# Prove set of sequences is a basis

1. Sep 30, 2013

### SMA_01

Let c_00 be the subspace of all sequences of complex numbers that are "eventually zero". i.e. for an element x∈c_00, ∃N∈N such that xn=0,∀n≥n.

Let {e_i}, i∈N be the set where e_i is the sequence in c_00 given by (e_i)_n =1 if n=i and (e_i)_n=0 if n≠i.

Show that (e_i), i∈N is a basis for c_00.

So I need to show it's linearly independent and that it spans c_00. I am not sure how to go about proving this makes it confusing is that it's an infinite set, so I can't use the usual method and take a finite number of vectors.

I have an idea of how to prove linear independence, but not spanning.

Any tips/hints?

Thanks

2. Sep 30, 2013

### Zondrina

Hmm, I think contradiction would be good here.

Suppose that $\{e_i\}$ is not a basis for $C_∞$.

What does that tell you about $\{e_i\}$?

3. Sep 30, 2013

### tiny-tim

Hi SMA_01!
I don't see the difficulty … for spanning, you need to prove that given any element, there's a finite number of basis elements that it is a linear combination of.

4. Sep 30, 2013

### SMA_01

What confused me was the fact that c_00 and {e_i} are infinite sets.

5. Sep 30, 2013

### tiny-tim

i] they're not sets

ii] all you have to do is add a finite number of them …

what difficulty would you have adding a finite number of decimal expansions?

6. Sep 30, 2013

### Zondrina

I would just like to make a side note that $\{e_i\}$ is a countably infinite set of sequences.

$C_∞$ is an infinite dimensional subspace.

7. Sep 30, 2013

### SMA_01

Sorry, c_00 is a subspace, but {e_i} is a set.
I understand now though how a finite number of the e_i's span any x in c_00, because x_n=0 for n≥N

8. Sep 30, 2013

### Zondrina

Yes, that's the idea.

Since you know any sequence in $C_∞$ converges to zero (eventually the sequence terminates), it will always be possible to find a finite basis. You can scale this finite basis accordingly to represent any element in $C_∞$.