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Let $S$ be a set and let $B(S)$ be the set of all bounded real-valued functions on $S$. Define a metric on $B(S)$ by $$d(f,g) = \sup_{s\in S}|f(s)-g(s)|.$$ The problem I am trying to solve is to show that $B(S)$ is separable if and only if $S$ is finite.

I have seen this question posed in the present forum. (When is a metric space of bounded real valued functions separable?) The assertion that is demonstrated well enough for me is the converse: If...separable, then the domain is finite. I am stuck on the "easy" part, which is passed off simply with "If it's finite, (then) it's trivially separable as you have a finite dense subset. Just pick all the functions", according to Aram. (I have edited Aram's statement a bit.) What is finite by hypothesis is the domain of the functions. Let's take such a domain: $S=\{s_1,s_2,s_3\}$. The number of bounded, real valued functions one can have on such a domain is not countable. To take all of them as the dense set does not make $B(S)$ a separable metric space. Is there more to the argument that is implied and I am missing? In any event, I have not been able to prove the assertion in the title. Could someone please assist?

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goedelite
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  • With your domain $S ={s_1,s_2,s_3},$ a dense subset of bounded functions would be the functions on $S$ with rational values. The set of such functions is in 1-1 correspondence with $\mathbb Q^3,$  which is countable. – zhw. Oct 27 '16 at 03:24

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Yeah, the claim that there's a finite dense subset is nonsense. However, there is a countable dense subset: namely, take the set of all rational-valued functions on $S$. If $S$ has $n$ elements, then the set of such functions is in bijection with the set $\mathbb{Q}^n$ of $n$-tuples of rational numbers, which is countable. To prove that this is dense, given $\epsilon>0$ and $f\in B(S)$, just pick a rational number $q_s$ such that $|f(s)-q_s|<\epsilon$ for each $s\in S$ and define $g(s)=q_s$. Then $g$ is rational-valued, and $d(g,f)<\epsilon$.

Eric Wofsey
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  • The theorem so nicely answered has a converse. If the subject space is separable, then the domain space is finite. It is stated in Gamilin and Greene's "Intro to Topology" along with a hint that the characteristic functions for the domain of the bounded functions of the metric space are disjoint. That follows from an employment of the triangle inequality of the uniform metric. I have not been able to use the hint to prove that the space is finite. May I have another hint? – goedelite Oct 02 '17 at 19:08