Prove weighted ball in $l^2$ space is compact

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Let $B subseteq l^2$, $B=left{xin l^2:sum_{ngeq1}n|x_n|^2leq1right}$, show that $B$ is compact.



My thought: $B$ is closed in $l^2$ which is complete. Then $B$ is complete. It suffices to show $B$ is totally bounded. I think we need to first get rid of the infinite tail sum, i.e., bound all sequences with balls centered at sequences that only have finitely many terms. And then find a ball cover for the finite sequences. But I don't know how to bound the tail sum and I'm stuck.






real-analysis general-topology compactness lp-spaces







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    4












    $begingroup$


    Let $B subseteq l^2$, $B=left{xin l^2:sum_{ngeq1}n|x_n|^2leq1right}$, show that $B$ is compact.



    My thought: $B$ is closed in $l^2$ which is complete. Then $B$ is complete. It suffices to show $B$ is totally bounded. I think we need to first get rid of the infinite tail sum, i.e., bound all sequences with balls centered at sequences that only have finitely many terms. And then find a ball cover for the finite sequences. But I don't know how to bound the tail sum and I'm stuck.






    real-analysis general-topology compactness lp-spaces







    share|cite|improve this question











    $endgroup$















      4












      4








      4





      $begingroup$


      Let $B subseteq l^2$, $B=left{xin l^2:sum_{ngeq1}n|x_n|^2leq1right}$, show that $B$ is compact.



      My thought: $B$ is closed in $l^2$ which is complete. Then $B$ is complete. It suffices to show $B$ is totally bounded. I think we need to first get rid of the infinite tail sum, i.e., bound all sequences with balls centered at sequences that only have finitely many terms. And then find a ball cover for the finite sequences. But I don't know how to bound the tail sum and I'm stuck.






      real-analysis general-topology compactness lp-spaces







      share|cite|improve this question











      $endgroup$




      Let $B subseteq l^2$, $B=left{xin l^2:sum_{ngeq1}n|x_n|^2leq1right}$, show that $B$ is compact.



      My thought: $B$ is closed in $l^2$ which is complete. Then $B$ is complete. It suffices to show $B$ is totally bounded. I think we need to first get rid of the infinite tail sum, i.e., bound all sequences with balls centered at sequences that only have finitely many terms. And then find a ball cover for the finite sequences. But I don't know how to bound the tail sum and I'm stuck.






      real-analysis general-topology compactness lp-spaces




      real-analysis general-topology compactness lp-spaces






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 17 at 20:51









      Davide Giraudo

      126k16150261




      126k16150261










      asked Jan 17 at 18:26









      Fluffy SkyeFluffy Skye

      1148




      1148






















          1 Answer
          1






          active

          oldest

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          2












          $begingroup$


          Theorem. Let $pin[1,+infty]$ and $Msubseteqell_p$ be a bounded subset such that
          $$
          limlimits_{Ntoinfty}sup{Vert (0,0,ldots,0,x_N,x_{N+1},ldots) Vert_p:xin M}=0
          tag{1}$$
          then $M$ is totally bounded.




          For the proof, see an answer of that question: How to show that this set is compact in $ell^2$.



          The theorem provides a characterization of totally bounded subsets of $ell_p$.



          Let’s apply it to $B$.



          For $xin B$, we have



          $$Vert xVert_2 = sum_{nge1} vert x_n vert^2 le sum_{nge1} nvert x_n vert^2 le 1$$ proving that $B$ in included in the closed ball centered on the origin with radius equal to one. Hence $B$ is bounded.



          We’ll be done if we prove that $B$ satisfies condition $(1)$ of theorem above.



          For $x in B$



          $$Nsum_{kge N} vert x_kvert^2 le sum_{kge N} kvert x_kvert^2 le sum_{kge 1} kvert x_kvert^2 le 1$$
          Therefore
          $$sup{Vert (0,0,ldots,0,x_N,x_{N+1},ldots) Vert_2:xin B} le 1/N$$ and condition $(1)$ is satisfied.



          $B$ is totally bounded. And also complete as being a closed subset of a complete space.



          Finally $B$ is compact.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you so much!
            $endgroup$
            – Fluffy Skye
            Jan 17 at 20:25










          • $begingroup$
            @FluffySkye You’re welcome! The ideas that you mentioned in your question are in fact applied behind the stated theorem.
            $endgroup$
            – mathcounterexamples.net
            Jan 17 at 20:27












          • $begingroup$
            @mathcounterexamples.net I think you made a typo in your second last inequality.
            $endgroup$
            – BigbearZzz
            Jan 17 at 20:34










          • $begingroup$
            @BigbearZzz Thanks for noticing. I corrected.
            $endgroup$
            – mathcounterexamples.net
            Jan 17 at 20:35











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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$


          Theorem. Let $pin[1,+infty]$ and $Msubseteqell_p$ be a bounded subset such that
          $$
          limlimits_{Ntoinfty}sup{Vert (0,0,ldots,0,x_N,x_{N+1},ldots) Vert_p:xin M}=0
          tag{1}$$
          then $M$ is totally bounded.




          For the proof, see an answer of that question: How to show that this set is compact in $ell^2$.



          The theorem provides a characterization of totally bounded subsets of $ell_p$.



          Let’s apply it to $B$.



          For $xin B$, we have



          $$Vert xVert_2 = sum_{nge1} vert x_n vert^2 le sum_{nge1} nvert x_n vert^2 le 1$$ proving that $B$ in included in the closed ball centered on the origin with radius equal to one. Hence $B$ is bounded.



          We’ll be done if we prove that $B$ satisfies condition $(1)$ of theorem above.



          For $x in B$



          $$Nsum_{kge N} vert x_kvert^2 le sum_{kge N} kvert x_kvert^2 le sum_{kge 1} kvert x_kvert^2 le 1$$
          Therefore
          $$sup{Vert (0,0,ldots,0,x_N,x_{N+1},ldots) Vert_2:xin B} le 1/N$$ and condition $(1)$ is satisfied.



          $B$ is totally bounded. And also complete as being a closed subset of a complete space.



          Finally $B$ is compact.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you so much!
            $endgroup$
            – Fluffy Skye
            Jan 17 at 20:25










          • $begingroup$
            @FluffySkye You’re welcome! The ideas that you mentioned in your question are in fact applied behind the stated theorem.
            $endgroup$
            – mathcounterexamples.net
            Jan 17 at 20:27












          • $begingroup$
            @mathcounterexamples.net I think you made a typo in your second last inequality.
            $endgroup$
            – BigbearZzz
            Jan 17 at 20:34










          • $begingroup$
            @BigbearZzz Thanks for noticing. I corrected.
            $endgroup$
            – mathcounterexamples.net
            Jan 17 at 20:35
















          2












          $begingroup$


          Theorem. Let $pin[1,+infty]$ and $Msubseteqell_p$ be a bounded subset such that
          $$
          limlimits_{Ntoinfty}sup{Vert (0,0,ldots,0,x_N,x_{N+1},ldots) Vert_p:xin M}=0
          tag{1}$$
          then $M$ is totally bounded.




          For the proof, see an answer of that question: How to show that this set is compact in $ell^2$.



          The theorem provides a characterization of totally bounded subsets of $ell_p$.



          Let’s apply it to $B$.



          For $xin B$, we have



          $$Vert xVert_2 = sum_{nge1} vert x_n vert^2 le sum_{nge1} nvert x_n vert^2 le 1$$ proving that $B$ in included in the closed ball centered on the origin with radius equal to one. Hence $B$ is bounded.



          We’ll be done if we prove that $B$ satisfies condition $(1)$ of theorem above.



          For $x in B$



          $$Nsum_{kge N} vert x_kvert^2 le sum_{kge N} kvert x_kvert^2 le sum_{kge 1} kvert x_kvert^2 le 1$$
          Therefore
          $$sup{Vert (0,0,ldots,0,x_N,x_{N+1},ldots) Vert_2:xin B} le 1/N$$ and condition $(1)$ is satisfied.



          $B$ is totally bounded. And also complete as being a closed subset of a complete space.



          Finally $B$ is compact.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you so much!
            $endgroup$
            – Fluffy Skye
            Jan 17 at 20:25










          • $begingroup$
            @FluffySkye You’re welcome! The ideas that you mentioned in your question are in fact applied behind the stated theorem.
            $endgroup$
            – mathcounterexamples.net
            Jan 17 at 20:27












          • $begingroup$
            @mathcounterexamples.net I think you made a typo in your second last inequality.
            $endgroup$
            – BigbearZzz
            Jan 17 at 20:34










          • $begingroup$
            @BigbearZzz Thanks for noticing. I corrected.
            $endgroup$
            – mathcounterexamples.net
            Jan 17 at 20:35














          2












          2








          2





          $begingroup$


          Theorem. Let $pin[1,+infty]$ and $Msubseteqell_p$ be a bounded subset such that
          $$
          limlimits_{Ntoinfty}sup{Vert (0,0,ldots,0,x_N,x_{N+1},ldots) Vert_p:xin M}=0
          tag{1}$$
          then $M$ is totally bounded.




          For the proof, see an answer of that question: How to show that this set is compact in $ell^2$.



          The theorem provides a characterization of totally bounded subsets of $ell_p$.



          Let’s apply it to $B$.



          For $xin B$, we have



          $$Vert xVert_2 = sum_{nge1} vert x_n vert^2 le sum_{nge1} nvert x_n vert^2 le 1$$ proving that $B$ in included in the closed ball centered on the origin with radius equal to one. Hence $B$ is bounded.



          We’ll be done if we prove that $B$ satisfies condition $(1)$ of theorem above.



          For $x in B$



          $$Nsum_{kge N} vert x_kvert^2 le sum_{kge N} kvert x_kvert^2 le sum_{kge 1} kvert x_kvert^2 le 1$$
          Therefore
          $$sup{Vert (0,0,ldots,0,x_N,x_{N+1},ldots) Vert_2:xin B} le 1/N$$ and condition $(1)$ is satisfied.



          $B$ is totally bounded. And also complete as being a closed subset of a complete space.



          Finally $B$ is compact.






          share|cite|improve this answer











          $endgroup$




          Theorem. Let $pin[1,+infty]$ and $Msubseteqell_p$ be a bounded subset such that
          $$
          limlimits_{Ntoinfty}sup{Vert (0,0,ldots,0,x_N,x_{N+1},ldots) Vert_p:xin M}=0
          tag{1}$$
          then $M$ is totally bounded.




          For the proof, see an answer of that question: How to show that this set is compact in $ell^2$.



          The theorem provides a characterization of totally bounded subsets of $ell_p$.



          Let’s apply it to $B$.



          For $xin B$, we have



          $$Vert xVert_2 = sum_{nge1} vert x_n vert^2 le sum_{nge1} nvert x_n vert^2 le 1$$ proving that $B$ in included in the closed ball centered on the origin with radius equal to one. Hence $B$ is bounded.



          We’ll be done if we prove that $B$ satisfies condition $(1)$ of theorem above.



          For $x in B$



          $$Nsum_{kge N} vert x_kvert^2 le sum_{kge N} kvert x_kvert^2 le sum_{kge 1} kvert x_kvert^2 le 1$$
          Therefore
          $$sup{Vert (0,0,ldots,0,x_N,x_{N+1},ldots) Vert_2:xin B} le 1/N$$ and condition $(1)$ is satisfied.



          $B$ is totally bounded. And also complete as being a closed subset of a complete space.



          Finally $B$ is compact.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 17 at 20:35

























          answered Jan 17 at 20:17









          mathcounterexamples.netmathcounterexamples.net

          26.7k22157




          26.7k22157












          • $begingroup$
            Thank you so much!
            $endgroup$
            – Fluffy Skye
            Jan 17 at 20:25










          • $begingroup$
            @FluffySkye You’re welcome! The ideas that you mentioned in your question are in fact applied behind the stated theorem.
            $endgroup$
            – mathcounterexamples.net
            Jan 17 at 20:27












          • $begingroup$
            @mathcounterexamples.net I think you made a typo in your second last inequality.
            $endgroup$
            – BigbearZzz
            Jan 17 at 20:34










          • $begingroup$
            @BigbearZzz Thanks for noticing. I corrected.
            $endgroup$
            – mathcounterexamples.net
            Jan 17 at 20:35


















          • $begingroup$
            Thank you so much!
            $endgroup$
            – Fluffy Skye
            Jan 17 at 20:25










          • $begingroup$
            @FluffySkye You’re welcome! The ideas that you mentioned in your question are in fact applied behind the stated theorem.
            $endgroup$
            – mathcounterexamples.net
            Jan 17 at 20:27












          • $begingroup$
            @mathcounterexamples.net I think you made a typo in your second last inequality.
            $endgroup$
            – BigbearZzz
            Jan 17 at 20:34










          • $begingroup$
            @BigbearZzz Thanks for noticing. I corrected.
            $endgroup$
            – mathcounterexamples.net
            Jan 17 at 20:35
















          $begingroup$
          Thank you so much!
          $endgroup$
          – Fluffy Skye
          Jan 17 at 20:25




          $begingroup$
          Thank you so much!
          $endgroup$
          – Fluffy Skye
          Jan 17 at 20:25












          $begingroup$
          @FluffySkye You’re welcome! The ideas that you mentioned in your question are in fact applied behind the stated theorem.
          $endgroup$
          – mathcounterexamples.net
          Jan 17 at 20:27






          $begingroup$
          @FluffySkye You’re welcome! The ideas that you mentioned in your question are in fact applied behind the stated theorem.
          $endgroup$
          – mathcounterexamples.net
          Jan 17 at 20:27














          $begingroup$
          @mathcounterexamples.net I think you made a typo in your second last inequality.
          $endgroup$
          – BigbearZzz
          Jan 17 at 20:34




          $begingroup$
          @mathcounterexamples.net I think you made a typo in your second last inequality.
          $endgroup$
          – BigbearZzz
          Jan 17 at 20:34












          $begingroup$
          @BigbearZzz Thanks for noticing. I corrected.
          $endgroup$
          – mathcounterexamples.net
          Jan 17 at 20:35




          $begingroup$
          @BigbearZzz Thanks for noticing. I corrected.
          $endgroup$
          – mathcounterexamples.net
          Jan 17 at 20:35


















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