Prove that ${x^n}$ is Cauchy in $Ssubseteq ell_infty$

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I'm kind of new into Functional analysis. So, I have the following question bothering me. Let $S$ be the set of sequences having only a finite number of non-zero terms. Clearly, $Ssubseteq ell_infty$. Take ${x^n}$ in $S$ where $x^n=(1,frac{1}{2},frac{1}{3},cdots,frac{1}{n},0,0,cdots).$ I want to prove that ${x^n}$ is Cauchy in $S.$



MY TRIAL



Let $mgeq n.$ Then,
begin{align}Vert x^n-x^m Vert_ {ell_infty}=suplimits_{m,nin Bbb{N}}left|frac{1}{m}-frac{1}{n} right|end{align}
By convergence of $frac{1}{n}$ and $frac{1}{m}$ for each $m$ and $n$, there exists $NinBbb{N}$ such that
begin{align}Vert x^n-x^m Vert_ {ell_infty}=suplimits_{m,nin Bbb{N}}left|frac{1}{m}-frac{1}{n} right|=frac{1}{n}-frac{1}{m}<epsilon,;;forall;mgeq ngeq N. end{align}



Please check if I'm right or wrong. If I'm wrong, kindly help out. Thanks










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    I'm kind of new into Functional analysis. So, I have the following question bothering me. Let $S$ be the set of sequences having only a finite number of non-zero terms. Clearly, $Ssubseteq ell_infty$. Take ${x^n}$ in $S$ where $x^n=(1,frac{1}{2},frac{1}{3},cdots,frac{1}{n},0,0,cdots).$ I want to prove that ${x^n}$ is Cauchy in $S.$



    MY TRIAL



    Let $mgeq n.$ Then,
    begin{align}Vert x^n-x^m Vert_ {ell_infty}=suplimits_{m,nin Bbb{N}}left|frac{1}{m}-frac{1}{n} right|end{align}
    By convergence of $frac{1}{n}$ and $frac{1}{m}$ for each $m$ and $n$, there exists $NinBbb{N}$ such that
    begin{align}Vert x^n-x^m Vert_ {ell_infty}=suplimits_{m,nin Bbb{N}}left|frac{1}{m}-frac{1}{n} right|=frac{1}{n}-frac{1}{m}<epsilon,;;forall;mgeq ngeq N. end{align}



    Please check if I'm right or wrong. If I'm wrong, kindly help out. Thanks










    share|cite|improve this question



























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      I'm kind of new into Functional analysis. So, I have the following question bothering me. Let $S$ be the set of sequences having only a finite number of non-zero terms. Clearly, $Ssubseteq ell_infty$. Take ${x^n}$ in $S$ where $x^n=(1,frac{1}{2},frac{1}{3},cdots,frac{1}{n},0,0,cdots).$ I want to prove that ${x^n}$ is Cauchy in $S.$



      MY TRIAL



      Let $mgeq n.$ Then,
      begin{align}Vert x^n-x^m Vert_ {ell_infty}=suplimits_{m,nin Bbb{N}}left|frac{1}{m}-frac{1}{n} right|end{align}
      By convergence of $frac{1}{n}$ and $frac{1}{m}$ for each $m$ and $n$, there exists $NinBbb{N}$ such that
      begin{align}Vert x^n-x^m Vert_ {ell_infty}=suplimits_{m,nin Bbb{N}}left|frac{1}{m}-frac{1}{n} right|=frac{1}{n}-frac{1}{m}<epsilon,;;forall;mgeq ngeq N. end{align}



      Please check if I'm right or wrong. If I'm wrong, kindly help out. Thanks










      share|cite|improve this question















      I'm kind of new into Functional analysis. So, I have the following question bothering me. Let $S$ be the set of sequences having only a finite number of non-zero terms. Clearly, $Ssubseteq ell_infty$. Take ${x^n}$ in $S$ where $x^n=(1,frac{1}{2},frac{1}{3},cdots,frac{1}{n},0,0,cdots).$ I want to prove that ${x^n}$ is Cauchy in $S.$



      MY TRIAL



      Let $mgeq n.$ Then,
      begin{align}Vert x^n-x^m Vert_ {ell_infty}=suplimits_{m,nin Bbb{N}}left|frac{1}{m}-frac{1}{n} right|end{align}
      By convergence of $frac{1}{n}$ and $frac{1}{m}$ for each $m$ and $n$, there exists $NinBbb{N}$ such that
      begin{align}Vert x^n-x^m Vert_ {ell_infty}=suplimits_{m,nin Bbb{N}}left|frac{1}{m}-frac{1}{n} right|=frac{1}{n}-frac{1}{m}<epsilon,;;forall;mgeq ngeq N. end{align}



      Please check if I'm right or wrong. If I'm wrong, kindly help out. Thanks







      functional-analysis analysis convergence normed-spaces cauchy-sequences






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      edited Dec 13 '18 at 9:54









      José Carlos Santos

      152k22123225




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      asked Dec 13 '18 at 9:52









      Omojola MichealOmojola Micheal

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          Actually.$$mgeqslant nimplieslVert x^n-x^mrVert_infty=begin{cases}0&text{ if }m=n\dfrac1{n+1}&text{ otherwise,}end{cases}$$since, if $m>n$,$$lVert x^n-x^mrVert_infty=supleft{0,frac1{n+1},frac1{n+2},ldots,frac1mright}=frac1{n+1}.$$So, given $varepsilon>0$, take $Ninmathbb{N}$ such that $frac1N<varepsilon$ and then$$mgeqslant ngeqslant NimplieslVert x^n-x^mrVert_infty<varepsilon.$$






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          • That was nice! I am grateful! +1!
            – Omojola Micheal
            Dec 13 '18 at 10:14













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          Actually.$$mgeqslant nimplieslVert x^n-x^mrVert_infty=begin{cases}0&text{ if }m=n\dfrac1{n+1}&text{ otherwise,}end{cases}$$since, if $m>n$,$$lVert x^n-x^mrVert_infty=supleft{0,frac1{n+1},frac1{n+2},ldots,frac1mright}=frac1{n+1}.$$So, given $varepsilon>0$, take $Ninmathbb{N}$ such that $frac1N<varepsilon$ and then$$mgeqslant ngeqslant NimplieslVert x^n-x^mrVert_infty<varepsilon.$$






          share|cite|improve this answer





















          • That was nice! I am grateful! +1!
            – Omojola Micheal
            Dec 13 '18 at 10:14


















          2














          Actually.$$mgeqslant nimplieslVert x^n-x^mrVert_infty=begin{cases}0&text{ if }m=n\dfrac1{n+1}&text{ otherwise,}end{cases}$$since, if $m>n$,$$lVert x^n-x^mrVert_infty=supleft{0,frac1{n+1},frac1{n+2},ldots,frac1mright}=frac1{n+1}.$$So, given $varepsilon>0$, take $Ninmathbb{N}$ such that $frac1N<varepsilon$ and then$$mgeqslant ngeqslant NimplieslVert x^n-x^mrVert_infty<varepsilon.$$






          share|cite|improve this answer





















          • That was nice! I am grateful! +1!
            – Omojola Micheal
            Dec 13 '18 at 10:14
















          2












          2








          2






          Actually.$$mgeqslant nimplieslVert x^n-x^mrVert_infty=begin{cases}0&text{ if }m=n\dfrac1{n+1}&text{ otherwise,}end{cases}$$since, if $m>n$,$$lVert x^n-x^mrVert_infty=supleft{0,frac1{n+1},frac1{n+2},ldots,frac1mright}=frac1{n+1}.$$So, given $varepsilon>0$, take $Ninmathbb{N}$ such that $frac1N<varepsilon$ and then$$mgeqslant ngeqslant NimplieslVert x^n-x^mrVert_infty<varepsilon.$$






          share|cite|improve this answer












          Actually.$$mgeqslant nimplieslVert x^n-x^mrVert_infty=begin{cases}0&text{ if }m=n\dfrac1{n+1}&text{ otherwise,}end{cases}$$since, if $m>n$,$$lVert x^n-x^mrVert_infty=supleft{0,frac1{n+1},frac1{n+2},ldots,frac1mright}=frac1{n+1}.$$So, given $varepsilon>0$, take $Ninmathbb{N}$ such that $frac1N<varepsilon$ and then$$mgeqslant ngeqslant NimplieslVert x^n-x^mrVert_infty<varepsilon.$$







          share|cite|improve this answer












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          answered Dec 13 '18 at 9:58









          José Carlos SantosJosé Carlos Santos

          152k22123225




          152k22123225












          • That was nice! I am grateful! +1!
            – Omojola Micheal
            Dec 13 '18 at 10:14




















          • That was nice! I am grateful! +1!
            – Omojola Micheal
            Dec 13 '18 at 10:14


















          That was nice! I am grateful! +1!
          – Omojola Micheal
          Dec 13 '18 at 10:14






          That was nice! I am grateful! +1!
          – Omojola Micheal
          Dec 13 '18 at 10:14




















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