Function Spaces from Compact Sets into Complete Metric Spaces

Let $X$ be a compact space, and let $(Y,d)$ be a complete metric space. We show that the set $F$ of continuous functions from $X$ to $Y$ under the metric $$h(f,g)=\max_{x∈X}~d(f(x),g(x))~~~~~\text{for }f,g∈F$$ induces a complete metric space.

Note that when $f,g : X→Y$ are continuous, then since $d : Y×Y→ℝ$ is continuous, we observe $d∘(f×g) : X→ℝ$ is continuous and due to compactness of $X$ indeed attains a maximum value, so that the function $h : F×F→ℝ$ is well defined. To verify that it is a metric, we see $h(f,g)=0$ iff $d(f(x),g(x))=0$ iff $f(x)=g(x)$ for all $x∈X$, i.e. $f=g$. As well, $$h(f,g)=\max_{x∈X}~d(f(x),g(x))=\max_{x∈X}~d(g(x),f(x))=h(g,f)$$ Finally, let $a,b,c∈F$. Then $$h(a,c) = \max_{x∈X}~d(a(x),c(x)) ≤ \max_{x∈X}~[d(a(x),b(x))+d(b(x),c(x))] ≤$$$$\max_{x∈X}~[d(a(x),b(x))]+\max_{x∈X}~[d(b(x),c(x))] = h(a,b)+h(b,c)$$ so that $h(a,c) ≤ h(a,b)+h(b,c)$ and $(F,h)$ is a metric space.

Now we show completeness. Let $(f_n)$ be a Cauchy sequence in $F$. We show that for each $x∈X$, the sequence $(f_n(x))$ is Cauchy in $Y$. To wit, given $ε > 0$, choose $N$ such that $n,m > N$ implies $h(f_n,f_m) < ε$. Then $d(f_n(x),f_m(x)) ≤ \max_{x∈X}~d(f_n(x),f_m(x)) = h(f_n,f_m) < ε$. Thus since $Y$ is complete write $f_n(x)→f(x)$ for each $x∈X$. It now suffices to show $f : X→Y$ is continuous and $f_n→f$. We shall prove the latter convergence is uniform and the former will follow by the uniform limit theorem.

Let $ε > 0$ be given. Since $(f_n)$ is Cauchy, let $N$ be such that $n,m > N$ imply $h(f_n,f_m) < ε/2$. Then $d(f_n(x),f_m(x)) < ε/2$ for all $x∈X$. Now let $n > N$ be given. Since $f_n(x)→f(x)$ choose $m > N$ such that $d(f_m(x),f(x)) < ε/2$. Then $d(f_n(x),f(x)) ≤ d(f_n(x),f_m(x))+d(f_m(x),f(x)) < ε$ so the convergence is uniform.