Prove a subspace is separable












2














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







share|cite|improve this question


















  • 1




    Hint: A compact metric space is separable.
    – Math1000
    Dec 9 at 0:13










  • @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.
    – whereamI
    Dec 9 at 0:59










  • $X$ won't be compact (it is unbounded). @Math1000 is referring to $K$ being the compact metric space.
    – user25959
    Dec 9 at 1:18










  • Reference for the fact that $K$ is separable here: math.stackexchange.com/questions/974233/…
    – user25959
    Dec 9 at 1:18
















2














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







share|cite|improve this question


















  • 1




    Hint: A compact metric space is separable.
    – Math1000
    Dec 9 at 0:13










  • @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.
    – whereamI
    Dec 9 at 0:59










  • $X$ won't be compact (it is unbounded). @Math1000 is referring to $K$ being the compact metric space.
    – user25959
    Dec 9 at 1:18










  • Reference for the fact that $K$ is separable here: math.stackexchange.com/questions/974233/…
    – user25959
    Dec 9 at 1:18














2












2








2


1





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







share|cite|improve this question













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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 8 at 23:31









whereamI

311114




311114








  • 1




    Hint: A compact metric space is separable.
    – Math1000
    Dec 9 at 0:13










  • @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.
    – whereamI
    Dec 9 at 0:59










  • $X$ won't be compact (it is unbounded). @Math1000 is referring to $K$ being the compact metric space.
    – user25959
    Dec 9 at 1:18










  • Reference for the fact that $K$ is separable here: math.stackexchange.com/questions/974233/…
    – user25959
    Dec 9 at 1:18














  • 1




    Hint: A compact metric space is separable.
    – Math1000
    Dec 9 at 0:13










  • @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.
    – whereamI
    Dec 9 at 0:59










  • $X$ won't be compact (it is unbounded). @Math1000 is referring to $K$ being the compact metric space.
    – user25959
    Dec 9 at 1:18










  • Reference for the fact that $K$ is separable here: math.stackexchange.com/questions/974233/…
    – user25959
    Dec 9 at 1:18








1




1




Hint: A compact metric space is separable.
– Math1000
Dec 9 at 0:13




Hint: A compact metric space is separable.
– Math1000
Dec 9 at 0:13












@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.
– whereamI
Dec 9 at 0:59




@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.
– whereamI
Dec 9 at 0:59












$X$ won't be compact (it is unbounded). @Math1000 is referring to $K$ being the compact metric space.
– user25959
Dec 9 at 1:18




$X$ won't be compact (it is unbounded). @Math1000 is referring to $K$ being the compact metric space.
– user25959
Dec 9 at 1:18












Reference for the fact that $K$ is separable here: math.stackexchange.com/questions/974233/…
– user25959
Dec 9 at 1:18




Reference for the fact that $K$ is separable here: math.stackexchange.com/questions/974233/…
– user25959
Dec 9 at 1:18










1 Answer
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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 $C$.






share|cite|improve this answer





















  • Thanks a lot! I forget that the smallest closed subspace can be written as the closure of linear combinations of elements of $K$.
    – whereamI
    Dec 9 at 1:22













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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 $C$.






share|cite|improve this answer





















  • Thanks a lot! I forget that the smallest closed subspace can be written as the closure of linear combinations of elements of $K$.
    – whereamI
    Dec 9 at 1:22


















3














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 $C$.






share|cite|improve this answer





















  • Thanks a lot! I forget that the smallest closed subspace can be written as the closure of linear combinations of elements of $K$.
    – whereamI
    Dec 9 at 1:22
















3












3








3






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 $C$.






share|cite|improve this answer












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 $C$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 9 at 1:14









user25959

1,563816




1,563816












  • Thanks a lot! I forget that the smallest closed subspace can be written as the closure of linear combinations of elements of $K$.
    – whereamI
    Dec 9 at 1:22




















  • Thanks a lot! I forget that the smallest closed subspace can be written as the closure of linear combinations of elements of $K$.
    – whereamI
    Dec 9 at 1:22


















Thanks a lot! I forget that the smallest closed subspace can be written as the closure of linear combinations of elements of $K$.
– whereamI
Dec 9 at 1:22






Thanks a lot! I forget that the smallest closed subspace can be written as the closure of linear combinations of elements of $K$.
– whereamI
Dec 9 at 1:22




















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