# Is the metric space of all the functions defined and bounded on a given set complete?

Let $S$ be a given non-empty set, and let $B(S)$ denote the metric space of all the bounded (real- or complex-valued) functions with set $S$ as domain, with the metric defined as follows: $$d(x,y) \colon= \sup_{s \in S} \left\vert x(s) - y(s) \right\vert \ \mbox{ for all } \ x, y \in B(S).$$ Is $B(S)$ a complete metric space?

If not, then can we impose any conditions on $S$ under which $B(S)$ does become complete?

My effort:

Let $(x_n)$ be a Cauchy sequence in $B(S)$. Then, given $\epsilon > 0$, we can find a natural number $N$ such that $$d(x_m, x_n) < \epsilon \ \mbox{ for all natural numbers } \ m \ \mbox{ and } \ n \ \mbox{ such that } \ m > N \ \mbox{ and } \ n>N.$$ Let $s_0 \in S$ be arbitrary. Then we must have $$\left\vert x_m(s_0) - x_n(s_0) \right\vert < \epsilon \ \mbox{ for all natural numbers } \ m \ \mbox{ and } \ n \ \mbox{ such that } \ m > N \ \mbox{ and } \ n>N.$$ Thus it follows that the sequence $\left(x_n(s_0)\right)$ is a Cauchy sequence (in the complete metric space $\mathbb{R}$ or $\mathbb{C}$ ) and hence convergent. Let $$x(s_0) \colon= \lim_{n\to \infty} x_n(s_0).$$ In this way, we define a (real- or complex-valued) function $x$ on the set $S$.

We need to show that this function $x$ is bounded, and furthermore that
$$\lim_{n\to\infty} d(x_n,x) = 0.$$

How to accomplish these two tasks?

Please also note that the non-empty set $S$ too is completely arbitrary.

• Your analysis is correct and your two tasks are the right ones. Go for it! – B. S. Thomson Oct 28 '15 at 19:22

$(x_n)$ is $d$-Cauchy, so $d$-bounded. This means that for some $L > 0$ we have that for all $n$: $d(x_n, x_1) = \sup_{s \in S} |x_n(s) - x_1(s)| < L$. Then for all $n$,$s$: $$|x_n(s)| = |x_1(s) + (x_n(s) - x_1(s)| \le |x_1(s)| + |x_n(s) - x_1(s)| \le \sup_{s \in S} |x_1(s)| + d(x_n, x_1) \le \sup_{s \in S} |x_1(s)| + L$$ which is finite. Taking the limit over $n$ implies that $|x(s)|$ has the same finite upper bound.

Now fix $\epsilon > 0$. There exists some $N$ such that $d(x_n, x_m) < {\epsilon \over 2}$ for all $n,m \ge N$. (Using Cauchyness). As $d(x_n, x_N) \le \frac{\epsilon}{2}$ for all $n \ge N$, passing the limit of $n$ to infinity we get that $d(x, x_N) \le {\epsilon \over 2}$ as well.

Then $d(x,x_n) \le d(x,x_N) + d(x_N, x_n) \le \epsilon$, again for all $n \ge N$. As $\epsilon$ was arbitrary, $d(x, x_n)$ converges to $0$ as required.

Note that your metric actually comes from a norm $\lVert x\rVert = \sup_S\lVert x(s)\rVert$, so we can regard $B(S)$ as a normed vector space. If we see $S$ as a space with the discrete topology, $B(S)$ is the normed space of all bounded continuous maps $S\to\mathbb K$.

Now to the proof: Since $(x_n)_n$ is Cauchy $$\textstyle \text{ for each \varepsilon>0 there is an N such that \sup_S\lVert x_m(s)-x_N(s)\rVert<\varepsilon for any m>N }\qquad (1)$$

If we put $\varepsilon=1$ and $N$ as in $(1)$, and we let $M$ be an upper bound for $x_N$, then $$\lVert x_m(s)\rVert = \lVert x_m(s)-x_N(s)+x_N(s)\rVert \le \lVert x_m(s)-x_N(s)\rVert + \lVert x_N(s)\rVert \le 1+M$$ for all $m>N$ and $s\in S$. Hence $x$ is bounded.
Now given an $\varepsilon>0$ there exists an $N$ as in $(1)$. For any $s\in S$, there is a $k>N$ such that $\lVert x(s)-x_k(s)\rVert<\varepsilon$. Then it follows that $\lVert x(s)-x_N(s)\rVert<2\varepsilon$. Thus $(x_n)_n$ converges uniformly to $x$ in the metric space of all bounded maps $S\to\Bbb K$.

If you replace $S$ by a topological space $(S,\tau)$ and assume that all $x_n$ are continuous, then $x$ must be continuous as well, since it is the uniform limit of a sequence of continuous maps. That means that the space $C_b((S,\tau))$ of bounded continous maps $(S,\tau)\to\Bbb K$ is a complete subspace of $B(S)$, thus a Banach space.