Is a metrizable and compact subset of $E^*$ is a metrizable complete subset ? for the weak* topology
$begingroup$
If we have a Banach space $E$. and we consider it's dual $E^*$, it is a Banach space. so we consider the weak* topology ($sigma(E^*,E)$) on $E^*$.
So My question is : If we have a set $Bsubset E^*$, that is compact and Metrizable (weak* topology).
Why is $B$ is Metrizable complete ?
general-topology functional-analysis weak-topology
asked Jan 14 at 11:25
Anas BOUALIIAnas BOUALII
1408
$endgroup$
add a comment |
$begingroup$
If we have a Banach space $E$. and we consider it's dual $E^*$, it is a Banach space. so we consider the weak* topology ($sigma(E^*,E)$) on $E^*$.
So My question is : If we have a set $Bsubset E^*$, that is compact and Metrizable (weak* topology).
Why is $B$ is Metrizable complete ?
general-topology functional-analysis weak-topology
asked Jan 14 at 11:25
Anas BOUALIIAnas BOUALII
1408
$endgroup$
add a comment |
$begingroup$
If we have a Banach space $E$. and we consider it's dual $E^*$, it is a Banach space. so we consider the weak* topology ($sigma(E^*,E)$) on $E^*$.
So My question is : If we have a set $Bsubset E^*$, that is compact and Metrizable (weak* topology).
Why is $B$ is Metrizable complete ?
general-topology functional-analysis weak-topology
asked Jan 14 at 11:25
Anas BOUALIIAnas BOUALII
1408
$endgroup$
If we have a Banach space $E$. and we consider it's dual $E^*$, it is a Banach space. so we consider the weak* topology ($sigma(E^*,E)$) on $E^*$.
So My question is : If we have a set $Bsubset E^*$, that is compact and Metrizable (weak* topology).
Why is $B$ is Metrizable complete ?
general-topology functional-analysis weak-topology
general-topology functional-analysis weak-topology
asked Jan 14 at 11:25
Anas BOUALIIAnas BOUALII
1408
asked Jan 14 at 11:25
Anas BOUALIIAnas BOUALII
1408
asked Jan 14 at 11:25
Anas BOUALIIAnas BOUALII
1408
asked Jan 14 at 11:25
Anas BOUALIIAnas BOUALII
1408
asked Jan 14 at 11:25
Anas BOUALIIAnas BOUALII
1408
1408
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
All compact metric spaces are sequentially compact. Thus, if you have a Cauchy sequence in a compact metric space, then this has a convergent subsequence. By the properties of a Cauchy sequence you can prove that the limit of this subsequence must also be the limit of the sequence. Therefore every compact metric space is complete.
answered Jan 14 at 11:30
SmileyCraftSmileyCraft
3,776519
$endgroup$
$begingroup$
It makes sense. Thank you.
$endgroup$
– Anas BOUALII
Jan 14 at 11:31
1
$begingroup$
Not every compact space is sequentially compact (see this thread: math.stackexchange.com/questions/220422/…). Of course, both properties are equivalent for metric spaces.
$endgroup$
– MaoWao
Jan 14 at 12:47
add a comment |
$begingroup$
Note that any compact metric space is complete and totally bounded (and vice versa). So if $B$ is a compact metrizable space, it must be completely metrizable.
answered Jan 14 at 11:31
SongSong
18.6k21651
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
All compact metric spaces are sequentially compact. Thus, if you have a Cauchy sequence in a compact metric space, then this has a convergent subsequence. By the properties of a Cauchy sequence you can prove that the limit of this subsequence must also be the limit of the sequence. Therefore every compact metric space is complete.
answered Jan 14 at 11:30
SmileyCraftSmileyCraft
3,776519
$endgroup$
$begingroup$
It makes sense. Thank you.
$endgroup$
– Anas BOUALII
Jan 14 at 11:31
1
$begingroup$
Not every compact space is sequentially compact (see this thread: math.stackexchange.com/questions/220422/…). Of course, both properties are equivalent for metric spaces.
$endgroup$
– MaoWao
Jan 14 at 12:47
add a comment |
$begingroup$
All compact metric spaces are sequentially compact. Thus, if you have a Cauchy sequence in a compact metric space, then this has a convergent subsequence. By the properties of a Cauchy sequence you can prove that the limit of this subsequence must also be the limit of the sequence. Therefore every compact metric space is complete.
answered Jan 14 at 11:30
SmileyCraftSmileyCraft
3,776519
$endgroup$
$begingroup$
It makes sense. Thank you.
$endgroup$
– Anas BOUALII
Jan 14 at 11:31
1
$begingroup$
Not every compact space is sequentially compact (see this thread: math.stackexchange.com/questions/220422/…). Of course, both properties are equivalent for metric spaces.
$endgroup$
– MaoWao
Jan 14 at 12:47
add a comment |
$begingroup$
All compact metric spaces are sequentially compact. Thus, if you have a Cauchy sequence in a compact metric space, then this has a convergent subsequence. By the properties of a Cauchy sequence you can prove that the limit of this subsequence must also be the limit of the sequence. Therefore every compact metric space is complete.
answered Jan 14 at 11:30
SmileyCraftSmileyCraft
3,776519
$endgroup$
All compact metric spaces are sequentially compact. Thus, if you have a Cauchy sequence in a compact metric space, then this has a convergent subsequence. By the properties of a Cauchy sequence you can prove that the limit of this subsequence must also be the limit of the sequence. Therefore every compact metric space is complete.
answered Jan 14 at 11:30
SmileyCraftSmileyCraft
3,776519
edited Jan 14 at 13:22
answered Jan 14 at 11:30
SmileyCraftSmileyCraft
3,776519
answered Jan 14 at 11:30
SmileyCraftSmileyCraft
3,776519
answered Jan 14 at 11:30
SmileyCraftSmileyCraft
3,776519
3,776519
$begingroup$
It makes sense. Thank you.
$endgroup$
– Anas BOUALII
Jan 14 at 11:31
1
$begingroup$
Not every compact space is sequentially compact (see this thread: math.stackexchange.com/questions/220422/…). Of course, both properties are equivalent for metric spaces.
$endgroup$
– MaoWao
Jan 14 at 12:47
add a comment |
$begingroup$
It makes sense. Thank you.
$endgroup$
– Anas BOUALII
Jan 14 at 11:31
1
$begingroup$
Not every compact space is sequentially compact (see this thread: math.stackexchange.com/questions/220422/…). Of course, both properties are equivalent for metric spaces.
$endgroup$
– MaoWao
Jan 14 at 12:47
$begingroup$
It makes sense. Thank you.
$endgroup$
– Anas BOUALII
Jan 14 at 11:31
$begingroup$
It makes sense. Thank you.
$endgroup$
– Anas BOUALII
Jan 14 at 11:31
1
1
$begingroup$
Not every compact space is sequentially compact (see this thread: math.stackexchange.com/questions/220422/…). Of course, both properties are equivalent for metric spaces.
$endgroup$
– MaoWao
Jan 14 at 12:47
$begingroup$
Not every compact space is sequentially compact (see this thread: math.stackexchange.com/questions/220422/…). Of course, both properties are equivalent for metric spaces.
$endgroup$
– MaoWao
Jan 14 at 12:47
add a comment |
$begingroup$
Note that any compact metric space is complete and totally bounded (and vice versa). So if $B$ is a compact metrizable space, it must be completely metrizable.
answered Jan 14 at 11:31
SongSong
18.6k21651
$endgroup$
add a comment |
$begingroup$
Note that any compact metric space is complete and totally bounded (and vice versa). So if $B$ is a compact metrizable space, it must be completely metrizable.
answered Jan 14 at 11:31
SongSong
18.6k21651
$endgroup$
add a comment |
$begingroup$
Note that any compact metric space is complete and totally bounded (and vice versa). So if $B$ is a compact metrizable space, it must be completely metrizable.
answered Jan 14 at 11:31
SongSong
18.6k21651
$endgroup$
Note that any compact metric space is complete and totally bounded (and vice versa). So if $B$ is a compact metrizable space, it must be completely metrizable.
answered Jan 14 at 11:31
SongSong
18.6k21651
answered Jan 14 at 11:31
SongSong
18.6k21651
answered Jan 14 at 11:31
SongSong
18.6k21651
answered Jan 14 at 11:31
SongSong
18.6k21651
18.6k21651
add a comment |
add a comment |
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