Complete metric for the topology of compact convergence

Question

This question has been posed in multiple forms on here, but there is always one bit of information lacking. Suppose one considers the space $C$ of continuous functions $f:\mathopen [ 0,+\infty \mathclose [ \rightarrow Y$ where $Y$ is a complete metric space with metric $\rho \leqslant 1$. Define $d_n(f,g)=\sup \left \{\rho(f(x),g(x))| \ x\in \mathopen [ 0,n\mathclose ] \right \}$ for $f,g \in C$, then set $d\left ( f,g \right )= \sum_{n=1}^{\infty} {2^{-n}}d_n\left ( f,g \right )$. One proves: $d$ is a metric on $C$ that induces the topology of compact convergence. However, why is $d$ complete? Supposedly, this is a trivial matter - it never gets explained - but I don't see why. If $\left ( f_k\right )$ is a Cauchy sequence in $C$ for the metric $d$, apparently, then one should prove that the sequence converges pointwise (that's easy enough) but what next?

Answer 1

First note that $$ \lim_{k\to\infty} d(f_k, f) = 0 \iff \forall n \ \lim_{k\to\infty} d_n(f_k, f) = 0 \tag1 $$ If $(f_k)$ is a Cauchy sequence with respect to $d$, then by (1) their restrictions to $[0, n]$ form a Cauchy sequence with respect to $d_n$. Since $C[0, n]$ is complete (reference), we have $g_n\in C[0, n]$ such that $f_k\to g_n$ uniformly on $[0, n]$.

Next, observe that the functions $g_n$ are such that the restriction of $g_n$ to $[0, m]$ with $m<n$ is exactly $g_m$. (This follows from the uniqueness of pointwise limits.) Thus, we can define $$ g(x) = g_n(x) \quad \text{where $n\in \mathbb{N}$ is any such that $n\ge x$} $$ which is a continuous function because each $g_n$ is continuous. Finally, $d(f_k, g)\to 0$ by (1).