Proving this set is dense in $\ell^2$ I found this weirdest question and was wondering how could this be proved.
This question is a part of a beautiful semi-constructive built of two dense disjoint convex sets in $\ell^2$, which I find amazing.
Let $A$ be all the sequences in $\ell^2$ which are constant from some point.
Let $\{e_n\}$ be the standard basis for $\ell^2$. Extend this base to a Hamel basis (it's obvious it can be done). Define $f:A\to \mathbb R$ by $f(x)=\sum_n x_n$.
This is when it gets trickier:
Extend $f$ to $\ell^2$, using the Hamel basis. Prove $B=f^{-1}((0,\infty))$ is dense. If you think it's easier, explain for $C=f^{-1}((-\infty,0])$
Intuitivly the proof that $\ell^2-B$ would also be dense should be the same. Convexity is obvious. Density? I found it VERY hard to prove.
Now to what I did so far:


*

*It's tricky how one would want to extend. I picked $f(x)$ to be the sum of the coefficients of the unique representation of $x$ as finite linear combination of elements of the Hamel basis.

*I realized that I absolutely have no idea. Of course, I'd have to take a converging series in $B$ and show that the limit is also in $B$, but actually showing that is hard for me. Any tips?
Please guide me a little bit, so I'd be able to continue on my own.
 A: First, $A\subset l^2$ is the collection of sequences that only take finitely many values, their entries are all zero after some point (you said constant, but the constant must be zero), and $A$ is dense in $l^2$. 
Now let us show $B$ is dense in $A$ (which would mean $B$ is also dense in $l^2$), let $a\in A$ and $\epsilon$ be given, we show there exists a $b\in B$ such that 
$$\|a-b\|^2 \leq \epsilon.$$
If $f(a) = \sum_{i=1}^n a_i >0$, we are done since $a\in B$. Else suppose $0\geq f(a) >-C$ for some positive constant $C$, now let us construct $b$.
For the first $n$ entries, let us take $a_i = b_i$ for $i=1,...,n$, we now need
$$\|b-a\|^2  = \sum_{i=n+1}^\infty b_i^2 \leq \epsilon$$
and at the same time
$$f(b) = \sum_i b_i = a_1+a_2,...,+a_n +\sum_{i=n+1}^\infty b_i >0 \; \text{ or to say } \sum_{i=n+1}^\infty b_i > C.$$
To construct the tail of $b$ such that 
$$\sum_{i=n+1}^\infty b_i > C \text{ and }\sum_{i=n+1}^\infty b_i^2 \leq \epsilon$$
isn't very hard, use the fact that $\sum \frac{1}{k^2} = \frac{\pi^2}{6}$ and $\sum \frac{1}{k} $ is unbounded, then $\sum \frac{6\epsilon}{\pi^2}\frac{1}{k^2} = \epsilon $ and $\sum \sqrt{\frac{6\epsilon}{\pi}}\frac{1}{k} $ is still unbounded. You stop  at $b_{n+m}$ when this sum $\sum_{k=1}^m \sqrt{\frac{6\epsilon}{\pi}}\frac{1}{k} $ becomes greater than $C$, and define the entries of $b$ to be zero after that, since you don't want $\sum_i b_i = +\infty$. 
For $l^2 - B = f^{-1}((-\infty,0])$, similar construction applies. 
Edit: Just a remark, 


*

*the subset $A\subset l^2$ where after some point all entries are zero

*the subset $B\subset l^2$ where after some point all entries are non-negative, and 

*the subset $C\subset l^2$ where after some point all entries are non-positive
The three sets are dense in $l^2$.
For density, let $x\in l^2$ and $\epsilon$ be given, there exists $N$ such that 
$$\sum_{i=N}^\infty x_i^2 \leq \epsilon.$$
because $\sum_{i=1}^\infty x_i^2 <\infty$.
$(x_1,x_2, ...x_N, 0, 0, ...)$ will give you $x_A\in A$ and $\|x_A - x\|^2\leq\epsilon$
$(x_1,x_2, ...x_N, |x_{N+1}|, |x_{N+2}|,...)$ will give you $x_B\in B$ and $\|x_B - x\|^2\leq 4\epsilon$
$(x_1,x_2, ...x_N, -|x_{N+1}|, -|x_{N+2}|,...)$ will give you $x_C\in C$ and $\|x_C - x\|^2\leq 4\epsilon$.
And with another argument, you can make the tail of the sequences in $B$ to be positive instead of non-negative, and the tail of sequences in $C$ to be negative instead of non-positive. Then $A, B, C,$ are all dense, convex, and disjoint.
