# A metric that makes $l^\infty$ separable

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?

• Separability is a topological concept. This means that if you're looking for a metric on $\ell^{\infty}$ other than the one induced by the supremum norm that still generates the same topology, then you won't find any. – triple_sec Dec 1 '14 at 4:34
• @triple_sec, you are right about separability and the same topology. But I'm not looking for a metric that generates the same topology, in fact I'm looking for a metric that gives me a different topology, a separable one. – Woria Dec 2 '14 at 2:56

with something like $$d(x,y)=\sum_n2^{-n}|x(n)-y(n)|,$$ the collection of bounded sequences is separable, for instance the rational span of the "standard basis" $e_k(n)=\delta_{kn}$ is dense.

as noted in the comments, the topology generated by this metric is that of pointwise convergence (i.e. bounded sequences as a subspace of $\mathbb{R}^{\mathbb{N}}$ with the product topology).

• I'm not sure that how we can find a sequence in "standard basis" that converges to the constant sequence $x=(x_n=2)_{n=1}^\infty$. Although it seems that $M=\{(\alpha_{n}):\alpha_{n}$'s are rational and only finitely many number of $\alpha_{n}$'s are nonzero$\}$ is a countable dense set. – Woria Dec 2 '14 at 2:52
• @Woria: Fix $k$ and consider the sequence $y$ with $y(n) = 2$ for $n \le k$ and $y(n)=0$ for $n > k$. Compute $d(x,y)$. Observe that it can be made as small as desired by taking $k$ sufficiently large. – Nate Eldredge Dec 2 '14 at 3:41
• Great question and great answer. Are there any other metrics that make $\ell^\infty$ separable or can there only be one? – user167889 Dec 2 '14 at 3:43
• @student: The cardinality of $\ell^\infty$ is $\mathfrak{c}$, so for any metric space $(X,\rho)$ of that cardinality, you can put a metric $d$ on $\ell^\infty$ that makes it isometric (in particular, homeomorphic) to $(X,\rho)$. For instance, there is a metric on $\ell^\infty$ that makes it isometric to $[0,1]$. It may take a bit of work to write it down. The question can get more interesting if there is other structure you want the metric to respect (e.g. the vector space structure). – Nate Eldredge Dec 2 '14 at 3:46
• Can you please explain what "the rational span of the "standard basis" $e_k(n)=\delta_{kn}$" is? – Viktor Glombik Apr 30 '19 at 19:14

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...

• Without using a deeper metrization theorem we can easily show that any subspace $Y$ of a metrizable space $Z$ is metrizable. I.e. if $d$ is a metric for $Z$ then the sub-space topology on $Y$ co-incides with the topology generated by $d|_{Y\times Y}.$ – DanielWainfleet Oct 30 '19 at 18:30