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\(\displaystyle x_{1}+x_{3}+x_{5}+.....=x_{2}+x_{4}+x_{6}+....\)

Let Y be the linear span of \(\displaystyle {{e_{n}+e_{n+1}:n=1,2,....}}\) where \(\displaystyle e_{n}\) is the sequence with a 1 in the nth position, zeroes elsewhere.

You are given that X is a closed subspace, and you use the following result:

Let F be a subspace of E and x be in E. Suppose that for all maps f in the dual space of E satisfying f(y)=0 for all y in F, we have that f(x)=0. Then x is in the closure of Y.

Prove that the closure of Y is equal to X.

My work: I show Y is a subset of X: an aritrary member of Y is of the form:\(\displaystyle t_{1}(e_{1}+e_{2})+.....+t_{k}(e_{k}+e_{k+1})\) and its easy to show that the sum of the odd positioned terms is equal to the even positioned ones.

So since X is closed, the closure of Y is contained in X. Now I use result with X in place of E and Y in place of F. So let f in the dual of X satisfy f(y)=0 for all y in Y.

taking k=1, then k=2, and so in the above, we see this implies that \(\displaystyle f(e_{n}+e_{n+1})=0\) for all n. Can anyone help me finish to show f(x)=0 for any x in X?

Thanks