# Completeness of the bounded-real-function-space B(S)

By | August 9, 2010

Let $$S$$ be a nonempty set and $$B(S)$$ a metric space of all the bounded real functions on $$S$$ with the supremum norm $\Vert f \Vert = \sup \left\{ | f(s) | ~|~ s\in S \right\}$ and the metric function $d(f,~g) = \Vert f-g \Vert .$ Then $$B(S)$$ becomes a complete metric space.

To prove this fact, let $$\left\{ f_n \right\}$$ be a Cauchy sequence in $$B(S).$$ We must show that $$\left\{ f_n \right\}$$ converges to a function $$f$$ in $$B(S).$$ This is done in two steps. Firstly, we identify a likely candidate for $$f,$$ and then secondly we show that $$f$$ is indeed the right choice. To identify the candidate take $$\epsilon > 0.$$ Since $$\left\{ f_n \right\}$$ is a Cauchy sequence, we can find $$N$$ such that whenever both $$p$$ and $$q$$ are greater than or equal to $$N,$$ then $$\Vert f_p - f_q \Vert \le \frac{\epsilon}{2} ,$$ or equivalently $$| f_p (s) - f_q (s) | \le \frac{\epsilon}{2}$$ for all $$s \in S.$$ This means that for each fixed value of $$s$$ in $$S,$$ $$\left\{ f_n (s) \right\}$$ is a Cauchy sequence of real numbers and so has a limit which is a real number. We will denote this number by $$f(s).$$ From the triangle inequality, $| \Vert f_p \Vert - \Vert f_q \Vert | \le \Vert f_p - f_q \Vert ,$ and so $$\left\{ \Vert f_n \Vert \right\}$$ is a Cauchy sequence of real numbers. It follows that this same sequence is bounded, and from this we may deduce that $$f$$ lies in $$B(S).$$ Hence $$\left\{ f_n \right\}$$ converges pointwise to $$f.$$ We show that in fact the convergence is uniform. Fix $$p \ge N$$ and $$s \in S.$$ Choose $$q \ge N$$ so that $$| f_p (s) - f_q (s)| \le \frac{\epsilon}{2} .$$ Since $$f_n (s) \to f(s)$$ as $$n \to \infty$$ we may further constrain $$q$$ to be sufficiently large to ensure that $$| f_q (s) - f(s)| < \frac{\epsilon}{2} .$$ Then $| f_p (s) - f(s)| = |f_p (s) - f_q (s)| + | f_q (s) - f(s)| \le \frac{\epsilon}{2} + \frac{\epsilon}{2} = ε.$ This establishes that for all $$s \in S$$ and $$p \ge N,$$ $$| f_p (s) - f(s) | \le \epsilon ,$$ i.e. for all $$p \ge N,$$ $$\Vert f-f_p \Vert \le \epsilon .$$ Thus $$\left\{ f_n \right\}$$ converges to $$f$$ in $$(B(S),~ \Vert \cdot \Vert ).$$