Countable dense subset of $\ell_2$? I have been given the following problem:
Show that there exists a countable dense subset $A=\{ x_n: n\ge 1\}$ of $\ell_2$ satisfying:
For each $n\ge 1$, the vector $x_n=(x_n(1), x_n(2), x_n(3),\dots )$ satisfies that $x_n(j)= 0$ if and only if $j>n$.
I know $\ell_2$ is separable, but I'm not sure about this particular subset. Any hints would be appreciated.
 A: Here is a possible approach. The idea is to approximate sequences with finite support (as those are dense in $\ell^2$) with rational sequences with finite support.
First consider sequences with only the first coordinate nonzero. Let $(q_n)_n$ be an enumeration of $\Bbb{Q}^\times$ and consider the sequence $(x_n)_n$ given by
$$x_1=\left(q_1, 0, 0, 0, 0, 0\ldots\right),$$
$$x_2=\left(q_2, 1, 0, 0,0, 0 \ldots\right),$$
$$x_3=\left(q_3, \frac12, \frac12, 0,0, 0 \ldots\right),$$
$$x_4=\left(q_4, \frac13, \frac13, \frac13, 0, 0\ldots\right),$$
$$\vdots$$
$$x_n=\left(q_n, \underbrace{\frac1{n}, \frac1{n}, \ldots, \frac1{n}}_{n-1}, 0, 0\ldots\right),$$
$$\vdots$$
This sequence approximates any sequence of the form
$$x = (\alpha,0,0, \ldots)$$
for some $\alpha \in \Bbb{R}$. Indeed, there is a subsequence $(q_{p(n)})_n$ of the rationals such that $q_{p(n)} \to \alpha$. Then
$$\|x-x_{p(n)}\|_2^2 = |\alpha-q_{p(n)}|^2 + \frac{p(n)-1}{p(n)^2} \xrightarrow{n\to\infty} 0.$$
Now the idea is to approximate sequences with only first or first two coordinates nonzero. Let $((q_{a(n)},q_{b(n)}))_n$ be an enumeration of $\Bbb{Q}^\times \times \Bbb{Q}^\times$. Consider $(x_n)_n$ given by
$$x_1=\left(q_1, 0, 0, 0, 0, 0\ldots\right),$$
$$x_2=\left(q_{a(1)}, q_{b(1)}, 0, 0, 0, 0\ldots\right),$$
$$x_3=\left(q_2, 1, 1, 0,0, 0 \ldots\right),$$
$$x_4=\left(q_{a(2)}, q_{b(2)}, \frac12, \frac12, 0, 0\ldots\right),$$
$$x_5=\left(q_3, \frac13, \frac13, \frac13, \frac13, 0\ldots\right),$$
$$x_6=\left(q_{a(3)}, q_{b(3)}, \frac14, \frac14, \frac14, \frac14, 0\ldots\right),$$
$$x_7=\left(q_4, \frac15, \frac15, \frac15, \frac15, \frac15, \frac15, 0\ldots\right),$$
$$x_8=\left(q_{a(4)}, q_{b(4)}, \frac16, \frac16, \frac16, \frac16, \frac16, \frac16, 0\ldots\right),$$
$$\vdots$$
which oscillates between lengths $1$ and $2$. (Perhaps the tails should decay quicker but you get the idea.) Now sequences of the form $$(\alpha, 0, 0, \ldots)$$
can be approximated by subsequences with odd indices and those of the form $$(\alpha, \beta, 0, 0\ldots)$$
can be approximated by subsequences with even indices.
Now we enumerate $\Bbb{Q}^\times \times \Bbb{Q}^\times \times \Bbb{Q}^\times$ and interlace. And so on. However, elements of certain length clearly cannot be evenly spaced as we need to make room for increasingly longer ones. Hence we can put those of length $1$ at positions with perfect squares and put the other ones after them. Since perfect squares grow further and further apart as $n\to\infty$, eventually we cover all finite lengths. To illustrate:
\begin{matrix}
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 &  18 & \ldots\\
1 & 2 & 3 & 1 & 2 & 3 & 4 & 5 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 1 & 2 & 3 & \ldots
\end{matrix}
