _{iin I}C_i$ is nonempty. Now does the following assertion hold:

If $(X,mathcal T)$ is compact and $(C_i)_{iin I}$ is a family of $textbf{compact}$ subsets satisfying the finite intersection property, then $bigcap_{iin I}C_i$ is compact.

Since compact subsets in non-Hausdorff spaces need not be closed, I am not sure if this holds. Is there a counterexample?

general-topology compactness

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edited 9 hours ago

Eric Wofsey

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asked 9 hours ago

lasik43lasik43

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add a comment | 

4

$begingroup$

Let $(X,mathcal T)$ be a (not necessarily Hausdorff) topological space. It is well-known that $X$ is compact (in the sense of every open cover has a finite subcover) if and only if for every family of closed subsets $(C_i)_{iin I}$ of $X$ satisfying the finite intersection property, $bigcap _{iin I}C_i$ is nonempty. Now does the following assertion hold:

If $(X,mathcal T)$ is compact and $(C_i)_{iin I}$ is a family of $textbf{compact}$ subsets satisfying the finite intersection property, then $bigcap_{iin I}C_i$ is compact.

Since compact subsets in non-Hausdorff spaces need not be closed, I am not sure if this holds. Is there a counterexample?

general-topology compactness

share|cite|improve this question

edited 9 hours ago

Eric Wofsey

205k1414 gold badges239239 silver badges373373 bronze badges

asked 9 hours ago

lasik43lasik43

33511 silver badge1212 bronze badges

$endgroup$

add a comment | 

$begingroup$

Let $(X,mathcal T)$ be a (not necessarily Hausdorff) topological space. It is well-known that $X$ is compact (in the sense of every open cover has a finite subcover) if and only if for every family of closed subsets $(C_i)_{iin I}$ of $X$ satisfying the finite intersection property, $bigcap _{iin I}C_i$ is nonempty. Now does the following assertion hold:

If $(X,mathcal T)$ is compact and $(C_i)_{iin I}$ is a family of $textbf{compact}$ subsets satisfying the finite intersection property, then $bigcap_{iin I}C_i$ is compact.

Since compact subsets in non-Hausdorff spaces need not be closed, I am not sure if this holds. Is there a counterexample?

general-topology compactness

share|cite|improve this question

edited 9 hours ago

Eric Wofsey

205k1414 gold badges239239 silver badges373373 bronze badges

asked 9 hours ago

lasik43lasik43

33511 silver badge1212 bronze badges

$endgroup$

share|cite|improve this question

edited 9 hours ago

Eric Wofsey

205k1414 gold badges239239 silver badges373373 bronze badges

asked 9 hours ago

lasik43lasik43

33511 silver badge1212 bronze badges

share|cite|improve this question

edited 9 hours ago

Eric Wofsey

205k1414 gold badges239239 silver badges373373 bronze badges

asked 9 hours ago

lasik43lasik43

33511 silver badge1212 bronze badges

edited 9 hours ago

Eric Wofsey

205k1414 gold badges239239 silver badges373373 bronze badges

edited 9 hours ago

Eric Wofsey

205k1414 gold badges239239 silver badges373373 bronze badges

asked 9 hours ago

lasik43lasik43

33511 silver badge1212 bronze badges

asked 9 hours ago

lasik43lasik43

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2 Answers
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5

$begingroup$

This is not true in general. For instance, let $X=[0,1]$ with the topology that a set is open iff it is downward closed (i.e., $xin U$ and $yleq x$ implies $yin U$). Then $X$ is compact since any open set containing $1$ is the whole space. Now for $iin (1/2,1]$, let $C_i=[0,1/2)cup(1/2,i]$. Each $C_i$ is similarly compact, and they have the finite intersection property. But their intersection $[0,1/2)$ is not compact, since it is covered by the open sets $[0,r)$ for $r<1/2$.

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

Eric WofseyEric Wofsey

205k1414 gold badges239239 silver badges373373 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice example. Thanks!
  $endgroup$
  – lasik43
  8 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

Take the set $X=mathbb{N}cup{a_1,a_2}$ where $a_1,a_2$ are some points outside of $mathbb{N}$. We define the following topology: a set is open if it is $mathbb{N}cup{a_1},mathbb{N}cup{a_2}$, $mathbb{N}cup{a_1,a_2}$ or any subset of $mathbb{N}$. It is easy to see that this is a compact topological space and $mathbb{N}cup{a_1},mathbb{N}cup{a_2}$ are compact subsets with non empty intersection. However, their intersection is $mathbb{N}$ which is not compact.

share|cite|improve this answer

answered 9 hours ago

MarkMark

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$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also an interesting example. Thanks. :)
  $endgroup$
  – lasik43
  8 hours ago
  

  
  
  
  

add a comment | 

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5

$begingroup$

This is not true in general. For instance, let $X=[0,1]$ with the topology that a set is open iff it is downward closed (i.e., $xin U$ and $yleq x$ implies $yin U$). Then $X$ is compact since any open set containing $1$ is the whole space. Now for $iin (1/2,1]$, let $C_i=[0,1/2)cup(1/2,i]$. Each $C_i$ is similarly compact, and they have the finite intersection property. But their intersection $[0,1/2)$ is not compact, since it is covered by the open sets $[0,r)$ for $r<1/2$.

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

Eric WofseyEric Wofsey

205k1414 gold badges239239 silver badges373373 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice example. Thanks!
  $endgroup$
  – lasik43
  8 hours ago
  

  
  
  
  

add a comment | 

5

$begingroup$

This is not true in general. For instance, let $X=[0,1]$ with the topology that a set is open iff it is downward closed (i.e., $xin U$ and $yleq x$ implies $yin U$). Then $X$ is compact since any open set containing $1$ is the whole space. Now for $iin (1/2,1]$, let $C_i=[0,1/2)cup(1/2,i]$. Each $C_i$ is similarly compact, and they have the finite intersection property. But their intersection $[0,1/2)$ is not compact, since it is covered by the open sets $[0,r)$ for $r<1/2$.

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

Eric WofseyEric Wofsey

205k1414 gold badges239239 silver badges373373 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice example. Thanks!
  $endgroup$
  – lasik43
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

This is not true in general. For instance, let $X=[0,1]$ with the topology that a set is open iff it is downward closed (i.e., $xin U$ and $yleq x$ implies $yin U$). Then $X$ is compact since any open set containing $1$ is the whole space. Now for $iin (1/2,1]$, let $C_i=[0,1/2)cup(1/2,i]$. Each $C_i$ is similarly compact, and they have the finite intersection property. But their intersection $[0,1/2)$ is not compact, since it is covered by the open sets $[0,r)$ for $r<1/2$.

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

Eric WofseyEric Wofsey

205k1414 gold badges239239 silver badges373373 bronze badges

$endgroup$

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

Eric WofseyEric Wofsey

205k1414 gold badges239239 silver badges373373 bronze badges

answered 9 hours ago

Eric WofseyEric Wofsey

205k1414 gold badges239239 silver badges373373 bronze badges

answered 9 hours ago

Eric WofseyEric Wofsey

205k1414 gold badges239239 silver badges373373 bronze badges

3

$begingroup$

Take the set $X=mathbb{N}cup{a_1,a_2}$ where $a_1,a_2$ are some points outside of $mathbb{N}$. We define the following topology: a set is open if it is $mathbb{N}cup{a_1},mathbb{N}cup{a_2}$, $mathbb{N}cup{a_1,a_2}$ or any subset of $mathbb{N}$. It is easy to see that this is a compact topological space and $mathbb{N}cup{a_1},mathbb{N}cup{a_2}$ are compact subsets with non empty intersection. However, their intersection is $mathbb{N}$ which is not compact.

share|cite|improve this answer

answered 9 hours ago

MarkMark

14.4k11 gold badge88 silver badges2525 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also an interesting example. Thanks. :)
  $endgroup$
  – lasik43
  8 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

Take the set $X=mathbb{N}cup{a_1,a_2}$ where $a_1,a_2$ are some points outside of $mathbb{N}$. We define the following topology: a set is open if it is $mathbb{N}cup{a_1},mathbb{N}cup{a_2}$, $mathbb{N}cup{a_1,a_2}$ or any subset of $mathbb{N}$. It is easy to see that this is a compact topological space and $mathbb{N}cup{a_1},mathbb{N}cup{a_2}$ are compact subsets with non empty intersection. However, their intersection is $mathbb{N}$ which is not compact.

share|cite|improve this answer

answered 9 hours ago

MarkMark

14.4k11 gold badge88 silver badges2525 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also an interesting example. Thanks. :)
  $endgroup$
  – lasik43
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Take the set $X=mathbb{N}cup{a_1,a_2}$ where $a_1,a_2$ are some points outside of $mathbb{N}$. We define the following topology: a set is open if it is $mathbb{N}cup{a_1},mathbb{N}cup{a_2}$, $mathbb{N}cup{a_1,a_2}$ or any subset of $mathbb{N}$. It is easy to see that this is a compact topological space and $mathbb{N}cup{a_1},mathbb{N}cup{a_2}$ are compact subsets with non empty intersection. However, their intersection is $mathbb{N}$ which is not compact.

share|cite|improve this answer

answered 9 hours ago

MarkMark

14.4k11 gold badge88 silver badges2525 bronze badges

$endgroup$

share|cite|improve this answer

answered 9 hours ago

MarkMark

14.4k11 gold badge88 silver badges2525 bronze badges

answered 9 hours ago

MarkMark

14.4k11 gold badge88 silver badges2525 bronze badges

answered 9 hours ago

MarkMark

14.4k11 gold badge88 silver badges2525 bronze badges