compactness and relative compactness
$begingroup$
Suppose that $M$ is compact subset of a Banach space X. Is $M$ relatively compact too?
As far as I know, there is a characterisation of compact sets via Hausdorff $varepsilon$ - net theorem and it's the same for compact and relatively compact sets. So the answer should be positive.
general-topology banach-spaces compactness normed-spaces
edited Dec 10 '18 at 12:09
José Carlos Santos
170k23132238
asked Dec 10 '18 at 12:05
zorro47zorro47
626514
$endgroup$
add a comment |
$begingroup$
Suppose that $M$ is compact subset of a Banach space X. Is $M$ relatively compact too?
As far as I know, there is a characterisation of compact sets via Hausdorff $varepsilon$ - net theorem and it's the same for compact and relatively compact sets. So the answer should be positive.
general-topology banach-spaces compactness normed-spaces
edited Dec 10 '18 at 12:09
José Carlos Santos
170k23132238
asked Dec 10 '18 at 12:05
zorro47zorro47
626514
$endgroup$
1
$begingroup$
A subset is relatively compact if its closure is compact. Every compact subset is closed. This should do it.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 10 '18 at 12:08
add a comment |
$begingroup$
Suppose that $M$ is compact subset of a Banach space X. Is $M$ relatively compact too?
As far as I know, there is a characterisation of compact sets via Hausdorff $varepsilon$ - net theorem and it's the same for compact and relatively compact sets. So the answer should be positive.
general-topology banach-spaces compactness normed-spaces
edited Dec 10 '18 at 12:09
José Carlos Santos
170k23132238
asked Dec 10 '18 at 12:05
zorro47zorro47
626514
$endgroup$
Suppose that $M$ is compact subset of a Banach space X. Is $M$ relatively compact too?
As far as I know, there is a characterisation of compact sets via Hausdorff $varepsilon$ - net theorem and it's the same for compact and relatively compact sets. So the answer should be positive.
general-topology banach-spaces compactness normed-spaces
general-topology banach-spaces compactness normed-spaces
edited Dec 10 '18 at 12:09
José Carlos Santos
170k23132238
asked Dec 10 '18 at 12:05
zorro47zorro47
626514
edited Dec 10 '18 at 12:09
José Carlos Santos
170k23132238
asked Dec 10 '18 at 12:05
zorro47zorro47
626514
edited Dec 10 '18 at 12:09
José Carlos Santos
170k23132238
edited Dec 10 '18 at 12:09
José Carlos Santos
170k23132238
edited Dec 10 '18 at 12:09
José Carlos Santos
170k23132238
170k23132238
asked Dec 10 '18 at 12:05
zorro47zorro47
626514
asked Dec 10 '18 at 12:05
zorro47zorro47
626514
asked Dec 10 '18 at 12:05
zorro47zorro47
626514
626514
1
$begingroup$
A subset is relatively compact if its closure is compact. Every compact subset is closed. This should do it.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 10 '18 at 12:08
add a comment |
1
$begingroup$
A subset is relatively compact if its closure is compact. Every compact subset is closed. This should do it.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 10 '18 at 12:08
1
1
$begingroup$
A subset is relatively compact if its closure is compact. Every compact subset is closed. This should do it.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 10 '18 at 12:08
$begingroup$
A subset is relatively compact if its closure is compact. Every compact subset is closed. This should do it.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 10 '18 at 12:08
add a comment |
2 Answers
2
active
oldest
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$begingroup$
I don't know what your definition of 'relatively compact' is but according to the usual definition a set is raltively compact if its closure is compact. Since compact sets are already closed it is obvious that they are also relatively compact.
answered Dec 10 '18 at 12:09
Kavi Rama MurthyKavi Rama Murthy
69.5k53170
$endgroup$
add a comment |
$begingroup$
For a subset $S$ of a topological space, being relatively compact means that $overline S$ is compact. So, if $M$ is compact, then, since $overline M=M$, which is compact, $M$ is relatively compact too.
answered Dec 10 '18 at 12:07
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
I don't know what your definition of 'relatively compact' is but according to the usual definition a set is raltively compact if its closure is compact. Since compact sets are already closed it is obvious that they are also relatively compact.
answered Dec 10 '18 at 12:09
Kavi Rama MurthyKavi Rama Murthy
69.5k53170
$endgroup$
add a comment |
$begingroup$
I don't know what your definition of 'relatively compact' is but according to the usual definition a set is raltively compact if its closure is compact. Since compact sets are already closed it is obvious that they are also relatively compact.
answered Dec 10 '18 at 12:09
Kavi Rama MurthyKavi Rama Murthy
69.5k53170
$endgroup$
add a comment |
$begingroup$
I don't know what your definition of 'relatively compact' is but according to the usual definition a set is raltively compact if its closure is compact. Since compact sets are already closed it is obvious that they are also relatively compact.
answered Dec 10 '18 at 12:09
Kavi Rama MurthyKavi Rama Murthy
69.5k53170
$endgroup$
I don't know what your definition of 'relatively compact' is but according to the usual definition a set is raltively compact if its closure is compact. Since compact sets are already closed it is obvious that they are also relatively compact.
answered Dec 10 '18 at 12:09
Kavi Rama MurthyKavi Rama Murthy
69.5k53170
answered Dec 10 '18 at 12:09
Kavi Rama MurthyKavi Rama Murthy
69.5k53170
answered Dec 10 '18 at 12:09
Kavi Rama MurthyKavi Rama Murthy
69.5k53170
answered Dec 10 '18 at 12:09
Kavi Rama MurthyKavi Rama Murthy
69.5k53170
69.5k53170
add a comment |
add a comment |
$begingroup$
For a subset $S$ of a topological space, being relatively compact means that $overline S$ is compact. So, if $M$ is compact, then, since $overline M=M$, which is compact, $M$ is relatively compact too.
answered Dec 10 '18 at 12:07
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
add a comment |
$begingroup$
For a subset $S$ of a topological space, being relatively compact means that $overline S$ is compact. So, if $M$ is compact, then, since $overline M=M$, which is compact, $M$ is relatively compact too.
answered Dec 10 '18 at 12:07
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
add a comment |
$begingroup$
For a subset $S$ of a topological space, being relatively compact means that $overline S$ is compact. So, if $M$ is compact, then, since $overline M=M$, which is compact, $M$ is relatively compact too.
answered Dec 10 '18 at 12:07
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
For a subset $S$ of a topological space, being relatively compact means that $overline S$ is compact. So, if $M$ is compact, then, since $overline M=M$, which is compact, $M$ is relatively compact too.
answered Dec 10 '18 at 12:07
José Carlos SantosJosé Carlos Santos
170k23132238
answered Dec 10 '18 at 12:07
José Carlos SantosJosé Carlos Santos
170k23132238
answered Dec 10 '18 at 12:07
José Carlos SantosJosé Carlos Santos
170k23132238
answered Dec 10 '18 at 12:07
José Carlos SantosJosé Carlos Santos
170k23132238
170k23132238
add a comment |
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$begingroup$
A subset is relatively compact if its closure is compact. Every compact subset is closed. This should do it.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 10 '18 at 12:08