# Thread: Metric space of bounded real functions is separable iff the space is finite. Prove.

1. ## Metric space of bounded real functions is separable iff the space is finite. Prove.

A metric space, B(S), with the "uniform metric" d(f,g) = sup {|f - g| }, of bounded, real valued functions on the set S is separable if and only if the space is finite. This is ex. 7 in the "Introduction to Topology" by Gamelin and Green (Second Edition) Part I. Section 5, which I am reading to keep my retired brain cells active. Regrettably, they are not active enough, because the hint supplied by the authors has not helped me. The hint defines a characteristic function for subsets of S as follows:
Chi-subT (s) = 1 for s belongs to T
Chi-subT (s) = 0 for s belongs to S-T
With this definition the open balls B(Chi-subT; 1/2) are disjoint, the authors state. That is the hint.

I see from the metric that the balls are indeed disjoint. I have not been able to use this property to show that S is finite if there is a countably dense subset of functions in B(S), nor given that the set S is finite that there is a countably dense set in B(S). I surmise that perhaps the zeros of a polynomial with real polynomial of degree N is involved in showing that S is finite; also that the polynomial is obtained from the limit that d(f, h-subj) = 0 where f is any bounded function in B(S) and {h-subj} is a countable set dense in B(S). I have not been able to carry through the argument, sadly.

Can anyone help?

2. ## Re: Metric space of bounded real functions is separable iff the space is finite. Prov

Please ignore this posting. I have re-submitted it with the same title using LaTex for easier reading.