a relationship between local compactness and closure The 2019 Stack Overflow Developer Survey Results Are InEquivalent 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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a relationship between local compactness and closure
The 2019 Stack Overflow Developer Survey Results Are InEquivalent 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
asked Apr 6 at 15:08
User12239User12239
364216
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add a comment |
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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
asked Apr 6 at 15:08
User12239User12239
364216
$endgroup$
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
asked Apr 6 at 15:08
User12239User12239
364216
$endgroup$
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
general-topology
asked Apr 6 at 15:08
User12239User12239
364216
asked Apr 6 at 15:08
User12239User12239
364216
asked Apr 6 at 15:08
User12239User12239
364216
asked Apr 6 at 15:08
User12239User12239
364216
asked Apr 6 at 15:08
User12239User12239
364216
364216
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
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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 $S$, $overline Vcap Sneq emptyset$ $implies Ucap Sneq emptyset$.
answered Apr 6 at 15:20
Thomas ShelbyThomas Shelby
4,7362727
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$begingroup$
Yes I was doubting if every neighborhood has a compact neighborhood so I need to prove this fact now
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– User12239
Apr 6 at 15:26
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You can find a proof in 'Topology' by Munkres.
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– Thomas Shelby
Apr 6 at 15:28
$begingroup$
Also see this.
$endgroup$
– Thomas Shelby
Apr 6 at 15:31
1
$begingroup$
Thanks I’m looking them up
$endgroup$
– User12239
Apr 6 at 15:33
add a comment |
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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 $xin barS$
answered Apr 6 at 15:18
MaksimMaksim
1,01719
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
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 $S$, $overline Vcap Sneq emptyset$ $implies Ucap Sneq emptyset$.
answered Apr 6 at 15:20
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
Apr 6 at 15:26
$begingroup$
You can find a proof in 'Topology' by Munkres.
$endgroup$
– Thomas Shelby
Apr 6 at 15:28
$begingroup$
Also see this.
$endgroup$
– Thomas Shelby
Apr 6 at 15:31
1
$begingroup$
Thanks I’m looking them up
$endgroup$
– User12239
Apr 6 at 15:33
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 $S$, $overline Vcap Sneq emptyset$ $implies Ucap Sneq emptyset$.
answered Apr 6 at 15:20
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
Apr 6 at 15:26
$begingroup$
You can find a proof in 'Topology' by Munkres.
$endgroup$
– Thomas Shelby
Apr 6 at 15:28
$begingroup$
Also see this.
$endgroup$
– Thomas Shelby
Apr 6 at 15:31
1
$begingroup$
Thanks I’m looking them up
$endgroup$
– User12239
Apr 6 at 15:33
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 $S$, $overline Vcap Sneq emptyset$ $implies Ucap Sneq emptyset$.
answered Apr 6 at 15:20
Thomas ShelbyThomas Shelby
4,7362727
$endgroup$
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 $S$, $overline Vcap Sneq emptyset$ $implies Ucap Sneq emptyset$.
answered Apr 6 at 15:20
Thomas ShelbyThomas Shelby
4,7362727
edited Apr 6 at 15:25
answered Apr 6 at 15:20
Thomas ShelbyThomas Shelby
4,7362727
answered Apr 6 at 15:20
Thomas ShelbyThomas Shelby
4,7362727
answered Apr 6 at 15:20
Thomas ShelbyThomas Shelby
4,7362727
4,7362727
$begingroup$
Yes I was doubting if every neighborhood has a compact neighborhood so I need to prove this fact now
$endgroup$
– User12239
Apr 6 at 15:26
$begingroup$
You can find a proof in 'Topology' by Munkres.
$endgroup$
– Thomas Shelby
Apr 6 at 15:28
$begingroup$
Also see this.
$endgroup$
– Thomas Shelby
Apr 6 at 15:31
1
$begingroup$
Thanks I’m looking them up
$endgroup$
– User12239
Apr 6 at 15:33
add a comment |
$begingroup$
Yes I was doubting if every neighborhood has a compact neighborhood so I need to prove this fact now
$endgroup$
– User12239
Apr 6 at 15:26
$begingroup$
You can find a proof in 'Topology' by Munkres.
$endgroup$
– Thomas Shelby
Apr 6 at 15:28
$begingroup$
Also see this.
$endgroup$
– Thomas Shelby
Apr 6 at 15:31
1
$begingroup$
Thanks I’m looking them up
$endgroup$
– User12239
Apr 6 at 15:33
$begingroup$
Yes I was doubting if every neighborhood has a compact neighborhood so I need to prove this fact now
$endgroup$
– User12239
Apr 6 at 15:26
$begingroup$
Yes I was doubting if every neighborhood has a compact neighborhood so I need to prove this fact now
$endgroup$
– User12239
Apr 6 at 15:26
$begingroup$
You can find a proof in 'Topology' by Munkres.
$endgroup$
– Thomas Shelby
Apr 6 at 15:28
$begingroup$
You can find a proof in 'Topology' by Munkres.
$endgroup$
– Thomas Shelby
Apr 6 at 15:28
$begingroup$
Also see this.
$endgroup$
– Thomas Shelby
Apr 6 at 15:31
$begingroup$
Also see this.
$endgroup$
– Thomas Shelby
Apr 6 at 15:31
1
1
$begingroup$
Thanks I’m looking them up
$endgroup$
– User12239
Apr 6 at 15:33
$begingroup$
Thanks I’m looking them up
$endgroup$
– User12239
Apr 6 at 15:33
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 $xin barS$
answered Apr 6 at 15:18
MaksimMaksim
1,01719
$endgroup$
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 $xin barS$
answered Apr 6 at 15:18
MaksimMaksim
1,01719
$endgroup$
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 $xin barS$
answered Apr 6 at 15:18
MaksimMaksim
1,01719
$endgroup$
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 $xin barS$
answered Apr 6 at 15:18
MaksimMaksim
1,01719
answered Apr 6 at 15:18
MaksimMaksim
1,01719
answered Apr 6 at 15:18
MaksimMaksim
1,01719
answered Apr 6 at 15:18
MaksimMaksim
1,01719
1,01719
add a comment |
add a comment |
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-general-topology
