# Thread: co is closed in l∞

1. ## 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.

2. 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 $\displaystyle n\geqslant N\,\Rightarrow\,\sup_k|x_n(k) - x(k)|<\varepsilon/2$.

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 $\displaystyle k\geqslant K\,\Rightarrow\,|x(k)|\leqslant|x_n(k)| + |x_n(k) - x(k)|<\varepsilon$. Therefore $\displaystyle x\in c_0$.

3. Thanks, it makes a lot of sense.