# Uniform Convergence of Cauchy Sequence of Continuous Functions

analysiscomplete-spacesfunctional-analysisgeneral-topologytopological-vector-spaces

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, not for the compact-open topology in general.