Prove that sequence space $\ell_p(\mathbb R)$ is separable Problem:

Prove that metric space $\left \langle \ell_p(\mathbb R), d_p(x,y)=(\sum_{i=1}^{\infty} |x_i|^p)^\frac{1}{p} \right \rangle$ is separable. Where $\ell_p(\mathbb R)=\left \{ (x_1,x_2,...,x_n,...):\sum_{i=1}^{\infty} |x_i|^p<\infty, p>1, x_i \in \mathbb R \right \}$

To show separability I need to find countable everywhere dense subset. I've already proved (with similar approach) that $\ell_p(\mathbb Q)$ is everywhere dense. However, I can't find a way to show $\ell_p(\mathbb Q)$ is countable.
 A: The sequences with finite support and rational entries are dense in $\ell_p(\mathbb R)$ for each $1\leqslant p\lt \infty$ (we can make the remainder of the series as small as we wish).
A: To elaborate on above answer, let $M$ be the set of all sequences $(r_{0}, \ldots, r_{n}, 0,\ldots)$ with $r_{i}$ being rational and $n \in \mathbb{N}$. (This is the set of sequences with finite support and rational entries).
Since $\mathbb{Q}$ is countable and finite product of countable sets is countable and then countable union of countable sets is countable, we see that $M$ is countable.
Now let us see that $M$ is dense in $\ell_{p}(\mathbb{R})$. To do this, given any $x = (x_{n}) \in \ell_{p}(\mathbb{R})$, and for $\epsilon > 0$ we must find an element $y \in M$ such that $d(x,y) <\epsilon$. We have
$$ \sum\limits_{n=0}^{\infty} |x_{n}|^{p} < \infty $$
Hence, given $\epsilon > 0$, there exists $m \in \mathbb{N}$ such that
$$ \sum\limits_{n=m+1}^{\infty} |x_{n}|^{p} < \epsilon/2 $$
Now for $0\leq i \leq m$, choose $r_{i} \in \mathbb{Q}$ such that $|r_{i} - x_{i}| < \left(\frac{\epsilon}{2m}\right)^{\frac{1}{p}}$ (using that the rationals are dense in reals).
Then the element $y = (r_{0}, r_{1}, \ldots, r_{m}, 0, 0, \ldots) \in M$ is the required element.
A: I think it takes some more work to show that the set of all sequences with rational entries and finite non-zero element is countable (denoted $A$).
Because Countable infinitely cartesian product of countable set may not be countable, otherwise we shall be able to prove that $\ell_p(Q)$ is also countable, however, its cardinality is $\mathbb{N}^{\mathbb{N}}=\mathcal{c}$, which is the cardinality of $\mathbb{R}$ and shall be uncountable.
Also, the above set $A$ can not be expressed as a finite product of a countable set. Consider any finite product $\underset{\text{M of Q here}}{\mathbb{Q} \times \mathbb{Q} \times \dots \mathbb{Q}}$, then a rational sequence with (M+1) non-zero entries still has finite support but does not belong to the above product.
Instead, I think we can prove it by constructing a one-to-one mapping from $A$ to $\mathbb{Q}$.
Consider a mapping that encodes the denominator and numerator into decimal spaces separated by 0, e.g. $f((1/3,3/5,\dots,100/101,\dots))=0.10303050\dots10001010\dots$.
Also, in order to take care of the negative entries, instead of using the irreducible fraction $\frac{p}{q}$, we use $\frac{2p}{2q}$ for a negative entry, so e.g. $f((-1/3,3/5,\dots,-100/101,\dots))=0.20603050\dots20002020\dots$.
Since each rational number can be expressed as its irreducible form $\frac{p}{q}$ where $p,q \in \mathbb{N}$ and since each sequence only has finite support, we know the image of $f$ is a subset of $\mathbb{Q}$. Suppose $f(x)=f(y)\Rightarrow 0.p_1^x0q_1^x0p_2^x0q_2^x \dots p_n^x0q_n^x \dots =0.p_1^y0q_1^y0p_2^y0q_2^y \dots p_n^y0q_n^y \dots$, then as there are only finite non-zero digits, we must have $p_i^x=p_i^y$ and $q_i^x = q_i^y$ for all i, hence $x=y$, and we know $f$ is one-to-one.
Since there exists an one-to-one mapping from $A$ to $\mathbb{Q}$, we know $A$ is also countable.
