Any finite dimensional normed linear space over a complete field is complete.












2












$begingroup$



Prove that any finite dimensional normed linear space over a complete field is complete.




After several comments and corrections, I present the correct proof in the answer section.






proof-verification normed-spaces complete-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    Looks fine to me. I would probably emphasize why we can choose $N$ uniformly in $i$ (after the line when you say ${alpha_{i,r}}^{n}_{i=1,rin Bbb{N}}$ is Cauchy) because this is the key place where the finite dimensionality is required.
    $endgroup$
    – Pratyush Sarkar
    Dec 28 '18 at 6:41












  • $begingroup$
    @Pratyush Sarkar: Thanks a lot for your comment! Can you please, elaborate on the use of finite dimension?
    $endgroup$
    – Omojola Micheal
    Dec 28 '18 at 6:43












  • $begingroup$
    You are using the fact that $alpha_{i,r}to alpha_{i}in Bbb{R};text{or};Bbb{C},;text{as};rto infty$. But there are multiple sequences here -- one for each $i$. For any fixed $i$ and $epsilon'$, you have a corresponding $N$ (depending on $epsilon'$ and $i$). But a priori these $N$ may be different for each $i$ (we can write $N_i$ to make the dependence explicit). You want to choose $N = max{N_i: i = 1, 2, dotsc, n}$ so that the same $N$ works for all $i$. Does that make sense? This also illuminates why things can go wrong in infinite dimensions.
    $endgroup$
    – Pratyush Sarkar
    Dec 28 '18 at 6:57












  • $begingroup$
    @Pratyush Sarkar: Oh, now I get you! That makes sense! Is it fine, now? I made some edits! I believe it should be.
    $endgroup$
    – Omojola Micheal
    Dec 28 '18 at 7:41








  • 1




    $begingroup$
    I removed some unnecessary parts which you forgot to delete. Looks fine now.
    $endgroup$
    – Pratyush Sarkar
    Dec 29 '18 at 1:01
















2












$begingroup$



Prove that any finite dimensional normed linear space over a complete field is complete.




After several comments and corrections, I present the correct proof in the answer section.






proof-verification normed-spaces complete-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    Looks fine to me. I would probably emphasize why we can choose $N$ uniformly in $i$ (after the line when you say ${alpha_{i,r}}^{n}_{i=1,rin Bbb{N}}$ is Cauchy) because this is the key place where the finite dimensionality is required.
    $endgroup$
    – Pratyush Sarkar
    Dec 28 '18 at 6:41












  • $begingroup$
    @Pratyush Sarkar: Thanks a lot for your comment! Can you please, elaborate on the use of finite dimension?
    $endgroup$
    – Omojola Micheal
    Dec 28 '18 at 6:43












  • $begingroup$
    You are using the fact that $alpha_{i,r}to alpha_{i}in Bbb{R};text{or};Bbb{C},;text{as};rto infty$. But there are multiple sequences here -- one for each $i$. For any fixed $i$ and $epsilon'$, you have a corresponding $N$ (depending on $epsilon'$ and $i$). But a priori these $N$ may be different for each $i$ (we can write $N_i$ to make the dependence explicit). You want to choose $N = max{N_i: i = 1, 2, dotsc, n}$ so that the same $N$ works for all $i$. Does that make sense? This also illuminates why things can go wrong in infinite dimensions.
    $endgroup$
    – Pratyush Sarkar
    Dec 28 '18 at 6:57












  • $begingroup$
    @Pratyush Sarkar: Oh, now I get you! That makes sense! Is it fine, now? I made some edits! I believe it should be.
    $endgroup$
    – Omojola Micheal
    Dec 28 '18 at 7:41








  • 1




    $begingroup$
    I removed some unnecessary parts which you forgot to delete. Looks fine now.
    $endgroup$
    – Pratyush Sarkar
    Dec 29 '18 at 1:01














2












2








2


1



$begingroup$



Prove that any finite dimensional normed linear space over a complete field is complete.




After several comments and corrections, I present the correct proof in the answer section.






proof-verification normed-spaces complete-spaces







share|cite|improve this question











$endgroup$





Prove that any finite dimensional normed linear space over a complete field is complete.




After several comments and corrections, I present the correct proof in the answer section.






proof-verification normed-spaces complete-spaces




proof-verification normed-spaces complete-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 11 at 6:17







Omojola Micheal

















asked Dec 28 '18 at 6:20









Omojola MichealOmojola Micheal

1,986324




1,986324












  • $begingroup$
    Looks fine to me. I would probably emphasize why we can choose $N$ uniformly in $i$ (after the line when you say ${alpha_{i,r}}^{n}_{i=1,rin Bbb{N}}$ is Cauchy) because this is the key place where the finite dimensionality is required.
    $endgroup$
    – Pratyush Sarkar
    Dec 28 '18 at 6:41












  • $begingroup$
    @Pratyush Sarkar: Thanks a lot for your comment! Can you please, elaborate on the use of finite dimension?
    $endgroup$
    – Omojola Micheal
    Dec 28 '18 at 6:43












  • $begingroup$
    You are using the fact that $alpha_{i,r}to alpha_{i}in Bbb{R};text{or};Bbb{C},;text{as};rto infty$. But there are multiple sequences here -- one for each $i$. For any fixed $i$ and $epsilon'$, you have a corresponding $N$ (depending on $epsilon'$ and $i$). But a priori these $N$ may be different for each $i$ (we can write $N_i$ to make the dependence explicit). You want to choose $N = max{N_i: i = 1, 2, dotsc, n}$ so that the same $N$ works for all $i$. Does that make sense? This also illuminates why things can go wrong in infinite dimensions.
    $endgroup$
    – Pratyush Sarkar
    Dec 28 '18 at 6:57












  • $begingroup$
    @Pratyush Sarkar: Oh, now I get you! That makes sense! Is it fine, now? I made some edits! I believe it should be.
    $endgroup$
    – Omojola Micheal
    Dec 28 '18 at 7:41








  • 1




    $begingroup$
    I removed some unnecessary parts which you forgot to delete. Looks fine now.
    $endgroup$
    – Pratyush Sarkar
    Dec 29 '18 at 1:01


















  • $begingroup$
    Looks fine to me. I would probably emphasize why we can choose $N$ uniformly in $i$ (after the line when you say ${alpha_{i,r}}^{n}_{i=1,rin Bbb{N}}$ is Cauchy) because this is the key place where the finite dimensionality is required.
    $endgroup$
    – Pratyush Sarkar
    Dec 28 '18 at 6:41












  • $begingroup$
    @Pratyush Sarkar: Thanks a lot for your comment! Can you please, elaborate on the use of finite dimension?
    $endgroup$
    – Omojola Micheal
    Dec 28 '18 at 6:43












  • $begingroup$
    You are using the fact that $alpha_{i,r}to alpha_{i}in Bbb{R};text{or};Bbb{C},;text{as};rto infty$. But there are multiple sequences here -- one for each $i$. For any fixed $i$ and $epsilon'$, you have a corresponding $N$ (depending on $epsilon'$ and $i$). But a priori these $N$ may be different for each $i$ (we can write $N_i$ to make the dependence explicit). You want to choose $N = max{N_i: i = 1, 2, dotsc, n}$ so that the same $N$ works for all $i$. Does that make sense? This also illuminates why things can go wrong in infinite dimensions.
    $endgroup$
    – Pratyush Sarkar
    Dec 28 '18 at 6:57












  • $begingroup$
    @Pratyush Sarkar: Oh, now I get you! That makes sense! Is it fine, now? I made some edits! I believe it should be.
    $endgroup$
    – Omojola Micheal
    Dec 28 '18 at 7:41








  • 1




    $begingroup$
    I removed some unnecessary parts which you forgot to delete. Looks fine now.
    $endgroup$
    – Pratyush Sarkar
    Dec 29 '18 at 1:01
















$begingroup$
Looks fine to me. I would probably emphasize why we can choose $N$ uniformly in $i$ (after the line when you say ${alpha_{i,r}}^{n}_{i=1,rin Bbb{N}}$ is Cauchy) because this is the key place where the finite dimensionality is required.
$endgroup$
– Pratyush Sarkar
Dec 28 '18 at 6:41






$begingroup$
Looks fine to me. I would probably emphasize why we can choose $N$ uniformly in $i$ (after the line when you say ${alpha_{i,r}}^{n}_{i=1,rin Bbb{N}}$ is Cauchy) because this is the key place where the finite dimensionality is required.
$endgroup$
– Pratyush Sarkar
Dec 28 '18 at 6:41














$begingroup$
@Pratyush Sarkar: Thanks a lot for your comment! Can you please, elaborate on the use of finite dimension?
$endgroup$
– Omojola Micheal
Dec 28 '18 at 6:43






$begingroup$
@Pratyush Sarkar: Thanks a lot for your comment! Can you please, elaborate on the use of finite dimension?
$endgroup$
– Omojola Micheal
Dec 28 '18 at 6:43














$begingroup$
You are using the fact that $alpha_{i,r}to alpha_{i}in Bbb{R};text{or};Bbb{C},;text{as};rto infty$. But there are multiple sequences here -- one for each $i$. For any fixed $i$ and $epsilon'$, you have a corresponding $N$ (depending on $epsilon'$ and $i$). But a priori these $N$ may be different for each $i$ (we can write $N_i$ to make the dependence explicit). You want to choose $N = max{N_i: i = 1, 2, dotsc, n}$ so that the same $N$ works for all $i$. Does that make sense? This also illuminates why things can go wrong in infinite dimensions.
$endgroup$
– Pratyush Sarkar
Dec 28 '18 at 6:57






$begingroup$
You are using the fact that $alpha_{i,r}to alpha_{i}in Bbb{R};text{or};Bbb{C},;text{as};rto infty$. But there are multiple sequences here -- one for each $i$. For any fixed $i$ and $epsilon'$, you have a corresponding $N$ (depending on $epsilon'$ and $i$). But a priori these $N$ may be different for each $i$ (we can write $N_i$ to make the dependence explicit). You want to choose $N = max{N_i: i = 1, 2, dotsc, n}$ so that the same $N$ works for all $i$. Does that make sense? This also illuminates why things can go wrong in infinite dimensions.
$endgroup$
– Pratyush Sarkar
Dec 28 '18 at 6:57














$begingroup$
@Pratyush Sarkar: Oh, now I get you! That makes sense! Is it fine, now? I made some edits! I believe it should be.
$endgroup$
– Omojola Micheal
Dec 28 '18 at 7:41






$begingroup$
@Pratyush Sarkar: Oh, now I get you! That makes sense! Is it fine, now? I made some edits! I believe it should be.
$endgroup$
– Omojola Micheal
Dec 28 '18 at 7:41






1




1




$begingroup$
I removed some unnecessary parts which you forgot to delete. Looks fine now.
$endgroup$
– Pratyush Sarkar
Dec 29 '18 at 1:01




$begingroup$
I removed some unnecessary parts which you forgot to delete. Looks fine now.
$endgroup$
– Pratyush Sarkar
Dec 29 '18 at 1:01










1 Answer
1






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$begingroup$

PROOF



Let $E$ be any finite dimensional normed linear space over a complete field, $Bbb{R}$ or $Bbb{C},$ say. Suppose $dim E=ngeq 1,$ and let ${e_i}^{n}_{i=1}$ be a basis for $E.$ Then, there exists scalars ${alpha_i}^{n}_{i=1}$ such that, for arbitrary $xin E$,
begin{align} x= sum^{n}_{i=1} alpha_i e_i .end{align}
Suppose ${x_r}_{rin Bbb{N}}$ is Cauchy in $E$ w.r.t $|cdot|$ norm and $epsilon'>0.$ Then, there exists $N$ such that for all $sgeq rgeq N,$
begin{align} |x_r-x_s|<epsilon'.end{align}
Now, $|cdot|_1$ defined by $|x|_1=sum^{n}_{i=1} |alpha_i|$ is a norm on $E$ and since All norms defined on a finite dimensional normed linear space are equivalent, we have that $|cdot|_1sim |cdot|,$ i.e., there exists $gamma,beta>0,$ such that
begin{align} gamma|x|_1leq|x|leq beta|x|_1,;forall,xin E.end{align}
Then,
begin{align}gamma|alpha_{i,r}-alpha_{i,s}|leqgammasum^{n}_{i=1}|alpha_{i,r}-alpha_{i,s}|=gamma|x_r-x_s|_1leq|x_r-x_s|<epsilon',;sgeq rgeq Nend{align}
and so the sequence ${alpha_{i,r}}^{n}_{i=1,rin Bbb{N}}$ is Cauchy in $Bbb{R}$ or $Bbb{C}$. Now, $alpha_{i,r}to alpha_{i}in Bbb{R};text{or};Bbb{C},;text{as};rto infty$ by completeness. This implies that for each $iin{1,2,cdots,n},$ there exists $N_i:=N(i,epsilon')$ such that
begin{align}|alpha_{i,r}-alpha_{i}|<epsilon',;rgeq N_i.end{align}
Taking $M=max{N_i:1leq ileq n},$ we have that
begin{align}|alpha_{i,r}-alpha_{i}|<epsilon',;rgeq M.end{align}



Let $epsilon>0$ and $ninBbb{N}$, then for $epsilon'=dfrac{epsilon}{nbeta},$ there exists $M$ such that begin{align}|alpha_{i}-alpha_{i,s}|leqdfrac{epsilon}{nbeta},;sgeq M.end{align}
Taking sums, we have
begin{align}|x-x_s|leqbeta|x-x_s|_1=betasum^{n}_{i=1}|alpha_{i}-alpha_{i,s}|leqsum^{n}_{i=1}dfrac{epsilon}{n},;sgeq M.end{align}
Hence, begin{align}|x_s-x|leqepsilon,;sgeq M,end{align}
and so, we have that $xin E,$ which implies that $E$ is a complete finite dimensional normed linear space and we are done!






share|cite|improve this answer









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    0












    $begingroup$

    PROOF



    Let $E$ be any finite dimensional normed linear space over a complete field, $Bbb{R}$ or $Bbb{C},$ say. Suppose $dim E=ngeq 1,$ and let ${e_i}^{n}_{i=1}$ be a basis for $E.$ Then, there exists scalars ${alpha_i}^{n}_{i=1}$ such that, for arbitrary $xin E$,
    begin{align} x= sum^{n}_{i=1} alpha_i e_i .end{align}
    Suppose ${x_r}_{rin Bbb{N}}$ is Cauchy in $E$ w.r.t $|cdot|$ norm and $epsilon'>0.$ Then, there exists $N$ such that for all $sgeq rgeq N,$
    begin{align} |x_r-x_s|<epsilon'.end{align}
    Now, $|cdot|_1$ defined by $|x|_1=sum^{n}_{i=1} |alpha_i|$ is a norm on $E$ and since All norms defined on a finite dimensional normed linear space are equivalent, we have that $|cdot|_1sim |cdot|,$ i.e., there exists $gamma,beta>0,$ such that
    begin{align} gamma|x|_1leq|x|leq beta|x|_1,;forall,xin E.end{align}
    Then,
    begin{align}gamma|alpha_{i,r}-alpha_{i,s}|leqgammasum^{n}_{i=1}|alpha_{i,r}-alpha_{i,s}|=gamma|x_r-x_s|_1leq|x_r-x_s|<epsilon',;sgeq rgeq Nend{align}
    and so the sequence ${alpha_{i,r}}^{n}_{i=1,rin Bbb{N}}$ is Cauchy in $Bbb{R}$ or $Bbb{C}$. Now, $alpha_{i,r}to alpha_{i}in Bbb{R};text{or};Bbb{C},;text{as};rto infty$ by completeness. This implies that for each $iin{1,2,cdots,n},$ there exists $N_i:=N(i,epsilon')$ such that
    begin{align}|alpha_{i,r}-alpha_{i}|<epsilon',;rgeq N_i.end{align}
    Taking $M=max{N_i:1leq ileq n},$ we have that
    begin{align}|alpha_{i,r}-alpha_{i}|<epsilon',;rgeq M.end{align}



    Let $epsilon>0$ and $ninBbb{N}$, then for $epsilon'=dfrac{epsilon}{nbeta},$ there exists $M$ such that begin{align}|alpha_{i}-alpha_{i,s}|leqdfrac{epsilon}{nbeta},;sgeq M.end{align}
    Taking sums, we have
    begin{align}|x-x_s|leqbeta|x-x_s|_1=betasum^{n}_{i=1}|alpha_{i}-alpha_{i,s}|leqsum^{n}_{i=1}dfrac{epsilon}{n},;sgeq M.end{align}
    Hence, begin{align}|x_s-x|leqepsilon,;sgeq M,end{align}
    and so, we have that $xin E,$ which implies that $E$ is a complete finite dimensional normed linear space and we are done!






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      PROOF



      Let $E$ be any finite dimensional normed linear space over a complete field, $Bbb{R}$ or $Bbb{C},$ say. Suppose $dim E=ngeq 1,$ and let ${e_i}^{n}_{i=1}$ be a basis for $E.$ Then, there exists scalars ${alpha_i}^{n}_{i=1}$ such that, for arbitrary $xin E$,
      begin{align} x= sum^{n}_{i=1} alpha_i e_i .end{align}
      Suppose ${x_r}_{rin Bbb{N}}$ is Cauchy in $E$ w.r.t $|cdot|$ norm and $epsilon'>0.$ Then, there exists $N$ such that for all $sgeq rgeq N,$
      begin{align} |x_r-x_s|<epsilon'.end{align}
      Now, $|cdot|_1$ defined by $|x|_1=sum^{n}_{i=1} |alpha_i|$ is a norm on $E$ and since All norms defined on a finite dimensional normed linear space are equivalent, we have that $|cdot|_1sim |cdot|,$ i.e., there exists $gamma,beta>0,$ such that
      begin{align} gamma|x|_1leq|x|leq beta|x|_1,;forall,xin E.end{align}
      Then,
      begin{align}gamma|alpha_{i,r}-alpha_{i,s}|leqgammasum^{n}_{i=1}|alpha_{i,r}-alpha_{i,s}|=gamma|x_r-x_s|_1leq|x_r-x_s|<epsilon',;sgeq rgeq Nend{align}
      and so the sequence ${alpha_{i,r}}^{n}_{i=1,rin Bbb{N}}$ is Cauchy in $Bbb{R}$ or $Bbb{C}$. Now, $alpha_{i,r}to alpha_{i}in Bbb{R};text{or};Bbb{C},;text{as};rto infty$ by completeness. This implies that for each $iin{1,2,cdots,n},$ there exists $N_i:=N(i,epsilon')$ such that
      begin{align}|alpha_{i,r}-alpha_{i}|<epsilon',;rgeq N_i.end{align}
      Taking $M=max{N_i:1leq ileq n},$ we have that
      begin{align}|alpha_{i,r}-alpha_{i}|<epsilon',;rgeq M.end{align}



      Let $epsilon>0$ and $ninBbb{N}$, then for $epsilon'=dfrac{epsilon}{nbeta},$ there exists $M$ such that begin{align}|alpha_{i}-alpha_{i,s}|leqdfrac{epsilon}{nbeta},;sgeq M.end{align}
      Taking sums, we have
      begin{align}|x-x_s|leqbeta|x-x_s|_1=betasum^{n}_{i=1}|alpha_{i}-alpha_{i,s}|leqsum^{n}_{i=1}dfrac{epsilon}{n},;sgeq M.end{align}
      Hence, begin{align}|x_s-x|leqepsilon,;sgeq M,end{align}
      and so, we have that $xin E,$ which implies that $E$ is a complete finite dimensional normed linear space and we are done!






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        PROOF



        Let $E$ be any finite dimensional normed linear space over a complete field, $Bbb{R}$ or $Bbb{C},$ say. Suppose $dim E=ngeq 1,$ and let ${e_i}^{n}_{i=1}$ be a basis for $E.$ Then, there exists scalars ${alpha_i}^{n}_{i=1}$ such that, for arbitrary $xin E$,
        begin{align} x= sum^{n}_{i=1} alpha_i e_i .end{align}
        Suppose ${x_r}_{rin Bbb{N}}$ is Cauchy in $E$ w.r.t $|cdot|$ norm and $epsilon'>0.$ Then, there exists $N$ such that for all $sgeq rgeq N,$
        begin{align} |x_r-x_s|<epsilon'.end{align}
        Now, $|cdot|_1$ defined by $|x|_1=sum^{n}_{i=1} |alpha_i|$ is a norm on $E$ and since All norms defined on a finite dimensional normed linear space are equivalent, we have that $|cdot|_1sim |cdot|,$ i.e., there exists $gamma,beta>0,$ such that
        begin{align} gamma|x|_1leq|x|leq beta|x|_1,;forall,xin E.end{align}
        Then,
        begin{align}gamma|alpha_{i,r}-alpha_{i,s}|leqgammasum^{n}_{i=1}|alpha_{i,r}-alpha_{i,s}|=gamma|x_r-x_s|_1leq|x_r-x_s|<epsilon',;sgeq rgeq Nend{align}
        and so the sequence ${alpha_{i,r}}^{n}_{i=1,rin Bbb{N}}$ is Cauchy in $Bbb{R}$ or $Bbb{C}$. Now, $alpha_{i,r}to alpha_{i}in Bbb{R};text{or};Bbb{C},;text{as};rto infty$ by completeness. This implies that for each $iin{1,2,cdots,n},$ there exists $N_i:=N(i,epsilon')$ such that
        begin{align}|alpha_{i,r}-alpha_{i}|<epsilon',;rgeq N_i.end{align}
        Taking $M=max{N_i:1leq ileq n},$ we have that
        begin{align}|alpha_{i,r}-alpha_{i}|<epsilon',;rgeq M.end{align}



        Let $epsilon>0$ and $ninBbb{N}$, then for $epsilon'=dfrac{epsilon}{nbeta},$ there exists $M$ such that begin{align}|alpha_{i}-alpha_{i,s}|leqdfrac{epsilon}{nbeta},;sgeq M.end{align}
        Taking sums, we have
        begin{align}|x-x_s|leqbeta|x-x_s|_1=betasum^{n}_{i=1}|alpha_{i}-alpha_{i,s}|leqsum^{n}_{i=1}dfrac{epsilon}{n},;sgeq M.end{align}
        Hence, begin{align}|x_s-x|leqepsilon,;sgeq M,end{align}
        and so, we have that $xin E,$ which implies that $E$ is a complete finite dimensional normed linear space and we are done!






        share|cite|improve this answer









        $endgroup$



        PROOF



        Let $E$ be any finite dimensional normed linear space over a complete field, $Bbb{R}$ or $Bbb{C},$ say. Suppose $dim E=ngeq 1,$ and let ${e_i}^{n}_{i=1}$ be a basis for $E.$ Then, there exists scalars ${alpha_i}^{n}_{i=1}$ such that, for arbitrary $xin E$,
        begin{align} x= sum^{n}_{i=1} alpha_i e_i .end{align}
        Suppose ${x_r}_{rin Bbb{N}}$ is Cauchy in $E$ w.r.t $|cdot|$ norm and $epsilon'>0.$ Then, there exists $N$ such that for all $sgeq rgeq N,$
        begin{align} |x_r-x_s|<epsilon'.end{align}
        Now, $|cdot|_1$ defined by $|x|_1=sum^{n}_{i=1} |alpha_i|$ is a norm on $E$ and since All norms defined on a finite dimensional normed linear space are equivalent, we have that $|cdot|_1sim |cdot|,$ i.e., there exists $gamma,beta>0,$ such that
        begin{align} gamma|x|_1leq|x|leq beta|x|_1,;forall,xin E.end{align}
        Then,
        begin{align}gamma|alpha_{i,r}-alpha_{i,s}|leqgammasum^{n}_{i=1}|alpha_{i,r}-alpha_{i,s}|=gamma|x_r-x_s|_1leq|x_r-x_s|<epsilon',;sgeq rgeq Nend{align}
        and so the sequence ${alpha_{i,r}}^{n}_{i=1,rin Bbb{N}}$ is Cauchy in $Bbb{R}$ or $Bbb{C}$. Now, $alpha_{i,r}to alpha_{i}in Bbb{R};text{or};Bbb{C},;text{as};rto infty$ by completeness. This implies that for each $iin{1,2,cdots,n},$ there exists $N_i:=N(i,epsilon')$ such that
        begin{align}|alpha_{i,r}-alpha_{i}|<epsilon',;rgeq N_i.end{align}
        Taking $M=max{N_i:1leq ileq n},$ we have that
        begin{align}|alpha_{i,r}-alpha_{i}|<epsilon',;rgeq M.end{align}



        Let $epsilon>0$ and $ninBbb{N}$, then for $epsilon'=dfrac{epsilon}{nbeta},$ there exists $M$ such that begin{align}|alpha_{i}-alpha_{i,s}|leqdfrac{epsilon}{nbeta},;sgeq M.end{align}
        Taking sums, we have
        begin{align}|x-x_s|leqbeta|x-x_s|_1=betasum^{n}_{i=1}|alpha_{i}-alpha_{i,s}|leqsum^{n}_{i=1}dfrac{epsilon}{n},;sgeq M.end{align}
        Hence, begin{align}|x_s-x|leqepsilon,;sgeq M,end{align}
        and so, we have that $xin E,$ which implies that $E$ is a complete finite dimensional normed linear space and we are done!







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        answered Jan 11 at 6:16









        Omojola MichealOmojola Micheal

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