My approach is the following:

Let $x \in c0$ be an arbitrary zero sequence in $\mathbb R$ i.e. $\lim_{i \to \infty} \Vert x_i\Vert $ = $0$, where here the norm is any norm on $\mathbb R$.

Then we can take a sequence $y_k$ in $\ell^1$ defined as follows:

$y_{k} := \left(x_1,x_2,...,x_k,0,0,..\right)$ ($\forall k \in \mathbb N$);

$y_k$ are obviously in $\ell^1$ and we observe:

$\lim_{k \to \infty} \Vert y_k - x\Vert_\infty = \lim_{k \to \infty} \sup_{i \geq k}\Vert x_i\Vert=\limsup_{k}\Vert x_k\Vert = lim_{k \to \infty} \Vert x_k\Vert = 0$

thus any sequence in $c0$ can be approximated by a sequence in $\ell^1$ with respect to our sup-norm $\Vert \cdot\Vert_\infty$;

  • $\begingroup$ I think that true $\endgroup$
    – Hamza
    Jan 13 '16 at 10:29
  • $\begingroup$ What is your question? $\endgroup$
    – MaoWao
    Jan 13 '16 at 10:31
  • $\begingroup$ I'm asking wether my proof is right or wrong. Other suggestion for a slicker, smarter or more enlightening proof are welcomed of course. $\endgroup$
    – noctusraid
    Jan 13 '16 at 10:46
  • $\begingroup$ Yes, It is correct. $\endgroup$
    – sinbadh
    Jan 13 '16 at 10:59
  • $\begingroup$ thanks guys! Am I supposed to do something now that my question has been answered? $\endgroup$
    – noctusraid
    Jan 14 '16 at 9:53

This proof is correct and hence the question is answered.


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