# Thread: Hard Proof

1. ## Hard Proof

Hi!

I have been struggling with an proof for days now and would like to see if there is someone in here who can help me out.

the excersise is as follows;
We have a Banach Space A. Let {e_alpha}_(alpha belongs to A) be a base when A is considered being a linear space and show with Baires category theorem that the index set A is either finite or not countable?

Someone who is indulged to try?

Peter

2. Originally Posted by Peter_L
We have a Banach Space A. Let {e_alpha}_(alpha belongs to A) be a base when A is considered being a linear space and show with Baires category theorem that the index set A is either finite or not countable?
I don't think you meant to use the same letter for the name of the space and the name of the index set, did you? I'll use E for the space and A for the index set.

If A is finite then E is finite-dimensional. Suppose, for a contradiction, that E is infinite-dimensional but has a countable base {e_n} (indexed by the natural numbers). For n=1, 2, ..., let E_n be the subspace of E spanned by the first n basis vectors e_1, ..., e_n. Let A_n be the complement of E_n in E. Then A_n is open and dense (easily proved). So by Baire's theorem the intersection of the A_n is dense. But every element of E is in E_n for n large enough, so the intersection of the A_n is empty. Contradiction.

3. thanks for the reply! I have two questions though,

1. is it ok to just take a space A_n which is the complement of the subspace E_n in E?

2. so, we have the complement A_n to E_n in the Banach Space E, and since E_n is a sub space spanned by the n first base vectors, A_n must be open and dense? is it because we know that E_n is closed?

4. Originally Posted by Peter_L
thanks for the reply! I have two questions though,

1. is it ok to just take a space A_n which is the complement of the subspace X_n in X?

2. so, we have the complement A_n to X_n in the Banach Space X, and since X_n is a sub space spanned by the n first base vectors, A_n must be open and dense? is it because we know that X_n is closed?
We seem to have another change of notation. The Banach space has become X, and presumably X_n is what I called E_n, the finite-dimensional subspace of X spanned by the first n basis vectors.

Just to be clear, the set A_n is not a subspace of X. It is the set-theoretic complement of X_n in X. So I prefer not to call it a "space": it's just a set. It is open because it's the complement of X_n, which is closed because it is finite-dimensional.

The reason that A_n is dense is that if a point x is not in A_n then it must be in X_n. Then we can approximate x by elements of the form $x+\epsilon e_{n+1}$ (where ε is small but nonzero). These points are not in X_n and are therefore in A_n.

5. oops, mixed the notation up with another exercise..hehe,

but thank you very much!! it was to a lot of help for me!! find it pretty hard with this kind of abstract math...