Show that the space [tex]c_0[/tex] of all sequences of real numbers that converge to(adsbygoogle = window.adsbygoogle || []).push({});

0 is a complete space with the [tex]l^\infty[/tex] norm. First I let [tex] A^j=\{a_k^j\}_{k=1}^\infty[/tex] be a sequence of sequences converging to zero and I assume that it is norm summable:

[tex]\sum \limits_{j=1}^\infty ||A^j||_\infty < \infty[/tex]

I argue that [tex] S= \sum \limits_{j=1}^\infty A^j[/tex] converges componentwise.

Then I want to show that the sequence S converges to 0 at infinity (and hence is in [tex]c_0[/tex]). Fix [tex]\varepsilon >0[/tex]. For each j, choose a [tex]K_j[/tex] such that [tex]k \geq K_j[/tex] implies [tex]|a_k^j| < \frac{\varepsilon}{2^j}[/tex]. Given [tex] N>0[/tex] let [tex]K=\sup\limits_{k\leq N} K_k[/tex]. Then for [tex]k\geq K[/tex] we have

[tex] |S_k|=|\sum \limits_{j=1}^{\infty}a_k^j| \leq \sum \limits_{j=1}^{N}|a_k^j|+\sum \limits_{j=N+1}^{\infty}|a_k^j|\leq \varepsilon + \sum \limits_{j=N+1}^{\infty}|a_k^j|[/tex].

Here is where I am stuck. I know the quantity on the right goes to 0 if I make N large enough, but putting that down rigorously presents a difficulty. Any ideas?

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# Homework Help: Completeness in l^\infty Norm

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