Show that $Bbb{R^n}$ with the $ell^2$-norm is complete












1












$begingroup$


I want to show that $Bbb{R^n}$ where $Vert x Vert_{R^n}=left(sum_{i=1}^{n}left| x_i right| ^2right)^{1/2}$ is complete



Here is what I've done.



Let ${x^{(s)}}subseteq (R^n,Vert cdot Vert_{R^n})$ be a Cauchy sequence and $epsilon>0.$ Then, $exists, Nin Bbb{N}$ s.t. $forall rgeq sgeq N,$
begin{align}Vert x^{(r)}-x^{(s)} Vert_{R^n}=left(sum_{i=1}^{n}left| x_{i}^{(r)}-x_{i}^{(s)} right|^2 right)<epsilon^{2}.end{align}
Hence, we have that $x_{i}^{(r)}to x_{i}^{*}in Bbb{R},;text{as};rtoinfty$, since $Bbb{R}$ is complete.



Fix $n,rin Bbb{N}$, then allow $ttoinfty.$ We have
begin{align}left(sum_{i=1}^{n}left| x_{i}^{(r)}-x_{i}^{*} right|^2 right)<epsilon^{2},;;forall ;rgeq N, nin Bbb{N}.end{align}
For $r=N,$
begin{align}left(sum_{i=1}^{n}left| x_{i}^{(N)}-x_{i}^{*} right| ^2right)<epsilon^{2},;;forall ; nin Bbb{N}.end{align}
Hence, $x^{N}-x^{*}in (R^n,Vert cdot Vert_{R^n})$ and since $(R^n,Vert cdot Vert_{R^n})$ is a linear vector space, then $x^{*}=x^{N}-(x^{N}-x^{*})in (R^n,Vert cdot Vert_{R^n}),$ and we are done!



Kindly check if I'm correct. Corrections and alternative proofs are welcome.






analysis normed-spaces







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  • $begingroup$
    The letter $n$ has been used for the dimension. Using it for other purposes in the proof is confusing.
    $endgroup$
    – user587192
    Dec 13 '18 at 15:25










  • $begingroup$
    @user587192: Ok, I will work on that!
    $endgroup$
    – Omojola Micheal
    Dec 13 '18 at 15:27










  • $begingroup$
    @user587192: I've done something on that! I hope it's okay now!
    $endgroup$
    – Omojola Micheal
    Dec 13 '18 at 15:31
















1












$begingroup$


I want to show that $Bbb{R^n}$ where $Vert x Vert_{R^n}=left(sum_{i=1}^{n}left| x_i right| ^2right)^{1/2}$ is complete



Here is what I've done.



Let ${x^{(s)}}subseteq (R^n,Vert cdot Vert_{R^n})$ be a Cauchy sequence and $epsilon>0.$ Then, $exists, Nin Bbb{N}$ s.t. $forall rgeq sgeq N,$
begin{align}Vert x^{(r)}-x^{(s)} Vert_{R^n}=left(sum_{i=1}^{n}left| x_{i}^{(r)}-x_{i}^{(s)} right|^2 right)<epsilon^{2}.end{align}
Hence, we have that $x_{i}^{(r)}to x_{i}^{*}in Bbb{R},;text{as};rtoinfty$, since $Bbb{R}$ is complete.



Fix $n,rin Bbb{N}$, then allow $ttoinfty.$ We have
begin{align}left(sum_{i=1}^{n}left| x_{i}^{(r)}-x_{i}^{*} right|^2 right)<epsilon^{2},;;forall ;rgeq N, nin Bbb{N}.end{align}
For $r=N,$
begin{align}left(sum_{i=1}^{n}left| x_{i}^{(N)}-x_{i}^{*} right| ^2right)<epsilon^{2},;;forall ; nin Bbb{N}.end{align}
Hence, $x^{N}-x^{*}in (R^n,Vert cdot Vert_{R^n})$ and since $(R^n,Vert cdot Vert_{R^n})$ is a linear vector space, then $x^{*}=x^{N}-(x^{N}-x^{*})in (R^n,Vert cdot Vert_{R^n}),$ and we are done!



Kindly check if I'm correct. Corrections and alternative proofs are welcome.






analysis normed-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    The letter $n$ has been used for the dimension. Using it for other purposes in the proof is confusing.
    $endgroup$
    – user587192
    Dec 13 '18 at 15:25










  • $begingroup$
    @user587192: Ok, I will work on that!
    $endgroup$
    – Omojola Micheal
    Dec 13 '18 at 15:27










  • $begingroup$
    @user587192: I've done something on that! I hope it's okay now!
    $endgroup$
    – Omojola Micheal
    Dec 13 '18 at 15:31














1












1








1


1



$begingroup$


I want to show that $Bbb{R^n}$ where $Vert x Vert_{R^n}=left(sum_{i=1}^{n}left| x_i right| ^2right)^{1/2}$ is complete



Here is what I've done.



Let ${x^{(s)}}subseteq (R^n,Vert cdot Vert_{R^n})$ be a Cauchy sequence and $epsilon>0.$ Then, $exists, Nin Bbb{N}$ s.t. $forall rgeq sgeq N,$
begin{align}Vert x^{(r)}-x^{(s)} Vert_{R^n}=left(sum_{i=1}^{n}left| x_{i}^{(r)}-x_{i}^{(s)} right|^2 right)<epsilon^{2}.end{align}
Hence, we have that $x_{i}^{(r)}to x_{i}^{*}in Bbb{R},;text{as};rtoinfty$, since $Bbb{R}$ is complete.



Fix $n,rin Bbb{N}$, then allow $ttoinfty.$ We have
begin{align}left(sum_{i=1}^{n}left| x_{i}^{(r)}-x_{i}^{*} right|^2 right)<epsilon^{2},;;forall ;rgeq N, nin Bbb{N}.end{align}
For $r=N,$
begin{align}left(sum_{i=1}^{n}left| x_{i}^{(N)}-x_{i}^{*} right| ^2right)<epsilon^{2},;;forall ; nin Bbb{N}.end{align}
Hence, $x^{N}-x^{*}in (R^n,Vert cdot Vert_{R^n})$ and since $(R^n,Vert cdot Vert_{R^n})$ is a linear vector space, then $x^{*}=x^{N}-(x^{N}-x^{*})in (R^n,Vert cdot Vert_{R^n}),$ and we are done!



Kindly check if I'm correct. Corrections and alternative proofs are welcome.






analysis normed-spaces







share|cite|improve this question











$endgroup$




I want to show that $Bbb{R^n}$ where $Vert x Vert_{R^n}=left(sum_{i=1}^{n}left| x_i right| ^2right)^{1/2}$ is complete



Here is what I've done.



Let ${x^{(s)}}subseteq (R^n,Vert cdot Vert_{R^n})$ be a Cauchy sequence and $epsilon>0.$ Then, $exists, Nin Bbb{N}$ s.t. $forall rgeq sgeq N,$
begin{align}Vert x^{(r)}-x^{(s)} Vert_{R^n}=left(sum_{i=1}^{n}left| x_{i}^{(r)}-x_{i}^{(s)} right|^2 right)<epsilon^{2}.end{align}
Hence, we have that $x_{i}^{(r)}to x_{i}^{*}in Bbb{R},;text{as};rtoinfty$, since $Bbb{R}$ is complete.



Fix $n,rin Bbb{N}$, then allow $ttoinfty.$ We have
begin{align}left(sum_{i=1}^{n}left| x_{i}^{(r)}-x_{i}^{*} right|^2 right)<epsilon^{2},;;forall ;rgeq N, nin Bbb{N}.end{align}
For $r=N,$
begin{align}left(sum_{i=1}^{n}left| x_{i}^{(N)}-x_{i}^{*} right| ^2right)<epsilon^{2},;;forall ; nin Bbb{N}.end{align}
Hence, $x^{N}-x^{*}in (R^n,Vert cdot Vert_{R^n})$ and since $(R^n,Vert cdot Vert_{R^n})$ is a linear vector space, then $x^{*}=x^{N}-(x^{N}-x^{*})in (R^n,Vert cdot Vert_{R^n}),$ and we are done!



Kindly check if I'm correct. Corrections and alternative proofs are welcome.






analysis normed-spaces




analysis normed-spaces






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edited Dec 13 '18 at 16:10









Did

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asked Dec 13 '18 at 15:15









Omojola MichealOmojola Micheal

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  • $begingroup$
    The letter $n$ has been used for the dimension. Using it for other purposes in the proof is confusing.
    $endgroup$
    – user587192
    Dec 13 '18 at 15:25










  • $begingroup$
    @user587192: Ok, I will work on that!
    $endgroup$
    – Omojola Micheal
    Dec 13 '18 at 15:27










  • $begingroup$
    @user587192: I've done something on that! I hope it's okay now!
    $endgroup$
    – Omojola Micheal
    Dec 13 '18 at 15:31


















  • $begingroup$
    The letter $n$ has been used for the dimension. Using it for other purposes in the proof is confusing.
    $endgroup$
    – user587192
    Dec 13 '18 at 15:25










  • $begingroup$
    @user587192: Ok, I will work on that!
    $endgroup$
    – Omojola Micheal
    Dec 13 '18 at 15:27










  • $begingroup$
    @user587192: I've done something on that! I hope it's okay now!
    $endgroup$
    – Omojola Micheal
    Dec 13 '18 at 15:31
















$begingroup$
The letter $n$ has been used for the dimension. Using it for other purposes in the proof is confusing.
$endgroup$
– user587192
Dec 13 '18 at 15:25




$begingroup$
The letter $n$ has been used for the dimension. Using it for other purposes in the proof is confusing.
$endgroup$
– user587192
Dec 13 '18 at 15:25












$begingroup$
@user587192: Ok, I will work on that!
$endgroup$
– Omojola Micheal
Dec 13 '18 at 15:27




$begingroup$
@user587192: Ok, I will work on that!
$endgroup$
– Omojola Micheal
Dec 13 '18 at 15:27












$begingroup$
@user587192: I've done something on that! I hope it's okay now!
$endgroup$
– Omojola Micheal
Dec 13 '18 at 15:31




$begingroup$
@user587192: I've done something on that! I hope it's okay now!
$endgroup$
– Omojola Micheal
Dec 13 '18 at 15:31










1 Answer
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Looks good to me.



By the way, since each $x^*_iinBbb R$ we already have $x^*=(x^*_1,dots,x^*_n)inBbb R^n$ by definition. You don't need that fact that $Bbb R^n$ is a linear space to conclude that.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Oh, thanks for that!
    $endgroup$
    – Omojola Micheal
    Dec 13 '18 at 20:56












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1 Answer
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1 Answer
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active

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active

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active

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1












$begingroup$

Looks good to me.



By the way, since each $x^*_iinBbb R$ we already have $x^*=(x^*_1,dots,x^*_n)inBbb R^n$ by definition. You don't need that fact that $Bbb R^n$ is a linear space to conclude that.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Oh, thanks for that!
    $endgroup$
    – Omojola Micheal
    Dec 13 '18 at 20:56
















1












$begingroup$

Looks good to me.



By the way, since each $x^*_iinBbb R$ we already have $x^*=(x^*_1,dots,x^*_n)inBbb R^n$ by definition. You don't need that fact that $Bbb R^n$ is a linear space to conclude that.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Oh, thanks for that!
    $endgroup$
    – Omojola Micheal
    Dec 13 '18 at 20:56














1












1








1





$begingroup$

Looks good to me.



By the way, since each $x^*_iinBbb R$ we already have $x^*=(x^*_1,dots,x^*_n)inBbb R^n$ by definition. You don't need that fact that $Bbb R^n$ is a linear space to conclude that.






share|cite|improve this answer









$endgroup$



Looks good to me.



By the way, since each $x^*_iinBbb R$ we already have $x^*=(x^*_1,dots,x^*_n)inBbb R^n$ by definition. You don't need that fact that $Bbb R^n$ is a linear space to conclude that.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 13 '18 at 16:16









BigbearZzzBigbearZzz

9,01021652




9,01021652












  • $begingroup$
    Oh, thanks for that!
    $endgroup$
    – Omojola Micheal
    Dec 13 '18 at 20:56


















  • $begingroup$
    Oh, thanks for that!
    $endgroup$
    – Omojola Micheal
    Dec 13 '18 at 20:56
















$begingroup$
Oh, thanks for that!
$endgroup$
– Omojola Micheal
Dec 13 '18 at 20:56




$begingroup$
Oh, thanks for that!
$endgroup$
– Omojola Micheal
Dec 13 '18 at 20:56


















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