Subset of $ell^2$ with distance property












8












$begingroup$


I'd been trying the following problem:




Prove that exists an infinite set $Asubset B(0,1)$ such that $|x-y|_2>sqrt{2}$ for all $x,y in A$.




My ideas always end in points with distance at most $sqrt{2}$.



I need hints, please.






functional-analysis hilbert-spaces lp-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    Is there something missing in the question? If $A$ contains only two points $(1,0,0, ldots)$ and $(-1, 0, 0, ldots)$, does this not work?
    $endgroup$
    – angryavian
    Dec 5 '18 at 20:40










  • $begingroup$
    My mistake, A must be infinite.
    $endgroup$
    – Ángela Flores
    Dec 5 '18 at 20:40










  • $begingroup$
    @AlexR. $l^2$ refers to the space of sequences $(a_1, a_2, ldots)$ such that $sum_n a_n^2 < infty$.
    $endgroup$
    – angryavian
    Dec 5 '18 at 20:47










  • $begingroup$
    Nitpick: You might want to add $x ne y$ in the formulation of your question. Otherwise, it is not possible (obviously).
    $endgroup$
    – gerw
    Dec 6 '18 at 7:52










  • $begingroup$
    An interesting spinoff: How large can $$inf_{x,y in A, x ne y} |x-y|$$ be? Can this be larger than $sqrt{2}$?
    $endgroup$
    – gerw
    Dec 6 '18 at 8:04


















8












$begingroup$


I'd been trying the following problem:




Prove that exists an infinite set $Asubset B(0,1)$ such that $|x-y|_2>sqrt{2}$ for all $x,y in A$.




My ideas always end in points with distance at most $sqrt{2}$.



I need hints, please.






functional-analysis hilbert-spaces lp-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    Is there something missing in the question? If $A$ contains only two points $(1,0,0, ldots)$ and $(-1, 0, 0, ldots)$, does this not work?
    $endgroup$
    – angryavian
    Dec 5 '18 at 20:40










  • $begingroup$
    My mistake, A must be infinite.
    $endgroup$
    – Ángela Flores
    Dec 5 '18 at 20:40










  • $begingroup$
    @AlexR. $l^2$ refers to the space of sequences $(a_1, a_2, ldots)$ such that $sum_n a_n^2 < infty$.
    $endgroup$
    – angryavian
    Dec 5 '18 at 20:47










  • $begingroup$
    Nitpick: You might want to add $x ne y$ in the formulation of your question. Otherwise, it is not possible (obviously).
    $endgroup$
    – gerw
    Dec 6 '18 at 7:52










  • $begingroup$
    An interesting spinoff: How large can $$inf_{x,y in A, x ne y} |x-y|$$ be? Can this be larger than $sqrt{2}$?
    $endgroup$
    – gerw
    Dec 6 '18 at 8:04
















8












8








8





$begingroup$


I'd been trying the following problem:




Prove that exists an infinite set $Asubset B(0,1)$ such that $|x-y|_2>sqrt{2}$ for all $x,y in A$.




My ideas always end in points with distance at most $sqrt{2}$.



I need hints, please.






functional-analysis hilbert-spaces lp-spaces







share|cite|improve this question











$endgroup$




I'd been trying the following problem:




Prove that exists an infinite set $Asubset B(0,1)$ such that $|x-y|_2>sqrt{2}$ for all $x,y in A$.




My ideas always end in points with distance at most $sqrt{2}$.



I need hints, please.






functional-analysis hilbert-spaces lp-spaces




functional-analysis hilbert-spaces lp-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 5 '18 at 22:43









Xander Henderson

14.3k103554




14.3k103554










asked Dec 5 '18 at 20:36









Ángela FloresÁngela Flores

26518




26518












  • $begingroup$
    Is there something missing in the question? If $A$ contains only two points $(1,0,0, ldots)$ and $(-1, 0, 0, ldots)$, does this not work?
    $endgroup$
    – angryavian
    Dec 5 '18 at 20:40










  • $begingroup$
    My mistake, A must be infinite.
    $endgroup$
    – Ángela Flores
    Dec 5 '18 at 20:40










  • $begingroup$
    @AlexR. $l^2$ refers to the space of sequences $(a_1, a_2, ldots)$ such that $sum_n a_n^2 < infty$.
    $endgroup$
    – angryavian
    Dec 5 '18 at 20:47










  • $begingroup$
    Nitpick: You might want to add $x ne y$ in the formulation of your question. Otherwise, it is not possible (obviously).
    $endgroup$
    – gerw
    Dec 6 '18 at 7:52










  • $begingroup$
    An interesting spinoff: How large can $$inf_{x,y in A, x ne y} |x-y|$$ be? Can this be larger than $sqrt{2}$?
    $endgroup$
    – gerw
    Dec 6 '18 at 8:04




















  • $begingroup$
    Is there something missing in the question? If $A$ contains only two points $(1,0,0, ldots)$ and $(-1, 0, 0, ldots)$, does this not work?
    $endgroup$
    – angryavian
    Dec 5 '18 at 20:40










  • $begingroup$
    My mistake, A must be infinite.
    $endgroup$
    – Ángela Flores
    Dec 5 '18 at 20:40










  • $begingroup$
    @AlexR. $l^2$ refers to the space of sequences $(a_1, a_2, ldots)$ such that $sum_n a_n^2 < infty$.
    $endgroup$
    – angryavian
    Dec 5 '18 at 20:47










  • $begingroup$
    Nitpick: You might want to add $x ne y$ in the formulation of your question. Otherwise, it is not possible (obviously).
    $endgroup$
    – gerw
    Dec 6 '18 at 7:52










  • $begingroup$
    An interesting spinoff: How large can $$inf_{x,y in A, x ne y} |x-y|$$ be? Can this be larger than $sqrt{2}$?
    $endgroup$
    – gerw
    Dec 6 '18 at 8:04


















$begingroup$
Is there something missing in the question? If $A$ contains only two points $(1,0,0, ldots)$ and $(-1, 0, 0, ldots)$, does this not work?
$endgroup$
– angryavian
Dec 5 '18 at 20:40




$begingroup$
Is there something missing in the question? If $A$ contains only two points $(1,0,0, ldots)$ and $(-1, 0, 0, ldots)$, does this not work?
$endgroup$
– angryavian
Dec 5 '18 at 20:40












$begingroup$
My mistake, A must be infinite.
$endgroup$
– Ángela Flores
Dec 5 '18 at 20:40




$begingroup$
My mistake, A must be infinite.
$endgroup$
– Ángela Flores
Dec 5 '18 at 20:40












$begingroup$
@AlexR. $l^2$ refers to the space of sequences $(a_1, a_2, ldots)$ such that $sum_n a_n^2 < infty$.
$endgroup$
– angryavian
Dec 5 '18 at 20:47




$begingroup$
@AlexR. $l^2$ refers to the space of sequences $(a_1, a_2, ldots)$ such that $sum_n a_n^2 < infty$.
$endgroup$
– angryavian
Dec 5 '18 at 20:47












$begingroup$
Nitpick: You might want to add $x ne y$ in the formulation of your question. Otherwise, it is not possible (obviously).
$endgroup$
– gerw
Dec 6 '18 at 7:52




$begingroup$
Nitpick: You might want to add $x ne y$ in the formulation of your question. Otherwise, it is not possible (obviously).
$endgroup$
– gerw
Dec 6 '18 at 7:52












$begingroup$
An interesting spinoff: How large can $$inf_{x,y in A, x ne y} |x-y|$$ be? Can this be larger than $sqrt{2}$?
$endgroup$
– gerw
Dec 6 '18 at 8:04






$begingroup$
An interesting spinoff: How large can $$inf_{x,y in A, x ne y} |x-y|$$ be? Can this be larger than $sqrt{2}$?
$endgroup$
– gerw
Dec 6 '18 at 8:04












1 Answer
1






active

oldest

votes


















2












$begingroup$

Recall that $ell^2$ is a Hilbert space with the natural inner product. Note that:
$$||x-y||^2=||x||^2+||y||^2-2langle x,yrangle,$$
in particular, if $x,yin S$ ($S$ the set of points with norm equal to $1$)
$$||x-y||^2=2(1-langle x,yrangle).$$



So it is enough to find an infinite family $B$ in $ell^2$ such that their inner product is $langle x,yrangle<0$ for all $x,yin B$, $xneq y$. Since you can consider
$$A=leftlbrace x/||x||;:;xin B, xneq 0rightrbrace,$$
and we have for $x/||x||$, $y/||y||$, $x,yin Bsetminus{0}$ the following
begin{align} ||x-y||^2 & = 2left( 1- frac{1}{||x||cdot||y||}langle x,yrangleright) \ & > 2.end{align}



Let $f:mathbb{N}tomathbb{N}:nmapsto n!$. Clearly $f(n)^2-nf(n-1)^2>0$. Now, let $(e_n)_n$ the standard basis of $ell^2$, and we define
$$u_n=f(n)e_n-sum_{k=1}^{n-1}f(k)e_k.$$
The set $B$ will be the set of the $u_n$'s. Let's check that for every pair of distinct natural numbers $n,m$ it is satisfied that $langle u_n,u_mrangle<0$. In fact
begin{align}langle u_n,u_mrangle & = leftlangle f(n)e_n-sum_{k=1}^{n-1}f(k)e_k,f(m)e_n-sum_{j=1}^{m-1}f(j)e_jrightrangle \ & = leftlangle f(n)e_n, f(m)e_mrightrangle-leftlangle f(n)e_n,sum_{j=1}^{m-1}f(j)e_jrightrangle \ & + leftlanglesum_{k=1}^{n-1}f(k)e_k,sum_{j=1}^{m-1}f(j)e_jrightrangle-leftlangle sum_{k=1}^{n-1}f(k)e_k, f(m)e_mrightrangle \ & = sum_{k=1}^{n-1}sum_{j=1}^{m-1}f(k)f(j)langle e_k,e_jrangle-f(m)^2 \ & = sum_{j=1}^{m-1}f(k)^2-f(m)^2 \ &leq mf(m-1)-f(m)^2 \ & < 0.end{align}






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    2












    $begingroup$

    Recall that $ell^2$ is a Hilbert space with the natural inner product. Note that:
    $$||x-y||^2=||x||^2+||y||^2-2langle x,yrangle,$$
    in particular, if $x,yin S$ ($S$ the set of points with norm equal to $1$)
    $$||x-y||^2=2(1-langle x,yrangle).$$



    So it is enough to find an infinite family $B$ in $ell^2$ such that their inner product is $langle x,yrangle<0$ for all $x,yin B$, $xneq y$. Since you can consider
    $$A=leftlbrace x/||x||;:;xin B, xneq 0rightrbrace,$$
    and we have for $x/||x||$, $y/||y||$, $x,yin Bsetminus{0}$ the following
    begin{align} ||x-y||^2 & = 2left( 1- frac{1}{||x||cdot||y||}langle x,yrangleright) \ & > 2.end{align}



    Let $f:mathbb{N}tomathbb{N}:nmapsto n!$. Clearly $f(n)^2-nf(n-1)^2>0$. Now, let $(e_n)_n$ the standard basis of $ell^2$, and we define
    $$u_n=f(n)e_n-sum_{k=1}^{n-1}f(k)e_k.$$
    The set $B$ will be the set of the $u_n$'s. Let's check that for every pair of distinct natural numbers $n,m$ it is satisfied that $langle u_n,u_mrangle<0$. In fact
    begin{align}langle u_n,u_mrangle & = leftlangle f(n)e_n-sum_{k=1}^{n-1}f(k)e_k,f(m)e_n-sum_{j=1}^{m-1}f(j)e_jrightrangle \ & = leftlangle f(n)e_n, f(m)e_mrightrangle-leftlangle f(n)e_n,sum_{j=1}^{m-1}f(j)e_jrightrangle \ & + leftlanglesum_{k=1}^{n-1}f(k)e_k,sum_{j=1}^{m-1}f(j)e_jrightrangle-leftlangle sum_{k=1}^{n-1}f(k)e_k, f(m)e_mrightrangle \ & = sum_{k=1}^{n-1}sum_{j=1}^{m-1}f(k)f(j)langle e_k,e_jrangle-f(m)^2 \ & = sum_{j=1}^{m-1}f(k)^2-f(m)^2 \ &leq mf(m-1)-f(m)^2 \ & < 0.end{align}






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Recall that $ell^2$ is a Hilbert space with the natural inner product. Note that:
      $$||x-y||^2=||x||^2+||y||^2-2langle x,yrangle,$$
      in particular, if $x,yin S$ ($S$ the set of points with norm equal to $1$)
      $$||x-y||^2=2(1-langle x,yrangle).$$



      So it is enough to find an infinite family $B$ in $ell^2$ such that their inner product is $langle x,yrangle<0$ for all $x,yin B$, $xneq y$. Since you can consider
      $$A=leftlbrace x/||x||;:;xin B, xneq 0rightrbrace,$$
      and we have for $x/||x||$, $y/||y||$, $x,yin Bsetminus{0}$ the following
      begin{align} ||x-y||^2 & = 2left( 1- frac{1}{||x||cdot||y||}langle x,yrangleright) \ & > 2.end{align}



      Let $f:mathbb{N}tomathbb{N}:nmapsto n!$. Clearly $f(n)^2-nf(n-1)^2>0$. Now, let $(e_n)_n$ the standard basis of $ell^2$, and we define
      $$u_n=f(n)e_n-sum_{k=1}^{n-1}f(k)e_k.$$
      The set $B$ will be the set of the $u_n$'s. Let's check that for every pair of distinct natural numbers $n,m$ it is satisfied that $langle u_n,u_mrangle<0$. In fact
      begin{align}langle u_n,u_mrangle & = leftlangle f(n)e_n-sum_{k=1}^{n-1}f(k)e_k,f(m)e_n-sum_{j=1}^{m-1}f(j)e_jrightrangle \ & = leftlangle f(n)e_n, f(m)e_mrightrangle-leftlangle f(n)e_n,sum_{j=1}^{m-1}f(j)e_jrightrangle \ & + leftlanglesum_{k=1}^{n-1}f(k)e_k,sum_{j=1}^{m-1}f(j)e_jrightrangle-leftlangle sum_{k=1}^{n-1}f(k)e_k, f(m)e_mrightrangle \ & = sum_{k=1}^{n-1}sum_{j=1}^{m-1}f(k)f(j)langle e_k,e_jrangle-f(m)^2 \ & = sum_{j=1}^{m-1}f(k)^2-f(m)^2 \ &leq mf(m-1)-f(m)^2 \ & < 0.end{align}






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Recall that $ell^2$ is a Hilbert space with the natural inner product. Note that:
        $$||x-y||^2=||x||^2+||y||^2-2langle x,yrangle,$$
        in particular, if $x,yin S$ ($S$ the set of points with norm equal to $1$)
        $$||x-y||^2=2(1-langle x,yrangle).$$



        So it is enough to find an infinite family $B$ in $ell^2$ such that their inner product is $langle x,yrangle<0$ for all $x,yin B$, $xneq y$. Since you can consider
        $$A=leftlbrace x/||x||;:;xin B, xneq 0rightrbrace,$$
        and we have for $x/||x||$, $y/||y||$, $x,yin Bsetminus{0}$ the following
        begin{align} ||x-y||^2 & = 2left( 1- frac{1}{||x||cdot||y||}langle x,yrangleright) \ & > 2.end{align}



        Let $f:mathbb{N}tomathbb{N}:nmapsto n!$. Clearly $f(n)^2-nf(n-1)^2>0$. Now, let $(e_n)_n$ the standard basis of $ell^2$, and we define
        $$u_n=f(n)e_n-sum_{k=1}^{n-1}f(k)e_k.$$
        The set $B$ will be the set of the $u_n$'s. Let's check that for every pair of distinct natural numbers $n,m$ it is satisfied that $langle u_n,u_mrangle<0$. In fact
        begin{align}langle u_n,u_mrangle & = leftlangle f(n)e_n-sum_{k=1}^{n-1}f(k)e_k,f(m)e_n-sum_{j=1}^{m-1}f(j)e_jrightrangle \ & = leftlangle f(n)e_n, f(m)e_mrightrangle-leftlangle f(n)e_n,sum_{j=1}^{m-1}f(j)e_jrightrangle \ & + leftlanglesum_{k=1}^{n-1}f(k)e_k,sum_{j=1}^{m-1}f(j)e_jrightrangle-leftlangle sum_{k=1}^{n-1}f(k)e_k, f(m)e_mrightrangle \ & = sum_{k=1}^{n-1}sum_{j=1}^{m-1}f(k)f(j)langle e_k,e_jrangle-f(m)^2 \ & = sum_{j=1}^{m-1}f(k)^2-f(m)^2 \ &leq mf(m-1)-f(m)^2 \ & < 0.end{align}






        share|cite|improve this answer









        $endgroup$



        Recall that $ell^2$ is a Hilbert space with the natural inner product. Note that:
        $$||x-y||^2=||x||^2+||y||^2-2langle x,yrangle,$$
        in particular, if $x,yin S$ ($S$ the set of points with norm equal to $1$)
        $$||x-y||^2=2(1-langle x,yrangle).$$



        So it is enough to find an infinite family $B$ in $ell^2$ such that their inner product is $langle x,yrangle<0$ for all $x,yin B$, $xneq y$. Since you can consider
        $$A=leftlbrace x/||x||;:;xin B, xneq 0rightrbrace,$$
        and we have for $x/||x||$, $y/||y||$, $x,yin Bsetminus{0}$ the following
        begin{align} ||x-y||^2 & = 2left( 1- frac{1}{||x||cdot||y||}langle x,yrangleright) \ & > 2.end{align}



        Let $f:mathbb{N}tomathbb{N}:nmapsto n!$. Clearly $f(n)^2-nf(n-1)^2>0$. Now, let $(e_n)_n$ the standard basis of $ell^2$, and we define
        $$u_n=f(n)e_n-sum_{k=1}^{n-1}f(k)e_k.$$
        The set $B$ will be the set of the $u_n$'s. Let's check that for every pair of distinct natural numbers $n,m$ it is satisfied that $langle u_n,u_mrangle<0$. In fact
        begin{align}langle u_n,u_mrangle & = leftlangle f(n)e_n-sum_{k=1}^{n-1}f(k)e_k,f(m)e_n-sum_{j=1}^{m-1}f(j)e_jrightrangle \ & = leftlangle f(n)e_n, f(m)e_mrightrangle-leftlangle f(n)e_n,sum_{j=1}^{m-1}f(j)e_jrightrangle \ & + leftlanglesum_{k=1}^{n-1}f(k)e_k,sum_{j=1}^{m-1}f(j)e_jrightrangle-leftlangle sum_{k=1}^{n-1}f(k)e_k, f(m)e_mrightrangle \ & = sum_{k=1}^{n-1}sum_{j=1}^{m-1}f(k)f(j)langle e_k,e_jrangle-f(m)^2 \ & = sum_{j=1}^{m-1}f(k)^2-f(m)^2 \ &leq mf(m-1)-f(m)^2 \ & < 0.end{align}







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 5 '18 at 23:24









        José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda

        802110




        802110






























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