Prove that $C_0$ is Banach.

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Let $x^n in C_0$ is Cauchy.



$rightarrow$ For $epsilon> 0$ there is N such that $n,m ge N $ $$ lVert x^m -x^nrVert< frac {epsilon} {2} $$
we know that for every k $$lvert x^m_k -x^n_krvert le sup_{ige 1} lvert x^m_i -x^n_irvert < frac {epsilon} {2} $$
So $x^n_k$ is Cauchy in $mathbb R$ which is Banach so $$x^n_k rightarrow x_k in mathbb R or lvert x^n_k -x_krvert le frac {epsilon} {2} $$
by this we can say that there is an N, $nge N$ such that $lVert x^n -xrVert = sup_{ige 1} lvert x^n_i -x_i lvert < frac {epsilon} {2} $ $ $



which means $x^n rightarrow x$



Since $x^m in c_0$,
there is some $N'$ such that $|x_i^m| < {1 over 2 } epsilon$ for $i ge N' (m=N)$



Now to show that $xin C_0$



$$lvert x_irvert le lvert x^m_i-x_irvert+ lvert x^m_irvert <frac {epsilon} {2}+frac {epsilon} {2}=epsilon for ige N' $$



This gives us $x_irightarrow 0$ for $ige N'$



So $xin C_0$



Is this Correct?






functional-analysis







share|cite|improve this question





























    1














    Let $x^n in C_0$ is Cauchy.



    $rightarrow$ For $epsilon> 0$ there is N such that $n,m ge N $ $$ lVert x^m -x^nrVert< frac {epsilon} {2} $$
    we know that for every k $$lvert x^m_k -x^n_krvert le sup_{ige 1} lvert x^m_i -x^n_irvert < frac {epsilon} {2} $$
    So $x^n_k$ is Cauchy in $mathbb R$ which is Banach so $$x^n_k rightarrow x_k in mathbb R or lvert x^n_k -x_krvert le frac {epsilon} {2} $$
    by this we can say that there is an N, $nge N$ such that $lVert x^n -xrVert = sup_{ige 1} lvert x^n_i -x_i lvert < frac {epsilon} {2} $ $ $



    which means $x^n rightarrow x$



    Since $x^m in c_0$,
    there is some $N'$ such that $|x_i^m| < {1 over 2 } epsilon$ for $i ge N' (m=N)$



    Now to show that $xin C_0$



    $$lvert x_irvert le lvert x^m_i-x_irvert+ lvert x^m_irvert <frac {epsilon} {2}+frac {epsilon} {2}=epsilon for ige N' $$



    This gives us $x_irightarrow 0$ for $ige N'$



    So $xin C_0$



    Is this Correct?






    functional-analysis







    share|cite|improve this question



























      1












      1








      1


      1





      Let $x^n in C_0$ is Cauchy.



      $rightarrow$ For $epsilon> 0$ there is N such that $n,m ge N $ $$ lVert x^m -x^nrVert< frac {epsilon} {2} $$
      we know that for every k $$lvert x^m_k -x^n_krvert le sup_{ige 1} lvert x^m_i -x^n_irvert < frac {epsilon} {2} $$
      So $x^n_k$ is Cauchy in $mathbb R$ which is Banach so $$x^n_k rightarrow x_k in mathbb R or lvert x^n_k -x_krvert le frac {epsilon} {2} $$
      by this we can say that there is an N, $nge N$ such that $lVert x^n -xrVert = sup_{ige 1} lvert x^n_i -x_i lvert < frac {epsilon} {2} $ $ $



      which means $x^n rightarrow x$



      Since $x^m in c_0$,
      there is some $N'$ such that $|x_i^m| < {1 over 2 } epsilon$ for $i ge N' (m=N)$



      Now to show that $xin C_0$



      $$lvert x_irvert le lvert x^m_i-x_irvert+ lvert x^m_irvert <frac {epsilon} {2}+frac {epsilon} {2}=epsilon for ige N' $$



      This gives us $x_irightarrow 0$ for $ige N'$



      So $xin C_0$



      Is this Correct?






      functional-analysis







      share|cite|improve this question















      Let $x^n in C_0$ is Cauchy.



      $rightarrow$ For $epsilon> 0$ there is N such that $n,m ge N $ $$ lVert x^m -x^nrVert< frac {epsilon} {2} $$
      we know that for every k $$lvert x^m_k -x^n_krvert le sup_{ige 1} lvert x^m_i -x^n_irvert < frac {epsilon} {2} $$
      So $x^n_k$ is Cauchy in $mathbb R$ which is Banach so $$x^n_k rightarrow x_k in mathbb R or lvert x^n_k -x_krvert le frac {epsilon} {2} $$
      by this we can say that there is an N, $nge N$ such that $lVert x^n -xrVert = sup_{ige 1} lvert x^n_i -x_i lvert < frac {epsilon} {2} $ $ $



      which means $x^n rightarrow x$



      Since $x^m in c_0$,
      there is some $N'$ such that $|x_i^m| < {1 over 2 } epsilon$ for $i ge N' (m=N)$



      Now to show that $xin C_0$



      $$lvert x_irvert le lvert x^m_i-x_irvert+ lvert x^m_irvert <frac {epsilon} {2}+frac {epsilon} {2}=epsilon for ige N' $$



      This gives us $x_irightarrow 0$ for $ige N'$



      So $xin C_0$



      Is this Correct?






      functional-analysis




      functional-analysis






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 2 days ago







      Hitman

















      asked Jan 5 at 21:53









      HitmanHitman

      1749




      1749






















          1 Answer
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          The question has been modified based on this answer. The answer below is for the first version of the question.
          First correction: $|a_n|<epsilon /2$ for all $n$ and $a_n to a$ does not imply $|a|<epsilon /2$. It implies $|a|leq epsilon /2$.
          Second correction: in the last part $|x|$ has no meaning for a sequence $x$. Use coordinates. You should write $|x_i| leq |x_i^{n}|+|x_i^{n}-x_i|$. Then you should make the argument more rigorous as follows. First choose $n$ such that $|x_i^{n}-x_i|<epsilon /2$ for all $i$. Fixing this $n$ choose $k$ such that $|x_i^{n}|<epsilon /2$ for all $i geq k$. Now you get $|x_i| <epsilon$ for all $i geq k$.






          share|cite|improve this answer























          • Could you please check again? I have edited the question.
            – Hitman
            2 days ago










          • Your last part is still not rigorous. There are two variables $i$ and $n$ and you have to state clearly that you first choose $n$ and then the final inequality becomes true for all $i$ sufficiently large.
            – Kavi Rama Murthy
            2 days ago










          • I think now it is okay.
            – Hitman
            2 days ago










          • what do you say?
            – Hitman
            2 days ago






          • 1




            In the last part you did not specify $m$. If you take $m=N$ your proof will be correct.
            – Kavi Rama Murthy
            2 days ago











          Your Answer





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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          1














          The question has been modified based on this answer. The answer below is for the first version of the question.
          First correction: $|a_n|<epsilon /2$ for all $n$ and $a_n to a$ does not imply $|a|<epsilon /2$. It implies $|a|leq epsilon /2$.
          Second correction: in the last part $|x|$ has no meaning for a sequence $x$. Use coordinates. You should write $|x_i| leq |x_i^{n}|+|x_i^{n}-x_i|$. Then you should make the argument more rigorous as follows. First choose $n$ such that $|x_i^{n}-x_i|<epsilon /2$ for all $i$. Fixing this $n$ choose $k$ such that $|x_i^{n}|<epsilon /2$ for all $i geq k$. Now you get $|x_i| <epsilon$ for all $i geq k$.






          share|cite|improve this answer























          • Could you please check again? I have edited the question.
            – Hitman
            2 days ago










          • Your last part is still not rigorous. There are two variables $i$ and $n$ and you have to state clearly that you first choose $n$ and then the final inequality becomes true for all $i$ sufficiently large.
            – Kavi Rama Murthy
            2 days ago










          • I think now it is okay.
            – Hitman
            2 days ago










          • what do you say?
            – Hitman
            2 days ago






          • 1




            In the last part you did not specify $m$. If you take $m=N$ your proof will be correct.
            – Kavi Rama Murthy
            2 days ago
















          1














          The question has been modified based on this answer. The answer below is for the first version of the question.
          First correction: $|a_n|<epsilon /2$ for all $n$ and $a_n to a$ does not imply $|a|<epsilon /2$. It implies $|a|leq epsilon /2$.
          Second correction: in the last part $|x|$ has no meaning for a sequence $x$. Use coordinates. You should write $|x_i| leq |x_i^{n}|+|x_i^{n}-x_i|$. Then you should make the argument more rigorous as follows. First choose $n$ such that $|x_i^{n}-x_i|<epsilon /2$ for all $i$. Fixing this $n$ choose $k$ such that $|x_i^{n}|<epsilon /2$ for all $i geq k$. Now you get $|x_i| <epsilon$ for all $i geq k$.






          share|cite|improve this answer























          • Could you please check again? I have edited the question.
            – Hitman
            2 days ago










          • Your last part is still not rigorous. There are two variables $i$ and $n$ and you have to state clearly that you first choose $n$ and then the final inequality becomes true for all $i$ sufficiently large.
            – Kavi Rama Murthy
            2 days ago










          • I think now it is okay.
            – Hitman
            2 days ago










          • what do you say?
            – Hitman
            2 days ago






          • 1




            In the last part you did not specify $m$. If you take $m=N$ your proof will be correct.
            – Kavi Rama Murthy
            2 days ago














          1












          1








          1






          The question has been modified based on this answer. The answer below is for the first version of the question.
          First correction: $|a_n|<epsilon /2$ for all $n$ and $a_n to a$ does not imply $|a|<epsilon /2$. It implies $|a|leq epsilon /2$.
          Second correction: in the last part $|x|$ has no meaning for a sequence $x$. Use coordinates. You should write $|x_i| leq |x_i^{n}|+|x_i^{n}-x_i|$. Then you should make the argument more rigorous as follows. First choose $n$ such that $|x_i^{n}-x_i|<epsilon /2$ for all $i$. Fixing this $n$ choose $k$ such that $|x_i^{n}|<epsilon /2$ for all $i geq k$. Now you get $|x_i| <epsilon$ for all $i geq k$.






          share|cite|improve this answer














          The question has been modified based on this answer. The answer below is for the first version of the question.
          First correction: $|a_n|<epsilon /2$ for all $n$ and $a_n to a$ does not imply $|a|<epsilon /2$. It implies $|a|leq epsilon /2$.
          Second correction: in the last part $|x|$ has no meaning for a sequence $x$. Use coordinates. You should write $|x_i| leq |x_i^{n}|+|x_i^{n}-x_i|$. Then you should make the argument more rigorous as follows. First choose $n$ such that $|x_i^{n}-x_i|<epsilon /2$ for all $i$. Fixing this $n$ choose $k$ such that $|x_i^{n}|<epsilon /2$ for all $i geq k$. Now you get $|x_i| <epsilon$ for all $i geq k$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 2 days ago

























          answered 2 days ago









          Kavi Rama MurthyKavi Rama Murthy

          51.8k32055




          51.8k32055












          • Could you please check again? I have edited the question.
            – Hitman
            2 days ago










          • Your last part is still not rigorous. There are two variables $i$ and $n$ and you have to state clearly that you first choose $n$ and then the final inequality becomes true for all $i$ sufficiently large.
            – Kavi Rama Murthy
            2 days ago










          • I think now it is okay.
            – Hitman
            2 days ago










          • what do you say?
            – Hitman
            2 days ago






          • 1




            In the last part you did not specify $m$. If you take $m=N$ your proof will be correct.
            – Kavi Rama Murthy
            2 days ago


















          • Could you please check again? I have edited the question.
            – Hitman
            2 days ago










          • Your last part is still not rigorous. There are two variables $i$ and $n$ and you have to state clearly that you first choose $n$ and then the final inequality becomes true for all $i$ sufficiently large.
            – Kavi Rama Murthy
            2 days ago










          • I think now it is okay.
            – Hitman
            2 days ago










          • what do you say?
            – Hitman
            2 days ago






          • 1




            In the last part you did not specify $m$. If you take $m=N$ your proof will be correct.
            – Kavi Rama Murthy
            2 days ago
















          Could you please check again? I have edited the question.
          – Hitman
          2 days ago




          Could you please check again? I have edited the question.
          – Hitman
          2 days ago












          Your last part is still not rigorous. There are two variables $i$ and $n$ and you have to state clearly that you first choose $n$ and then the final inequality becomes true for all $i$ sufficiently large.
          – Kavi Rama Murthy
          2 days ago




          Your last part is still not rigorous. There are two variables $i$ and $n$ and you have to state clearly that you first choose $n$ and then the final inequality becomes true for all $i$ sufficiently large.
          – Kavi Rama Murthy
          2 days ago












          I think now it is okay.
          – Hitman
          2 days ago




          I think now it is okay.
          – Hitman
          2 days ago












          what do you say?
          – Hitman
          2 days ago




          what do you say?
          – Hitman
          2 days ago




          1




          1




          In the last part you did not specify $m$. If you take $m=N$ your proof will be correct.
          – Kavi Rama Murthy
          2 days ago




          In the last part you did not specify $m$. If you take $m=N$ your proof will be correct.
          – Kavi Rama Murthy
          2 days ago


















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