# co is closed in l∞

• Apr 2nd 2009, 08:24 AM
math8
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.

• Apr 2nd 2009, 11:57 AM
Opalg
I think you have essentially done this. It's just a matter of piecing the bits together.

You know that $x_n\to x$ in the sup norm. So given $\varepsilon>0$ there exists N such that $n\geqslant N\,\Rightarrow\,\sup_k|x_n(k) - x(k)|<\varepsilon/2$.

Choose and fix a value of n greater than or equal to N. Then $x_n\in c_0$ and so $\lim_{k\to\infty}x_n(k) = 0$. Therefore there exists K such that $k\geqslant K\,\Rightarrow\,|x_n(k)|<\varepsilon/2$.

Hence $k\geqslant K\,\Rightarrow\,|x(k)|\leqslant|x_n(k)| + |x_n(k) - x(k)|<\varepsilon$. Therefore $x\in c_0$.
• Apr 2nd 2009, 12:44 PM
math8
Thanks, it makes a lot of sense.