Proof check: the inclusion of $\ell^1$ into $c0$ in dense with respect to the sup-norm on $c0$

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 $\Vert \cdot\Vert_\infty$;

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