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a relationship between local compactness and closure

Equivalent definition of locally compact when $X$ is Hausdorff. How did they get $overlineV cap C$ is empty?Lifting local compactness in covering spacesLocal compactness exerciseLocal compactness in an open subsetcompact and locally Hausdorff, but not locally compactWhat is the relationship between completeness and local compactness?Definition of local compactnessTwo (maybe nonequivalent) definitions of local compactnessCompact Hausdorff space, local compactnessCompact Hausdorff Spaces and their local compactnessThe local compactness and being Hausdorff

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Suppose that $X$ is a Hausdorff locally compact space and $S$ a subset of $X$. Let $xin X$ and suppose that every compact neighborhood of $x$ intersects $S$. Does it follow that $x$ lies in the closure of $S$?

general-topology

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asked 13 hours ago

User12239User12239

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3

1

$begingroup$

Suppose that $X$ is a Hausdorff locally compact space and $S$ a subset of $X$. Let $xin X$ and suppose that every compact neighborhood of $x$ intersects $S$. Does it follow that $x$ lies in the closure of $S$?

general-topology

share|cite|improve this question

asked 13 hours ago

User12239User12239

367216

$endgroup$

add a comment | 

$begingroup$

Suppose that $X$ is a Hausdorff locally compact space and $S$ a subset of $X$. Let $xin X$ and suppose that every compact neighborhood of $x$ intersects $S$. Does it follow that $x$ lies in the closure of $S$?

general-topology

share|cite|improve this question

asked 13 hours ago

User12239User12239

367216

$endgroup$

share|cite|improve this question

asked 13 hours ago

User12239User12239

367216

share|cite|improve this question

asked 13 hours ago

User12239User12239

367216

asked 13 hours ago

User12239User12239

367216

asked 13 hours ago

User12239User12239

367216

2 Answers
2

active

oldest

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2

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Yes, it is true. Since $X$ is locally compact and Hausdorff, given a neighborhood $U$ of $x$ in $X$, we can find a neighborhood $V$ such that $overline V$ is compact and $overline Vsubseteq U$. As every compact neighborhood of $x$ intersects with $S$, $overline Vcap Sneq emptyset$ $implies Ucap Sneq emptyset$.

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edited 13 hours ago

answered 13 hours ago

Thomas ShelbyThomas Shelby

4,7362727

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes I was doubting if every neighborhood has a compact neighborhood so I need to prove this fact now
  $endgroup$
  – User12239
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can find a proof in 'Topology' by Munkres.
  $endgroup$
  – Thomas Shelby
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also see this.
  $endgroup$
  – Thomas Shelby
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks I’m looking them up
  $endgroup$
  – User12239
  12 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

In this case ($X$ being locally compact Hausdorff) for every point $x$ of $X$, every neighbourhood of $x$ contains a compact neighbourhood of $x$ - see.

Then any neighbourhood of $x$ contains a compact neighbourhood which intersects $S$, so $xin barS$

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answered 13 hours ago

MaksimMaksim

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2

$begingroup$

Yes, it is true. Since $X$ is locally compact and Hausdorff, given a neighborhood $U$ of $x$ in $X$, we can find a neighborhood $V$ such that $overline V$ is compact and $overline Vsubseteq U$. As every compact neighborhood of $x$ intersects with $S$, $overline Vcap Sneq emptyset$ $implies Ucap Sneq emptyset$.

share|cite|improve this answer

edited 13 hours ago

answered 13 hours ago

Thomas ShelbyThomas Shelby

4,7362727

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes I was doubting if every neighborhood has a compact neighborhood so I need to prove this fact now
  $endgroup$
  – User12239
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can find a proof in 'Topology' by Munkres.
  $endgroup$
  – Thomas Shelby
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also see this.
  $endgroup$
  – Thomas Shelby
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks I’m looking them up
  $endgroup$
  – User12239
  12 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

Yes, it is true. Since $X$ is locally compact and Hausdorff, given a neighborhood $U$ of $x$ in $X$, we can find a neighborhood $V$ such that $overline V$ is compact and $overline Vsubseteq U$. As every compact neighborhood of $x$ intersects with $S$, $overline Vcap Sneq emptyset$ $implies Ucap Sneq emptyset$.

share|cite|improve this answer

edited 13 hours ago

answered 13 hours ago

Thomas ShelbyThomas Shelby

4,7362727

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes I was doubting if every neighborhood has a compact neighborhood so I need to prove this fact now
  $endgroup$
  – User12239
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can find a proof in 'Topology' by Munkres.
  $endgroup$
  – Thomas Shelby
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also see this.
  $endgroup$
  – Thomas Shelby
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks I’m looking them up
  $endgroup$
  – User12239
  12 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Yes, it is true. Since $X$ is locally compact and Hausdorff, given a neighborhood $U$ of $x$ in $X$, we can find a neighborhood $V$ such that $overline V$ is compact and $overline Vsubseteq U$. As every compact neighborhood of $x$ intersects with $S$, $overline Vcap Sneq emptyset$ $implies Ucap Sneq emptyset$.

share|cite|improve this answer

edited 13 hours ago

answered 13 hours ago

Thomas ShelbyThomas Shelby

4,7362727

$endgroup$

share|cite|improve this answer

edited 13 hours ago

answered 13 hours ago

Thomas ShelbyThomas Shelby

4,7362727

answered 13 hours ago

Thomas ShelbyThomas Shelby

4,7362727

answered 13 hours ago

Thomas ShelbyThomas Shelby

4,7362727

2

$begingroup$

In this case ($X$ being locally compact Hausdorff) for every point $x$ of $X$, every neighbourhood of $x$ contains a compact neighbourhood of $x$ - see.

Then any neighbourhood of $x$ contains a compact neighbourhood which intersects $S$, so $xin barS$

share|cite|improve this answer

answered 13 hours ago

MaksimMaksim

1,00719

$endgroup$

add a comment | 

2

$begingroup$

In this case ($X$ being locally compact Hausdorff) for every point $x$ of $X$, every neighbourhood of $x$ contains a compact neighbourhood of $x$ - see.

Then any neighbourhood of $x$ contains a compact neighbourhood which intersects $S$, so $xin barS$

share|cite|improve this answer

answered 13 hours ago

MaksimMaksim

1,00719

$endgroup$

add a comment | 

$begingroup$

In this case ($X$ being locally compact Hausdorff) for every point $x$ of $X$, every neighbourhood of $x$ contains a compact neighbourhood of $x$ - see.

Then any neighbourhood of $x$ contains a compact neighbourhood which intersects $S$, so $xin barS$

share|cite|improve this answer

answered 13 hours ago

MaksimMaksim

1,00719

$endgroup$

share|cite|improve this answer

answered 13 hours ago

MaksimMaksim

1,00719

answered 13 hours ago

MaksimMaksim

1,00719

answered 13 hours ago

MaksimMaksim

1,00719