Prove a subspace is separable
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Given a Hilbert space $H$ and let $K$ be a compact set in $H$. Let $X$ be the smallest closed subspace containing $K$. Prove that $X$ is separable.
Compact implies that $K$ is totally bounded. But how to use this prove $X$ is separable? And how to make use of the smallest closed subspace?
functional-analysis
asked Dec 8 '18 at 23:31
whereamIwhereamI
320115
$endgroup$
add a comment |
$begingroup$
Given a Hilbert space $H$ and let $K$ be a compact set in $H$. Let $X$ be the smallest closed subspace containing $K$. Prove that $X$ is separable.
Compact implies that $K$ is totally bounded. But how to use this prove $X$ is separable? And how to make use of the smallest closed subspace?
functional-analysis
asked Dec 8 '18 at 23:31
whereamIwhereamI
320115
$endgroup$
1
$begingroup$
Hint: A compact metric space is separable.
$endgroup$
– Math1000
Dec 9 '18 at 0:13
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@Math1000 So it suffices to prove that $X$ is compact? would you mind explain a bit more how to prove that? It is not so obvious to me.
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– whereamI
Dec 9 '18 at 0:59
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$X$ won't be compact (it is unbounded). @Math1000 is referring to $K$ being the compact metric space.
$endgroup$
– user25959
Dec 9 '18 at 1:18
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Reference for the fact that $K$ is separable here: math.stackexchange.com/questions/974233/…
$endgroup$
– user25959
Dec 9 '18 at 1:18
add a comment |
$begingroup$
Given a Hilbert space $H$ and let $K$ be a compact set in $H$. Let $X$ be the smallest closed subspace containing $K$. Prove that $X$ is separable.
Compact implies that $K$ is totally bounded. But how to use this prove $X$ is separable? And how to make use of the smallest closed subspace?
functional-analysis
asked Dec 8 '18 at 23:31
whereamIwhereamI
320115
$endgroup$
Given a Hilbert space $H$ and let $K$ be a compact set in $H$. Let $X$ be the smallest closed subspace containing $K$. Prove that $X$ is separable.
Compact implies that $K$ is totally bounded. But how to use this prove $X$ is separable? And how to make use of the smallest closed subspace?
functional-analysis
functional-analysis
asked Dec 8 '18 at 23:31
whereamIwhereamI
320115
asked Dec 8 '18 at 23:31
whereamIwhereamI
320115
asked Dec 8 '18 at 23:31
whereamIwhereamI
320115
asked Dec 8 '18 at 23:31
whereamIwhereamI
320115
asked Dec 8 '18 at 23:31
whereamIwhereamI
320115
320115
1
$begingroup$
Hint: A compact metric space is separable.
$endgroup$
– Math1000
Dec 9 '18 at 0:13
$begingroup$
@Math1000 So it suffices to prove that $X$ is compact? would you mind explain a bit more how to prove that? It is not so obvious to me.
$endgroup$
– whereamI
Dec 9 '18 at 0:59
$begingroup$
$X$ won't be compact (it is unbounded). @Math1000 is referring to $K$ being the compact metric space.
$endgroup$
– user25959
Dec 9 '18 at 1:18
$begingroup$
Reference for the fact that $K$ is separable here: math.stackexchange.com/questions/974233/…
$endgroup$
– user25959
Dec 9 '18 at 1:18
add a comment |
1
$begingroup$
Hint: A compact metric space is separable.
$endgroup$
– Math1000
Dec 9 '18 at 0:13
$begingroup$
@Math1000 So it suffices to prove that $X$ is compact? would you mind explain a bit more how to prove that? It is not so obvious to me.
$endgroup$
– whereamI
Dec 9 '18 at 0:59
$begingroup$
$X$ won't be compact (it is unbounded). @Math1000 is referring to $K$ being the compact metric space.
$endgroup$
– user25959
Dec 9 '18 at 1:18
$begingroup$
Reference for the fact that $K$ is separable here: math.stackexchange.com/questions/974233/…
$endgroup$
– user25959
Dec 9 '18 at 1:18
1
1
$begingroup$
Hint: A compact metric space is separable.
$endgroup$
– Math1000
Dec 9 '18 at 0:13
$begingroup$
Hint: A compact metric space is separable.
$endgroup$
– Math1000
Dec 9 '18 at 0:13
$begingroup$
@Math1000 So it suffices to prove that $X$ is compact? would you mind explain a bit more how to prove that? It is not so obvious to me.
$endgroup$
– whereamI
Dec 9 '18 at 0:59
$begingroup$
@Math1000 So it suffices to prove that $X$ is compact? would you mind explain a bit more how to prove that? It is not so obvious to me.
$endgroup$
– whereamI
Dec 9 '18 at 0:59
$begingroup$
$X$ won't be compact (it is unbounded). @Math1000 is referring to $K$ being the compact metric space.
$endgroup$
– user25959
Dec 9 '18 at 1:18
$begingroup$
$X$ won't be compact (it is unbounded). @Math1000 is referring to $K$ being the compact metric space.
$endgroup$
– user25959
Dec 9 '18 at 1:18
$begingroup$
Reference for the fact that $K$ is separable here: math.stackexchange.com/questions/974233/…
$endgroup$
– user25959
Dec 9 '18 at 1:18
$begingroup$
Reference for the fact that $K$ is separable here: math.stackexchange.com/questions/974233/…
$endgroup$
– user25959
Dec 9 '18 at 1:18
add a comment |
1 Answer
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$begingroup$
Once you have proven the hint in the comments, let $(x_1,x_2,ldots)$ be a countable dense sequence of $K$. Let $Q$ denote the rationals (or if you are working over $mathbb{C}$, the set of complex numbers of the form $p+qi$ for $p,q$ rational). Observe that the set $Qx_1 := {qx_1: qin Q}$ is countable. As is $Qx_1+Qx_2 := {u+v:uin Qx_1, vin Qx_2}$, and more generally, the set $C:=$ finite $Q$-linear combinations of ${x_i}$ is countable (this follows because a countable union of countable sets is countable).
Side claim: $X = $ closure of set of all finite linear combinations of elements of $K$, i.e. $X=overline{{x in H: x=sum_{i=1}^{N} alpha_i k_i text{ for some } Nin mathbb{N},alpha_iin mathbb{F},k_iin K}}$ (this is an easy proof).
Now what remains is to show that $C$ is dense in $X$. You can do this by showing that within $varepsilon/2$ of any $vin X$, there is some $u=$ a finite $mathbb{F}$-linear combination of elements of $K$; and within $varepsilon/2$ of $u$ there is an element $c$ of $C$.
answered Dec 9 '18 at 1:14
user25959user25959
1,573816
$endgroup$
$begingroup$
Thanks a lot! I forget that the smallest closed subspace can be written as the closure of linear combinations of elements of $K$.
$endgroup$
– whereamI
Dec 9 '18 at 1:22
add a comment |
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$begingroup$
Once you have proven the hint in the comments, let $(x_1,x_2,ldots)$ be a countable dense sequence of $K$. Let $Q$ denote the rationals (or if you are working over $mathbb{C}$, the set of complex numbers of the form $p+qi$ for $p,q$ rational). Observe that the set $Qx_1 := {qx_1: qin Q}$ is countable. As is $Qx_1+Qx_2 := {u+v:uin Qx_1, vin Qx_2}$, and more generally, the set $C:=$ finite $Q$-linear combinations of ${x_i}$ is countable (this follows because a countable union of countable sets is countable).
Side claim: $X = $ closure of set of all finite linear combinations of elements of $K$, i.e. $X=overline{{x in H: x=sum_{i=1}^{N} alpha_i k_i text{ for some } Nin mathbb{N},alpha_iin mathbb{F},k_iin K}}$ (this is an easy proof).
Now what remains is to show that $C$ is dense in $X$. You can do this by showing that within $varepsilon/2$ of any $vin X$, there is some $u=$ a finite $mathbb{F}$-linear combination of elements of $K$; and within $varepsilon/2$ of $u$ there is an element $c$ of $C$.
answered Dec 9 '18 at 1:14
user25959user25959
1,573816
$endgroup$
$begingroup$
Thanks a lot! I forget that the smallest closed subspace can be written as the closure of linear combinations of elements of $K$.
$endgroup$
– whereamI
Dec 9 '18 at 1:22
add a comment |
$begingroup$
Once you have proven the hint in the comments, let $(x_1,x_2,ldots)$ be a countable dense sequence of $K$. Let $Q$ denote the rationals (or if you are working over $mathbb{C}$, the set of complex numbers of the form $p+qi$ for $p,q$ rational). Observe that the set $Qx_1 := {qx_1: qin Q}$ is countable. As is $Qx_1+Qx_2 := {u+v:uin Qx_1, vin Qx_2}$, and more generally, the set $C:=$ finite $Q$-linear combinations of ${x_i}$ is countable (this follows because a countable union of countable sets is countable).
Side claim: $X = $ closure of set of all finite linear combinations of elements of $K$, i.e. $X=overline{{x in H: x=sum_{i=1}^{N} alpha_i k_i text{ for some } Nin mathbb{N},alpha_iin mathbb{F},k_iin K}}$ (this is an easy proof).
Now what remains is to show that $C$ is dense in $X$. You can do this by showing that within $varepsilon/2$ of any $vin X$, there is some $u=$ a finite $mathbb{F}$-linear combination of elements of $K$; and within $varepsilon/2$ of $u$ there is an element $c$ of $C$.
answered Dec 9 '18 at 1:14
user25959user25959
1,573816
$endgroup$
$begingroup$
Thanks a lot! I forget that the smallest closed subspace can be written as the closure of linear combinations of elements of $K$.
$endgroup$
– whereamI
Dec 9 '18 at 1:22
add a comment |
$begingroup$
Once you have proven the hint in the comments, let $(x_1,x_2,ldots)$ be a countable dense sequence of $K$. Let $Q$ denote the rationals (or if you are working over $mathbb{C}$, the set of complex numbers of the form $p+qi$ for $p,q$ rational). Observe that the set $Qx_1 := {qx_1: qin Q}$ is countable. As is $Qx_1+Qx_2 := {u+v:uin Qx_1, vin Qx_2}$, and more generally, the set $C:=$ finite $Q$-linear combinations of ${x_i}$ is countable (this follows because a countable union of countable sets is countable).
Side claim: $X = $ closure of set of all finite linear combinations of elements of $K$, i.e. $X=overline{{x in H: x=sum_{i=1}^{N} alpha_i k_i text{ for some } Nin mathbb{N},alpha_iin mathbb{F},k_iin K}}$ (this is an easy proof).
Now what remains is to show that $C$ is dense in $X$. You can do this by showing that within $varepsilon/2$ of any $vin X$, there is some $u=$ a finite $mathbb{F}$-linear combination of elements of $K$; and within $varepsilon/2$ of $u$ there is an element $c$ of $C$.
answered Dec 9 '18 at 1:14
user25959user25959
1,573816
$endgroup$
Once you have proven the hint in the comments, let $(x_1,x_2,ldots)$ be a countable dense sequence of $K$. Let $Q$ denote the rationals (or if you are working over $mathbb{C}$, the set of complex numbers of the form $p+qi$ for $p,q$ rational). Observe that the set $Qx_1 := {qx_1: qin Q}$ is countable. As is $Qx_1+Qx_2 := {u+v:uin Qx_1, vin Qx_2}$, and more generally, the set $C:=$ finite $Q$-linear combinations of ${x_i}$ is countable (this follows because a countable union of countable sets is countable).
Side claim: $X = $ closure of set of all finite linear combinations of elements of $K$, i.e. $X=overline{{x in H: x=sum_{i=1}^{N} alpha_i k_i text{ for some } Nin mathbb{N},alpha_iin mathbb{F},k_iin K}}$ (this is an easy proof).
Now what remains is to show that $C$ is dense in $X$. You can do this by showing that within $varepsilon/2$ of any $vin X$, there is some $u=$ a finite $mathbb{F}$-linear combination of elements of $K$; and within $varepsilon/2$ of $u$ there is an element $c$ of $C$.
answered Dec 9 '18 at 1:14
user25959user25959
1,573816
answered Dec 9 '18 at 1:14
user25959user25959
1,573816
answered Dec 9 '18 at 1:14
user25959user25959
1,573816
answered Dec 9 '18 at 1:14
user25959user25959
1,573816
1,573816
$begingroup$
Thanks a lot! I forget that the smallest closed subspace can be written as the closure of linear combinations of elements of $K$.
$endgroup$
– whereamI
Dec 9 '18 at 1:22
add a comment |
$begingroup$
Thanks a lot! I forget that the smallest closed subspace can be written as the closure of linear combinations of elements of $K$.
$endgroup$
– whereamI
Dec 9 '18 at 1:22
$begingroup$
Thanks a lot! I forget that the smallest closed subspace can be written as the closure of linear combinations of elements of $K$.
$endgroup$
– whereamI
Dec 9 '18 at 1:22
$begingroup$
Thanks a lot! I forget that the smallest closed subspace can be written as the closure of linear combinations of elements of $K$.
$endgroup$
– whereamI
Dec 9 '18 at 1:22
add a comment |
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$begingroup$
Hint: A compact metric space is separable.
$endgroup$
– Math1000
Dec 9 '18 at 0:13
$begingroup$
@Math1000 So it suffices to prove that $X$ is compact? would you mind explain a bit more how to prove that? It is not so obvious to me.
$endgroup$
– whereamI
Dec 9 '18 at 0:59
$begingroup$
$X$ won't be compact (it is unbounded). @Math1000 is referring to $K$ being the compact metric space.
$endgroup$
– user25959
Dec 9 '18 at 1:18
$begingroup$
Reference for the fact that $K$ is separable here: math.stackexchange.com/questions/974233/…
$endgroup$
– user25959
Dec 9 '18 at 1:18