- #1
math8
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- 0
I want to show the Banach space co is closed in l∞ .
So, I pick a convergent sequence x_n 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.
So, I pick a convergent sequence x_n 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.