# Show subspace of normed vector is closed under sup norm.

1. Oct 15, 2012

### Greger

http://imageshack.us/a/img141/4963/92113198.jpg [Broken]

hey,

I'm having some trouble with this question,

For part a) I know that in order for c_0 to be closed every sequence in c_0 must converge to a limit in c_0 but I am having trouble actually showing that formally with the use of the norm given.

Say x_n is a sequence in c_0 which converges to x then for any ε>0 there exists an N such that whenever n>N |x_n - x| < ε.

So that's just the definition for the limit of a sequence, but I'm having trouble how to use this to show that the limit x is in c_0 for every x_n in c_0.

Would anyone be able to help me out?

Last edited by a moderator: May 6, 2017
2. Oct 15, 2012

### micromass

So, you wish to show that $x=(x^0,x^1,x^2,x^3,...)\in c_0$. Take $\varepsilon>0$. You'll need to find an N>0 such that for all n>N holds that $|x^n|<\varepsilon$.

So, what you need to do is make $|x^n|$ smaller than something that is smaller than $\varepsilon$. What could you make $|x^n|$ smaller as? Hint: use the triangle inequality to introduce the $x_k$.

3. Oct 15, 2012

### Greger

Oh so using the trick where you add 0,

|x_n|=|x_n - x_k + x_k|≤ |x_n - x_k|+|x_k|

Then for x_k in c_0,

For ε>0 there is an N > 0 such that for all n>N, |x_k| <ε/4

So

|x_n - x_k|+|x_k| < |x_n - x_k|+ε/4

Then |x_n - x_k| will definitely be smaller then ε so

|x_n - x_k|+ε/4 < ε

I am having trouble seeing where the sup norm comes in,

Have I missed something?

4. Oct 17, 2012

### Greger

Does that look ok?

That shows that x_n is Cauchy, since you have

||x_n - x_k|| < 3e/4 < e

for n,k > N = max(N_n, N_K)

Last edited: Oct 17, 2012
5. Oct 17, 2012

### dirk_mec1

Do you mean by $$x= \{ x_1,x_2,..\}$$ that x_1 is a sequence of itself?

So for example $$x_1 (n) = 0.5^n, n =1, 2, 3,...$$
$$x_2(n) = 0.4^n, n =1, 2, 3,...$$

Or is x just ONE sequence?