Proof that $c_0$ is separable (with respect to the $l_\infty$ norm)

Question

I'm looking at the proof that $c_0$ is separable, but I don't understand the proof.In the proof below, it first shows that $S$ is separable, where $S$ is:

Next, it shows that $S$ is dense in $c_0$. However, $S$ is not a countable set, so how does this show that $c_0$ has a countable dense subset? I would greatly appreciate any explanation.

$\mathcal{S}$ is not countable, but the set of sequences in $\mathcal{S}$ with rational entries is. That should be good enough. — carmichael561, Feb 25 '16 at 21:52
Short version: If $A$ is separable, so is $\overline{A}$ (a countable dense subset of $A$ is dense in $\overline{A}$). A countable union of separable sets is separable. Finite-dimensional spaces are separable. — Daniel Fischer, Feb 25 '16 at 21:54
It's good enough because any element of $c_0$ is eventually small, and the finitely many entries at the beginning can be approximated by rational numbers. — carmichael561, Feb 25 '16 at 21:56
So you mean the set of rational terms of $S$ is also dense in $c_0$? — nomadicmathematician, Feb 25 '16 at 21:56
Assuming that $\ell^{\infty}$ consists of real sequences, yes. If it is complex, then choose a countable dense subset of $\mathbb{C}$ and use the same argument. — carmichael561, Feb 25 '16 at 21:59

score 9 · Answer 1 · answered Nov 12 '18 at 02:27

Define $S_k = \{(x_1, x_2, ..., x_k, 0, 0, 0,...): x_i\in \mathcal{Q}, \forall 1\leq i\leq k \}$, and $S= \bigcup_{k\geq 1} S_k$. For each $k$, $S_k$ is countable and so is their countable union $S$. Now you want to show that $S$ is dense in $c_0$.

Take any element $y\in c_0$, $y=(y_1, y_2, y_3, ...)$, $\forall \epsilon >0, \exists N$, s.t. $|y_n|<\epsilon$ for $n\geq N$. Since we know $\mathcal{Q}$ is dense in $\mathcal{R}$, we can take $x=(x_1, ..., x_N, 0, 0, ...)\in S$ with $|x_i - y_i|<\epsilon, \forall 1\leq i\leq N$. In this way, we can verify that $||x-y||_{\infty} <\epsilon$. Therefore $S$ is dense in $c_0$, and $c_0$ admits a countable dense subset, $c_0$ is separable by definition.

Proof that $c_0$ is separable (with respect to the $l_\infty$ norm)

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