Show that the space of all real sequences with finitely many nonzero is not complete with a specific norm












1












$begingroup$



Consider the space $ell_{0}^{2}$ of all real sequences with only a
finite number of nonzero terms. Show that the space is not complete,
with the norm $|x|={left(sum|x|^{2}right)}^{1/2}$.




My attempt: I've had a break for some time, so I am a bit rusty. But, I really just have to find a Cauchy sequence that does not converge. I was thinking about a sequence where $x_n=1/n$, the first $n$ terms and then zero (it has helped me before, but I'm not sure whether it works). Assume $m>n$, then



$$
d^2(x_m,x_n)=|x_m-x_n|^2=sum_{j=n}^mfrac{1}{j^2}leqfrac{m-n+1}{n^2}
$$



which we can make arbitrarily small, and therefore it is a Cauchy sequence. (right?)



So, I'm thinking that obviously the limit doesn't exist, since it would not have finitely many zeros, and therefore it does not converge. Is my reasoning OK? If so, how do I continue?



Alternative approach: If you have some alternative approach I'm happy to hear about it.






real-analysis







share|cite|improve this question









$endgroup$












  • $begingroup$
    Yes, it's OK. (Perhaps, explain why the "limit" would not have only finitely many non-zero coordinates.)
    $endgroup$
    – David Mitra
    Oct 1 '14 at 10:38












  • $begingroup$
    Well, as $ntoinfty$ so does the nonezero elements. But perhaps I could pick an easier sequence?
    $endgroup$
    – user136475
    Oct 1 '14 at 15:33










  • $begingroup$
    What happens with the sequence $a_n = (1, ldots, 1/n, 0, ldots)$?
    $endgroup$
    – Will M.
    Jan 13 at 2:33
















1












$begingroup$



Consider the space $ell_{0}^{2}$ of all real sequences with only a
finite number of nonzero terms. Show that the space is not complete,
with the norm $|x|={left(sum|x|^{2}right)}^{1/2}$.




My attempt: I've had a break for some time, so I am a bit rusty. But, I really just have to find a Cauchy sequence that does not converge. I was thinking about a sequence where $x_n=1/n$, the first $n$ terms and then zero (it has helped me before, but I'm not sure whether it works). Assume $m>n$, then



$$
d^2(x_m,x_n)=|x_m-x_n|^2=sum_{j=n}^mfrac{1}{j^2}leqfrac{m-n+1}{n^2}
$$



which we can make arbitrarily small, and therefore it is a Cauchy sequence. (right?)



So, I'm thinking that obviously the limit doesn't exist, since it would not have finitely many zeros, and therefore it does not converge. Is my reasoning OK? If so, how do I continue?



Alternative approach: If you have some alternative approach I'm happy to hear about it.






real-analysis







share|cite|improve this question









$endgroup$












  • $begingroup$
    Yes, it's OK. (Perhaps, explain why the "limit" would not have only finitely many non-zero coordinates.)
    $endgroup$
    – David Mitra
    Oct 1 '14 at 10:38












  • $begingroup$
    Well, as $ntoinfty$ so does the nonezero elements. But perhaps I could pick an easier sequence?
    $endgroup$
    – user136475
    Oct 1 '14 at 15:33










  • $begingroup$
    What happens with the sequence $a_n = (1, ldots, 1/n, 0, ldots)$?
    $endgroup$
    – Will M.
    Jan 13 at 2:33














1












1








1





$begingroup$



Consider the space $ell_{0}^{2}$ of all real sequences with only a
finite number of nonzero terms. Show that the space is not complete,
with the norm $|x|={left(sum|x|^{2}right)}^{1/2}$.




My attempt: I've had a break for some time, so I am a bit rusty. But, I really just have to find a Cauchy sequence that does not converge. I was thinking about a sequence where $x_n=1/n$, the first $n$ terms and then zero (it has helped me before, but I'm not sure whether it works). Assume $m>n$, then



$$
d^2(x_m,x_n)=|x_m-x_n|^2=sum_{j=n}^mfrac{1}{j^2}leqfrac{m-n+1}{n^2}
$$



which we can make arbitrarily small, and therefore it is a Cauchy sequence. (right?)



So, I'm thinking that obviously the limit doesn't exist, since it would not have finitely many zeros, and therefore it does not converge. Is my reasoning OK? If so, how do I continue?



Alternative approach: If you have some alternative approach I'm happy to hear about it.






real-analysis







share|cite|improve this question









$endgroup$





Consider the space $ell_{0}^{2}$ of all real sequences with only a
finite number of nonzero terms. Show that the space is not complete,
with the norm $|x|={left(sum|x|^{2}right)}^{1/2}$.




My attempt: I've had a break for some time, so I am a bit rusty. But, I really just have to find a Cauchy sequence that does not converge. I was thinking about a sequence where $x_n=1/n$, the first $n$ terms and then zero (it has helped me before, but I'm not sure whether it works). Assume $m>n$, then



$$
d^2(x_m,x_n)=|x_m-x_n|^2=sum_{j=n}^mfrac{1}{j^2}leqfrac{m-n+1}{n^2}
$$



which we can make arbitrarily small, and therefore it is a Cauchy sequence. (right?)



So, I'm thinking that obviously the limit doesn't exist, since it would not have finitely many zeros, and therefore it does not converge. Is my reasoning OK? If so, how do I continue?



Alternative approach: If you have some alternative approach I'm happy to hear about it.






real-analysis




real-analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Oct 1 '14 at 10:30









user136475user136475

325




325












  • $begingroup$
    Yes, it's OK. (Perhaps, explain why the "limit" would not have only finitely many non-zero coordinates.)
    $endgroup$
    – David Mitra
    Oct 1 '14 at 10:38












  • $begingroup$
    Well, as $ntoinfty$ so does the nonezero elements. But perhaps I could pick an easier sequence?
    $endgroup$
    – user136475
    Oct 1 '14 at 15:33










  • $begingroup$
    What happens with the sequence $a_n = (1, ldots, 1/n, 0, ldots)$?
    $endgroup$
    – Will M.
    Jan 13 at 2:33


















  • $begingroup$
    Yes, it's OK. (Perhaps, explain why the "limit" would not have only finitely many non-zero coordinates.)
    $endgroup$
    – David Mitra
    Oct 1 '14 at 10:38












  • $begingroup$
    Well, as $ntoinfty$ so does the nonezero elements. But perhaps I could pick an easier sequence?
    $endgroup$
    – user136475
    Oct 1 '14 at 15:33










  • $begingroup$
    What happens with the sequence $a_n = (1, ldots, 1/n, 0, ldots)$?
    $endgroup$
    – Will M.
    Jan 13 at 2:33
















$begingroup$
Yes, it's OK. (Perhaps, explain why the "limit" would not have only finitely many non-zero coordinates.)
$endgroup$
– David Mitra
Oct 1 '14 at 10:38






$begingroup$
Yes, it's OK. (Perhaps, explain why the "limit" would not have only finitely many non-zero coordinates.)
$endgroup$
– David Mitra
Oct 1 '14 at 10:38














$begingroup$
Well, as $ntoinfty$ so does the nonezero elements. But perhaps I could pick an easier sequence?
$endgroup$
– user136475
Oct 1 '14 at 15:33




$begingroup$
Well, as $ntoinfty$ so does the nonezero elements. But perhaps I could pick an easier sequence?
$endgroup$
– user136475
Oct 1 '14 at 15:33












$begingroup$
What happens with the sequence $a_n = (1, ldots, 1/n, 0, ldots)$?
$endgroup$
– Will M.
Jan 13 at 2:33




$begingroup$
What happens with the sequence $a_n = (1, ldots, 1/n, 0, ldots)$?
$endgroup$
– Will M.
Jan 13 at 2:33










2 Answers
2






active

oldest

votes


















2












$begingroup$

I think your approach is fine. One way to finish is to observe that the space $l^{2}_{0}$ is contained in $l^2$. Your limit sequence is in $l^2$ but not $l^{2}_{0}.$ Since a sequence has a unique limit in $l^2$ it can't have a limit in $l^{2}_{0}$ so $l^{2}_{0}$ is not complete.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Yes. Take any sequence $(a_j)_j$ of positive reals such that $sum_{jin Bbb N}a_j^2<infty.$ Let $x_n=(x_{n,j})_j$ where $x_{n,j}=0$ if $j>n$ and $x_{n,j}=a_j$ if $jle n.$ Then $$lim_{nto infty}sup_{m>n}|x_n-x_m|^2=lim_{nto infty}sum_{j=n+1}^{infty}a_j^2=0.$$ So $(x_n)_n$ is a Cauchy sequence.



    If $y=(y_j)_jin l^2_0$ and $lim_{nto infty}|x_n-y|=0,$ then for each $j,$ we have $lim_{nto infty}|x_{n,j}-y_j|=0.$



    Otherwise, for at least one $j$, there would be some $r>0$ such that $|x_{n,j}-y_j|>r$ for infinitely many $n.$ But then $|x_n-y|ge |x_{n,j}-y_j|ge r$ for infinitely many $n.$



    So if $y=lim_{nto infty}x_n$ exists in $l^2_0$ then $y_j=a_jne 0$ for all $j,$ so $ynot in l^2_0,$ a contradiction.



    Remark: Set-theoretically, a sequence is a function $f$ with domain $Bbb N$ (or some other suitable countably infinite domain). So let $x_n=f_n$ where $f_n(j)=x_{n,j}$ and let $y_j=g(j).$ If $(f_n)_n$ converges in norm to $g$ then $f_n$ converges point-wise to $g.$ That is, $f_n(j)to g(j)$ for each $j,$ Because $|f_n(j)-g(j)|=|x_{n,j}-y_j|le |x_n-y|to 0 $ as $nto infty.$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      It is sufficient that $sum_{jin Bbb N}a_j^2<infty$ and that $a_jne 0$ for infinitely many $n.$ Then in the last line in this answer we could say "... then $y_j=a_jne 0$ for infinitely many $n$..."
      $endgroup$
      – DanielWainfleet
      Jan 13 at 2:20













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    2 Answers
    2






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    2 Answers
    2






    active

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    active

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    2












    $begingroup$

    I think your approach is fine. One way to finish is to observe that the space $l^{2}_{0}$ is contained in $l^2$. Your limit sequence is in $l^2$ but not $l^{2}_{0}.$ Since a sequence has a unique limit in $l^2$ it can't have a limit in $l^{2}_{0}$ so $l^{2}_{0}$ is not complete.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      I think your approach is fine. One way to finish is to observe that the space $l^{2}_{0}$ is contained in $l^2$. Your limit sequence is in $l^2$ but not $l^{2}_{0}.$ Since a sequence has a unique limit in $l^2$ it can't have a limit in $l^{2}_{0}$ so $l^{2}_{0}$ is not complete.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        I think your approach is fine. One way to finish is to observe that the space $l^{2}_{0}$ is contained in $l^2$. Your limit sequence is in $l^2$ but not $l^{2}_{0}.$ Since a sequence has a unique limit in $l^2$ it can't have a limit in $l^{2}_{0}$ so $l^{2}_{0}$ is not complete.






        share|cite|improve this answer









        $endgroup$



        I think your approach is fine. One way to finish is to observe that the space $l^{2}_{0}$ is contained in $l^2$. Your limit sequence is in $l^2$ but not $l^{2}_{0}.$ Since a sequence has a unique limit in $l^2$ it can't have a limit in $l^{2}_{0}$ so $l^{2}_{0}$ is not complete.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Feb 26 '15 at 5:14









        Mark JoshiMark Joshi

        4,9131712




        4,9131712























            1












            $begingroup$

            Yes. Take any sequence $(a_j)_j$ of positive reals such that $sum_{jin Bbb N}a_j^2<infty.$ Let $x_n=(x_{n,j})_j$ where $x_{n,j}=0$ if $j>n$ and $x_{n,j}=a_j$ if $jle n.$ Then $$lim_{nto infty}sup_{m>n}|x_n-x_m|^2=lim_{nto infty}sum_{j=n+1}^{infty}a_j^2=0.$$ So $(x_n)_n$ is a Cauchy sequence.



            If $y=(y_j)_jin l^2_0$ and $lim_{nto infty}|x_n-y|=0,$ then for each $j,$ we have $lim_{nto infty}|x_{n,j}-y_j|=0.$



            Otherwise, for at least one $j$, there would be some $r>0$ such that $|x_{n,j}-y_j|>r$ for infinitely many $n.$ But then $|x_n-y|ge |x_{n,j}-y_j|ge r$ for infinitely many $n.$



            So if $y=lim_{nto infty}x_n$ exists in $l^2_0$ then $y_j=a_jne 0$ for all $j,$ so $ynot in l^2_0,$ a contradiction.



            Remark: Set-theoretically, a sequence is a function $f$ with domain $Bbb N$ (or some other suitable countably infinite domain). So let $x_n=f_n$ where $f_n(j)=x_{n,j}$ and let $y_j=g(j).$ If $(f_n)_n$ converges in norm to $g$ then $f_n$ converges point-wise to $g.$ That is, $f_n(j)to g(j)$ for each $j,$ Because $|f_n(j)-g(j)|=|x_{n,j}-y_j|le |x_n-y|to 0 $ as $nto infty.$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              It is sufficient that $sum_{jin Bbb N}a_j^2<infty$ and that $a_jne 0$ for infinitely many $n.$ Then in the last line in this answer we could say "... then $y_j=a_jne 0$ for infinitely many $n$..."
              $endgroup$
              – DanielWainfleet
              Jan 13 at 2:20


















            1












            $begingroup$

            Yes. Take any sequence $(a_j)_j$ of positive reals such that $sum_{jin Bbb N}a_j^2<infty.$ Let $x_n=(x_{n,j})_j$ where $x_{n,j}=0$ if $j>n$ and $x_{n,j}=a_j$ if $jle n.$ Then $$lim_{nto infty}sup_{m>n}|x_n-x_m|^2=lim_{nto infty}sum_{j=n+1}^{infty}a_j^2=0.$$ So $(x_n)_n$ is a Cauchy sequence.



            If $y=(y_j)_jin l^2_0$ and $lim_{nto infty}|x_n-y|=0,$ then for each $j,$ we have $lim_{nto infty}|x_{n,j}-y_j|=0.$



            Otherwise, for at least one $j$, there would be some $r>0$ such that $|x_{n,j}-y_j|>r$ for infinitely many $n.$ But then $|x_n-y|ge |x_{n,j}-y_j|ge r$ for infinitely many $n.$



            So if $y=lim_{nto infty}x_n$ exists in $l^2_0$ then $y_j=a_jne 0$ for all $j,$ so $ynot in l^2_0,$ a contradiction.



            Remark: Set-theoretically, a sequence is a function $f$ with domain $Bbb N$ (or some other suitable countably infinite domain). So let $x_n=f_n$ where $f_n(j)=x_{n,j}$ and let $y_j=g(j).$ If $(f_n)_n$ converges in norm to $g$ then $f_n$ converges point-wise to $g.$ That is, $f_n(j)to g(j)$ for each $j,$ Because $|f_n(j)-g(j)|=|x_{n,j}-y_j|le |x_n-y|to 0 $ as $nto infty.$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              It is sufficient that $sum_{jin Bbb N}a_j^2<infty$ and that $a_jne 0$ for infinitely many $n.$ Then in the last line in this answer we could say "... then $y_j=a_jne 0$ for infinitely many $n$..."
              $endgroup$
              – DanielWainfleet
              Jan 13 at 2:20
















            1












            1








            1





            $begingroup$

            Yes. Take any sequence $(a_j)_j$ of positive reals such that $sum_{jin Bbb N}a_j^2<infty.$ Let $x_n=(x_{n,j})_j$ where $x_{n,j}=0$ if $j>n$ and $x_{n,j}=a_j$ if $jle n.$ Then $$lim_{nto infty}sup_{m>n}|x_n-x_m|^2=lim_{nto infty}sum_{j=n+1}^{infty}a_j^2=0.$$ So $(x_n)_n$ is a Cauchy sequence.



            If $y=(y_j)_jin l^2_0$ and $lim_{nto infty}|x_n-y|=0,$ then for each $j,$ we have $lim_{nto infty}|x_{n,j}-y_j|=0.$



            Otherwise, for at least one $j$, there would be some $r>0$ such that $|x_{n,j}-y_j|>r$ for infinitely many $n.$ But then $|x_n-y|ge |x_{n,j}-y_j|ge r$ for infinitely many $n.$



            So if $y=lim_{nto infty}x_n$ exists in $l^2_0$ then $y_j=a_jne 0$ for all $j,$ so $ynot in l^2_0,$ a contradiction.



            Remark: Set-theoretically, a sequence is a function $f$ with domain $Bbb N$ (or some other suitable countably infinite domain). So let $x_n=f_n$ where $f_n(j)=x_{n,j}$ and let $y_j=g(j).$ If $(f_n)_n$ converges in norm to $g$ then $f_n$ converges point-wise to $g.$ That is, $f_n(j)to g(j)$ for each $j,$ Because $|f_n(j)-g(j)|=|x_{n,j}-y_j|le |x_n-y|to 0 $ as $nto infty.$






            share|cite|improve this answer











            $endgroup$



            Yes. Take any sequence $(a_j)_j$ of positive reals such that $sum_{jin Bbb N}a_j^2<infty.$ Let $x_n=(x_{n,j})_j$ where $x_{n,j}=0$ if $j>n$ and $x_{n,j}=a_j$ if $jle n.$ Then $$lim_{nto infty}sup_{m>n}|x_n-x_m|^2=lim_{nto infty}sum_{j=n+1}^{infty}a_j^2=0.$$ So $(x_n)_n$ is a Cauchy sequence.



            If $y=(y_j)_jin l^2_0$ and $lim_{nto infty}|x_n-y|=0,$ then for each $j,$ we have $lim_{nto infty}|x_{n,j}-y_j|=0.$



            Otherwise, for at least one $j$, there would be some $r>0$ such that $|x_{n,j}-y_j|>r$ for infinitely many $n.$ But then $|x_n-y|ge |x_{n,j}-y_j|ge r$ for infinitely many $n.$



            So if $y=lim_{nto infty}x_n$ exists in $l^2_0$ then $y_j=a_jne 0$ for all $j,$ so $ynot in l^2_0,$ a contradiction.



            Remark: Set-theoretically, a sequence is a function $f$ with domain $Bbb N$ (or some other suitable countably infinite domain). So let $x_n=f_n$ where $f_n(j)=x_{n,j}$ and let $y_j=g(j).$ If $(f_n)_n$ converges in norm to $g$ then $f_n$ converges point-wise to $g.$ That is, $f_n(j)to g(j)$ for each $j,$ Because $|f_n(j)-g(j)|=|x_{n,j}-y_j|le |x_n-y|to 0 $ as $nto infty.$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 13 at 2:35

























            answered Jan 13 at 2:15









            DanielWainfleetDanielWainfleet

            35.1k31648




            35.1k31648












            • $begingroup$
              It is sufficient that $sum_{jin Bbb N}a_j^2<infty$ and that $a_jne 0$ for infinitely many $n.$ Then in the last line in this answer we could say "... then $y_j=a_jne 0$ for infinitely many $n$..."
              $endgroup$
              – DanielWainfleet
              Jan 13 at 2:20




















            • $begingroup$
              It is sufficient that $sum_{jin Bbb N}a_j^2<infty$ and that $a_jne 0$ for infinitely many $n.$ Then in the last line in this answer we could say "... then $y_j=a_jne 0$ for infinitely many $n$..."
              $endgroup$
              – DanielWainfleet
              Jan 13 at 2:20


















            $begingroup$
            It is sufficient that $sum_{jin Bbb N}a_j^2<infty$ and that $a_jne 0$ for infinitely many $n.$ Then in the last line in this answer we could say "... then $y_j=a_jne 0$ for infinitely many $n$..."
            $endgroup$
            – DanielWainfleet
            Jan 13 at 2:20






            $begingroup$
            It is sufficient that $sum_{jin Bbb N}a_j^2<infty$ and that $a_jne 0$ for infinitely many $n.$ Then in the last line in this answer we could say "... then $y_j=a_jne 0$ for infinitely many $n$..."
            $endgroup$
            – DanielWainfleet
            Jan 13 at 2:20




















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            How to fix TextFormField cause rebuild widget in Flutter

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            .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I've got the same issue at #9280. I tried all solutions from that topic with/without GlobalKey for Scaffold and Form but the problem wasn't fixed. I tried to build a authentication form with login and signup. But the keyboard doesn't show when I tap the TextFormField at the first time. Then I try to input some letter from physical keyboard, the widget is rebuilt. Like this: login screen Nothing to be log when I run flutter run --verbose, so I logged manually. Every tapping on field, the widget called dispose -> init -> build. [√] Flutter (Channel beta, v1.0.0, on Microsoft Windows [Version 10.0.17763.195], ...
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            in spring boot 2.1 many test slices are not allowed anymore due to multiple @BootstrapWith

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            1 I tried to upgrade a yummy sandwich made of two test slices (@JsonTest and @JdbcTest in my case, crunchy test code in between) adding spring boot 2.1 flavour to it. But it seems it was not much of a success. I cannot annotate my tests with many @...Test since they are now each bringing their own XxxTestContextBootstrapper. It used to work when they all used same SpringBootTestContextBootstrapper. @RunWith(SpringRunner.class) @JdbcTest @JsonTest public class Test { @Test public void test() { System.out.printn("Hello, World !"); } } The error I get from BootstrapUtils is illegalStateException : Configuration error: found multiple declarations of @BootstrapWith for test class I understand I might be doing something wrong here but is there an easy way I could load both Json and Jdbc contex...
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