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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 6 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 6 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 6 hours ago

User12239User12239

367216

$endgroup$

share|cite|improve this question

asked 6 hours ago

User12239User12239

367216

share|cite|improve this question

asked 6 hours ago

User12239User12239

367216

asked 6 hours ago

User12239User12239

367216

asked 6 hours ago

User12239User12239

367216

2 Answers
2

active

oldest

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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 $S$, $overline Vcap Sneq emptyset$ $implies Ucap Sneq emptyset$.

share|cite|improve this answer

edited 6 hours ago

answered 6 hours ago

Thomas ShelbyThomas Shelby

4,7112727

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks I’m looking them up
  $endgroup$
  – User12239
  5 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 6 hours ago

MaksimMaksim

1,00719

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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 6 hours ago

answered 6 hours ago

Thomas ShelbyThomas Shelby

4,7112727

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks I’m looking them up
  $endgroup$
  – User12239
  5 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 6 hours ago

answered 6 hours ago

Thomas ShelbyThomas Shelby

4,7112727

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks I’m looking them up
  $endgroup$
  – User12239
  5 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 6 hours ago

answered 6 hours ago

Thomas ShelbyThomas Shelby

4,7112727

$endgroup$

share|cite|improve this answer

edited 6 hours ago

answered 6 hours ago

Thomas ShelbyThomas Shelby

4,7112727

answered 6 hours ago

Thomas ShelbyThomas Shelby

4,7112727

answered 6 hours ago

Thomas ShelbyThomas Shelby

4,7112727

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 6 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 6 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 6 hours ago

MaksimMaksim

1,00719

$endgroup$

share|cite|improve this answer

answered 6 hours ago

MaksimMaksim

1,00719

answered 6 hours ago

MaksimMaksim

1,00719

answered 6 hours ago

MaksimMaksim

1,00719