The definition of Cauchy Sequences in topological spaces is the following:

Let $\{x_n\}_{n \in \mathbb{N}}$ denote a Cauchy sequence. For any neighborhood $N$ of $0$, there exists a natural number $N$ such that $m, n \geq N$ gives $x_m – x_n \in N$.

Let $X$ denote a locally compact space. Consider $C(X, \mathbb{R})$ with the compact-open topology. Suppose that $N \subset C(X, \mathbb{R})$ is a compact neighborhood of $0$. Given a Cauchy sequence of continuous functions $\{f_n\}_{n \in \mathbb{N}} \subset C(X, \mathbb{R})$, does it follow that $\{f_n\}_{n \in \mathbb{N}}$ converges uniformly to the function $f(x) = \lim_{n \to \infty} f_n(x)$?

The function $f$ is well-defined due the fact that $\{f_n(x)\}$ is Cauchy for each $x \in X$. However, I do not know how to proceed from here. I suspect that there is a relation between my doubt and the following result (taken from Uniformly Cauchy Sequence article), but I was unable to locate a proof.

Let S be a topological space and M a complete metric space. Then any uniformly Cauchy sequence of continuous functions $f_n: S \to M$ tends uniformly to a unique continuous function $f: S \to M$.

## Best Answer

If $N \subseteq C(X,\Bbb R)$ is a compact neighbourhood of $0$ it follows by standard results that $C(X,\Bbb R)$ is finite-dimensional and $X$ is a finite space and $C(X,\Bbb R) \simeq $\Bbb R^k$ for some $k \in \Bbb N$.

In that case $f_n \to f$ uniformly is trivial.

The final statement of uniformly Cauchy sequences holds for $C(X,\Bbb R)$ in the uniform metric,

notfor the compact-open topology in general.