TITLE: A metric that makes $l^\infty$ separable
QUESTION [7 upvotes]: I know that "The metric space $l^\infty$ is not separable with the metric defined between two sequences $\{a_1,a_2,a_3\dots\}$ and $\{b_1,b_2,b_3,\dots\}$ as $\sup\limits_{i\in\Bbb{N}}|{a_i-b_i}|$.
Now I want to know: is there any other metric on $l^\infty$ that makes this space separable, or, more general, is there a metric that makes any space separable?
Thanks in advance!

REPLY [4 votes]: This answer concerns the more general problem.
Claim: A non-empty set $X$ can be endowed with a separable and metrizable topology if and only if $\# X\leq\mathfrak \#\mathbb R$.
Proof: My comment above explains necessity. As for sufficiency, suppose that $\# X\leq\mathfrak \#\mathbb R$ and let $f:X\to \mathbb R$ be an injective function. Let $Y\equiv f(X)$. Consider the topology $\tau$ on $X$ generated by sets of the form $\{f^{-1}(U)\,|\,U\subseteq Y,\text{ $U$ open in the relative topology}\}$. Then, $f$ becomes a homeomorphism between $X$ and $Y$, where $Y$ is endowed with the relative topology induced by the standard Euclidean topology on $\mathbb R$. It is clear that this topology on $Y$ is $T_1$, regular, and second countable and thus so is the topology $\tau$ on $X$. Now invoke Urysohn's metrization theorem to conclude that $(X,\tau)$ is a separable and metrizable topological space. $\blacksquare$

In particular, $\#\ell^{\infty}=\#\mathbb R$, so...