This statement has been given as an example in the book "Introductory real analysis" written by Kolmogorov and Fomin:
The set of all points $x=(x_1,x_2,\cdots,x_n,\cdots)$ with only finitely many non-zero coordinates, each a rational number, is dense in the space $\ell_2$.
and $\ell_2$ is defined this way:
$$\ell_2 = \{ (x_1,x_2, \cdots, x_n, \cdots): \sum_{n=1}^\infty x_n^2 <\infty \text{ and } x_n \in \mathbb{R} \} $$
Analysis is one of my weak subjects in math. So, please bear with me.
I know what I need to show. Suppose that I define: $$A = \{ (x_1,x_2, \cdots, x_n, \cdots): \text{only finitely many terms are non-zero and } x_i \in \mathbb{Q} \}$$
I need to show that if $(x_n)_{n \in\mathbb{N}}$ is in $l_2$ then one can find a "sequence of sequences in $A$" such that the limit approaches to $(x_n)_{n \in \mathbb{N}}$.
So, if I was allowed to use sequences like $(x_1,x_2, \cdots, x_n, \cdots)$ where only finitely many terms are non-zero but the terms were allowed to be in $\mathbb{R}$ I was done because for any sequence $(x_n)$ the sequence $(y_k)$ where $y_k = (x_1, x_2, \cdots, x_k, 0, 0, 0, \cdots, 0, \cdots)$ works. Am I right?
Now the problem is that I should find a sequence like $y_k$ where this time all terms must be chosen from rational numbers.
This is my solution for this:
Since $\mathbb{Q}$ is dense in $\mathbb{R}$ I can find a sequence of rational numbers that approaches any given real number $x_i$. therefore:
$$x_1 = \lim (q_{11},q_{12},q_{13}, \cdots, q_{1n}, \cdots)$$ $$x_2 = \lim (q_{21},q_{22},q_{23}, \cdots, q_{2n}, \cdots)$$ $$\vdots$$ $$x_i = \lim (q_{i1},q_{i2},q_{i3}, \cdots, q_{in}, \cdots)$$ $$\vdots$$
Now I create my new sequence this way:
$$y_1 = (q_{11},0,0,0,0,0,0,\cdots)$$ $$y_2 = (q_{12},q_{22},0,0,0,0,\cdots)$$ $$y_3 = (q_{13},q_{23},q_{33},0,0,\cdots)$$
and $y_k$ is formed the similar way. I mean you go the $k$-th column and choose $q_{1k},q_{2k},\cdots,q_{kk}$ on the rows and put zero for all other entries beyond the $k$-th coordinate.
I want to claim that given any $x=(x_1,x_2, \cdots, x_n, \cdots)$ in $\ell_2$, the sequence $(y_k)_{n \in \mathbb{N}}$ constructed in the way just explained approaches $x \in \ell_2$. I think it is obvious that this is true because:
$$\lim_{k \to \infty} y_k = (\lim_{k \to \infty} q_{nk})_{n \in \mathbb{N}} = (x_n)_{n \in \mathbb{N}}$$
Does what I'm saying makes sense? Is my solution correct or it needs to be modified?