Every convergent sequence is relatively compact
$begingroup$
Let $X$ be an Hilbert space. It is known that:
Every bounded set is relatively compact (i.e., has compact closure) iff $X$ is finite dimensional
Every convergent sequence is bounded.
Every totally bounded set is relatively compact, see here.
My question is: Does there exist an infinite dimensional $X$ such that every convergent sequence is relatively compact?
hilbert-spaces compactness
asked Jan 4 at 10:37
NduccioNduccio
518313
$endgroup$
add a comment |
$begingroup$
Let $X$ be an Hilbert space. It is known that:
Every bounded set is relatively compact (i.e., has compact closure) iff $X$ is finite dimensional
Every convergent sequence is bounded.
Every totally bounded set is relatively compact, see here.
My question is: Does there exist an infinite dimensional $X$ such that every convergent sequence is relatively compact?
hilbert-spaces compactness
asked Jan 4 at 10:37
NduccioNduccio
518313
$endgroup$
add a comment |
$begingroup$
Let $X$ be an Hilbert space. It is known that:
Every bounded set is relatively compact (i.e., has compact closure) iff $X$ is finite dimensional
Every convergent sequence is bounded.
Every totally bounded set is relatively compact, see here.
My question is: Does there exist an infinite dimensional $X$ such that every convergent sequence is relatively compact?
hilbert-spaces compactness
asked Jan 4 at 10:37
NduccioNduccio
518313
$endgroup$
Let $X$ be an Hilbert space. It is known that:
Every bounded set is relatively compact (i.e., has compact closure) iff $X$ is finite dimensional
Every convergent sequence is bounded.
Every totally bounded set is relatively compact, see here.
My question is: Does there exist an infinite dimensional $X$ such that every convergent sequence is relatively compact?
hilbert-spaces compactness
hilbert-spaces compactness
asked Jan 4 at 10:37
NduccioNduccio
518313
asked Jan 4 at 10:37
NduccioNduccio
518313
asked Jan 4 at 10:37
NduccioNduccio
518313
asked Jan 4 at 10:37
NduccioNduccio
518313
asked Jan 4 at 10:37
NduccioNduccio
518313
518313
add a comment |
add a comment |
1 Answer
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$begingroup$
In any metric space, if $(x_n)_{ninmathbb N}$ is a convergent sequence and if $x=lim_{ntoinfty}x_n$, then ${x_n,|,ninmathbb{N}}$ is relatively compact, since its closure is ${x_n,|,ninmathbb{N}}cup{x}$, which is compact.
answered Jan 4 at 10:41
José Carlos SantosJosé Carlos Santos
158k22126228
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
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active
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$begingroup$
In any metric space, if $(x_n)_{ninmathbb N}$ is a convergent sequence and if $x=lim_{ntoinfty}x_n$, then ${x_n,|,ninmathbb{N}}$ is relatively compact, since its closure is ${x_n,|,ninmathbb{N}}cup{x}$, which is compact.
answered Jan 4 at 10:41
José Carlos SantosJosé Carlos Santos
158k22126228
$endgroup$
add a comment |
$begingroup$
In any metric space, if $(x_n)_{ninmathbb N}$ is a convergent sequence and if $x=lim_{ntoinfty}x_n$, then ${x_n,|,ninmathbb{N}}$ is relatively compact, since its closure is ${x_n,|,ninmathbb{N}}cup{x}$, which is compact.
answered Jan 4 at 10:41
José Carlos SantosJosé Carlos Santos
158k22126228
$endgroup$
add a comment |
$begingroup$
In any metric space, if $(x_n)_{ninmathbb N}$ is a convergent sequence and if $x=lim_{ntoinfty}x_n$, then ${x_n,|,ninmathbb{N}}$ is relatively compact, since its closure is ${x_n,|,ninmathbb{N}}cup{x}$, which is compact.
answered Jan 4 at 10:41
José Carlos SantosJosé Carlos Santos
158k22126228
$endgroup$
In any metric space, if $(x_n)_{ninmathbb N}$ is a convergent sequence and if $x=lim_{ntoinfty}x_n$, then ${x_n,|,ninmathbb{N}}$ is relatively compact, since its closure is ${x_n,|,ninmathbb{N}}cup{x}$, which is compact.
answered Jan 4 at 10:41
José Carlos SantosJosé Carlos Santos
158k22126228
answered Jan 4 at 10:41
José Carlos SantosJosé Carlos Santos
158k22126228
answered Jan 4 at 10:41
José Carlos SantosJosé Carlos Santos
158k22126228
answered Jan 4 at 10:41
José Carlos SantosJosé Carlos Santos
158k22126228
158k22126228
add a comment |
add a comment |
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