Proving a subset of $l^2$ is not closed

Question

We consider $\ell^2$ equiped with its usual inner product and let $ u = (1/2, 1/4, 1/8 , \dots , 1/2^n, \dots) \in \ell^2 $ and $$F:= \{ x = (x_1,x_2, \dots ) \in \ell^2 : \langle u,x \rangle = 0, \text{ and } \exists N : \forall n > N : x_n = 0 \}$$

Now I need to show that $F$ is not closed. I tried constructing a sequence $x^n_i$ given by

$$ x^n_i = (-1, 0 , \dots, 0, 2^n , 0 , \dots) ,$$ with the $2^n$ term as the $(n+1)$th entry in the sequence. This gives a sequence in $F$ that does not have a limit in $F$, but the problem is that it does not converge in $\ell^2$. I can't seem to think of a sequence that accomplishes the latter; am I on the right track here?

Answer 1

I think the following element of $\ell^2$ $$ v = \left(\frac14,-\frac12,\frac1{16},-\frac18,\ldots\right) $$ is a limit point of $F$ but it doesn't belong to it.

Answer 2

Let $T_{2k}\colon \ell_2 \to \ell_2$ denotes the map given by $T_{2k}((x_n)_n) = (x_1,\ldots,x_{2k},0,\ldots)$ (so that $T_{2k}$ replaces all the entries of $(x_n)$ with $n>2k$ with zeros). One way to show that $F$ is not closed would be to find a sequence $\mathbf a = (a_1,a_2,\ldots) \in \ell_2$ such that $\mathbf a \notin F$ but $T_{2k}(\mathbf a) \in F$ for all $k \geq 1$.

The sequences $x^n = (x^n_i)_i = (-1,0,\ldots,2^n,0,\ldots)$ lie in $F$ because you weight the $n$-th so that it cancels with the first when paired with $u$. There is no need to "pair" all the terms with the first term this way -- indeed for any $a_1,\ldots,a_{2k-1}$ if we $a_{2k} = -2^{2k}.\sum_{n=1}^{2k-1} a_n/2^n$ then $\sum_{n=1}^{2k} a_n/2^n =0$. But if we have inductively ensured that $\langle T_{2m}(\mathbf a),u\rangle =0$ for $m=1,\ldots,k-1$, then the condition on $a_{2k}$ simplifies considerably.

Can you see how to construct an example from this?