Prove that ${x^n}$ is Cauchy in $Ssubseteq ell_infty$
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
edited Dec 13 '18 at 9:54
José Carlos Santos
152k22123225
asked Dec 13 '18 at 9:52
Omojola MichealOmojola Micheal
1,692324
add a comment |
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
edited Dec 13 '18 at 9:54
José Carlos Santos
152k22123225
asked Dec 13 '18 at 9:52
Omojola MichealOmojola Micheal
1,692324
add a comment |
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
edited Dec 13 '18 at 9:54
José Carlos Santos
152k22123225
asked Dec 13 '18 at 9:52
Omojola MichealOmojola Micheal
1,692324
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
functional-analysis analysis convergence normed-spaces cauchy-sequences
edited Dec 13 '18 at 9:54
José Carlos Santos
152k22123225
asked Dec 13 '18 at 9:52
Omojola MichealOmojola Micheal
1,692324
edited Dec 13 '18 at 9:54
José Carlos Santos
152k22123225
asked Dec 13 '18 at 9:52
Omojola MichealOmojola Micheal
1,692324
edited Dec 13 '18 at 9:54
José Carlos Santos
152k22123225
edited Dec 13 '18 at 9:54
José Carlos Santos
152k22123225
edited Dec 13 '18 at 9:54
José Carlos Santos
152k22123225
152k22123225
asked Dec 13 '18 at 9:52
Omojola MichealOmojola Micheal
1,692324
asked Dec 13 '18 at 9:52
Omojola MichealOmojola Micheal
1,692324
asked Dec 13 '18 at 9:52
Omojola MichealOmojola Micheal
1,692324
1,692324
add a comment |
add a comment |
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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$$mgeqslant ngeqslant NimplieslVert x^n-x^mrVert_infty<varepsilon.$$
answered Dec 13 '18 at 9:58
José Carlos SantosJosé Carlos Santos
152k22123225
That was nice! I am grateful! +1!
– Omojola Micheal
Dec 13 '18 at 10:14
add a comment |
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1 Answer
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1 Answer
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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$$mgeqslant ngeqslant NimplieslVert x^n-x^mrVert_infty<varepsilon.$$
answered Dec 13 '18 at 9:58
José Carlos SantosJosé Carlos Santos
152k22123225
That was nice! I am grateful! +1!
– Omojola Micheal
Dec 13 '18 at 10:14
add a comment |
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$$mgeqslant ngeqslant NimplieslVert x^n-x^mrVert_infty<varepsilon.$$
answered Dec 13 '18 at 9:58
José Carlos SantosJosé Carlos Santos
152k22123225
That was nice! I am grateful! +1!
– Omojola Micheal
Dec 13 '18 at 10:14
add a comment |
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$$mgeqslant ngeqslant NimplieslVert x^n-x^mrVert_infty<varepsilon.$$
answered Dec 13 '18 at 9:58
José Carlos SantosJosé Carlos Santos
152k22123225
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$$mgeqslant ngeqslant NimplieslVert x^n-x^mrVert_infty<varepsilon.$$
answered Dec 13 '18 at 9:58
José Carlos SantosJosé Carlos Santos
152k22123225
answered Dec 13 '18 at 9:58
José Carlos SantosJosé Carlos Santos
152k22123225
answered Dec 13 '18 at 9:58
José Carlos SantosJosé Carlos Santos
152k22123225
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
add a comment |
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
add a comment |
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