# completeness of metric space $C_b(X,Y)$

Let $$(X,d_X)$$ and $$(Y,d_Y)$$ be metric spaces. Given $$\emptyset \neq B\subseteq Y,$$ let $$\mathrm{diam}_Y(B) = \sup_{y_1, y_2\in B} d_Y(y_1, y_2).$$ Let $$C_b(X,Y) = \{f\in Y^X : \text{ f is continuous and } \mathrm{diam}_Y (f(X)) < \infty\}$$ where $$f(X)$$ is the range of $$f.$$

1. Show that $$d_C : C_b(X,Y) \times C_b(X,Y) \to [0,\infty)$$ given by $$d_C(f,g) = \sup_{x\in X}d_Y (f(x), g(x))$$ defines a metric on $$C_b(X,Y).$$
2. Show that if $$(Y,d_Y)$$ is complete, then $$(C_b(X,Y) , d_C)$$ is also complete.

$$1$$ seems fairly straightforward; most of the properties of a metric (e.g. positive definiteness, symmetry, etc.) are fairly easy to verify. Also, the triangle inequality follows from properties of the supremum.

But I'm not sure how to justify why $$d_C(f,g) < \infty$$ for all $$f,g$$. I tried to come up with a contradiction, but it seems that that method isn't working too well.

For $$2,$$ I think the proof is similar to the one that $$C([0,1])$$ is a Banach space. So I need to show that every Cauchy sequence $$(f_n)\subseteq C_b(X,Y)$$ converges to some $$f\in C_b(X,Y)$$. Let $$\epsilon>0.$$ It'll be useful to use the fact that $$(f_n(x))$$ is Cauchy for each $$x.$$ So since $$Y$$ is complete, we can define $$f(x):= \lim_{n} f_n(x)$$ for each $$x$$. By the Cauchyness of $$(f_n),\exists N\in\mathbb{N}$$ so that $$n,m\geq N$$ implies $$d_C(f_n, f_m) < \frac{\epsilon}3$$. Then for $$n\geq N, d_Y(f(x), f_n(x)) = \lim_{m\to \infty}d_Y(f_m(x), f_n(x)) \leq \lim_{m\to\infty} d_C(f_m, f_n) \leq \frac{\epsilon}3 <\frac{\epsilon}2$$.

Hence $$d_C(f, f_n) \leq \frac{\epsilon}2 < \epsilon$$ for $$n\geq N$$ so $$f_n \to f.$$ Also, $$\mathrm{diam}_Y(f(X)) < \infty$$ as $$\sup_{x,y\in X}d_Y(f(x), f(y)) \leq \sup_{x\in X} d_Y(f_N(x), f(x)) + \sup_{x,y \in X}d_Y(f_N(x), f_N(y)) + \sup_{y\in X} d_Y(f_N(y), f(y)) = 2d_C(f_N, f) + \mathrm{diam}_Y(f_N(X)) < \infty$$ and $$f$$ can be shown to be continuous as $$f_n$$ converges uniformly to $$f$$. Indeed if $$(a_n)\subseteq Y, a_n\to x \in Y,$$ since $$f_n\to f,$$ there exists $$N_2\in \mathbb{N}$$ so that $$n\geq N_2$$ implies $$d_C(f_n, f) < \epsilon/3.$$ Also, $$\exists N_3\in\mathbb{N}$$ so that $$n\geq N_3\Rightarrow d_Y(a_n, x) < \epsilon/3.$$ Then for $$n\geq N, d_Y(f(a_n), f(x)) \leq d_Y(f(a_n), f_{N_2}(a_n)) + d_Y(f_{N_2}(a_n), f_{N_2}(a)) + d_Y(f_{N_2}(a), f_n(a)) < 2d_C(f,f_{N_2}) + d_Y(f_{N_2}(a_n), f_{N_2}(a)) < \epsilon.$$ So $$f$$ is continuous.

For 1) fix any point $$x_0 \in X$$.We have $$d_Y(f(x),g(x))$$ $$\leq d_Y(f(x),f(x_0))+d_Y(f(x_0),g(x_0))+d_Y(g(x_0), g(x))$$ $$\leq diam (f(X))+d_Y(f(x_0,g(x_0))+diam (g(X))<\infty$$.