
co is closed in l∞
I want to show the Banach space co is closed in l∞ .
So, I pick a convergent sequence xn in co that converges to x in l∞
Now, x_n > x: given e>0, there is an N_e s.t. for all n>N_e,
x_n x = Sup x_n(k)x(k)<e (we're supping over k).
Since x_n is a sequence in co , for each fixed n, x_n(k)>0 as k> infinity.
So, given e>0, there is a K depending on n and e, such that for all k> K, we have x_n(k)<e.
We want to show x is in co
so we show there is a Ko such that for all k>Ko, x(k)<e
I am having trouble getting this Ko.
I know x(k) ≤ x_n(k) + x_n(k)x(k) ≤ x_n(k) + Sup (over k) x_n(k)x(k)
we have x_n(k)<e as k>K, but Sup (over k) x_n(k)x(k)<e for n>N_e.
So I am not so sure how to get this Ko.

I think you have essentially done this. It's just a matter of piecing the bits together.
You know that $\displaystyle x_n\to x$ in the sup norm. So given $\displaystyle \varepsilon>0$ there exists N such that $\displaystyle n\geqslant N\,\Rightarrow\,\sup_kx_n(k)  x(k)<\varepsilon/2$.
Choose and fix a value of n greater than or equal to N. Then $\displaystyle x_n\in c_0$ and so $\displaystyle \lim_{k\to\infty}x_n(k) = 0$. Therefore there exists K such that $\displaystyle k\geqslant K\,\Rightarrow\,x_n(k)<\varepsilon/2$.
Hence $\displaystyle k\geqslant K\,\Rightarrow\,x(k)\leqslantx_n(k) + x_n(k)  x(k)<\varepsilon$. Therefore $\displaystyle x\in c_0$.

Thanks, it makes a lot of sense.