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I was reading up and it says that $(\ell^2,\|.\|_2)$ is complete. I know that a metric space $X$ in which every Cauchy sequence converges to an element of $X$ is called complete. And I know that a sequence is Cauchy if given $\epsilon$, there exists $N \in \mathbb{N}$ such that $|x_m-x_n|<\epsilon$ for all $n,m>N$.

I was reading the proof that $(\ell^2,\|.\|_2)$ is complete, but I don't understand where does the $x_k^n$ comes from. What does it mean when it writes $x_k^n$?


Let $(x_n)$ be Cauchy in $\ell^2$, i.e. $\forall \epsilon>0$ there exists $N \in \mathbb{N}$ such that $\sum_{k=1}^\infty|x_k^n-x_k^m|^2 <\epsilon^2$ for $n,m>N$.

For any fixed $k_0$, $|x_{k_0}^n-x_{k_0}^m|<\epsilon$ for $n,m >N$. So $(x_{k_0}^n)$ is Cauchy in $\mathbb{K}$ ($\mathbb{R}$ or $\mathbb{C})$ and converges to say $y_{k_0}$.

Also, why did they square $\sum_{k=1}^\infty|x_k^n-x_k^m|^2 <\epsilon^2$?

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$\epsilon^2$ appears because $\|x\|<\epsilon$ is equivalent to $\sum |x_k|^2<\epsilon^2$. –  user31373 May 20 '12 at 16:22
    
Thanks Leonid, appreciated! –  Steven May 20 '12 at 16:41
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up vote 2 down vote accepted

Each element in $l^2$ is again a sequence (of real numbers), i.e. if $x\in l^2$, then $x=(x_k)_k$. Now if $(x^n)$ is a sequence in $l^2$, then we might write $x^n=(x^n_k)_k=(x^n_1,x^n_2,\ldots)$ for each $n$. Then for a Cauchy sequence in $l^2$, we have $$ \epsilon^2 > \|x^n-x^m\|_{l^2}^2 = \sum_{k=1}^\infty |x_k^n-x_k^m|^2. $$ Using this, you see that for any fixed $k$ (or $k_0$), the sequence $(x_k^n)$ is a Cauchy sequence of real numbers and then apply completeness of $\mathbb{R}$ to see that it converges. Then you must show that the resulting sequence $y=(y_k)_k$ is again in $l^2$, and that $x^n\to y$ in the $l^2$ norm.

As for the $\epsilon^2$, this is presumably because they are looking at at the square of the norm, rather than the norm itself.

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You forgot taking square roots in the definition of the norm. –  azarel May 20 '12 at 16:23
    
Thank you, fixed. Also explains epsilon square! –  Eepzy May 20 '12 at 16:27
    
Ahh okay, thanks, I understand it now. Most appreciated! –  Steven May 20 '12 at 16:41
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