# Completeness of the space of continuous functions

Theorem

If $$(E, d)$$ is a bounded metric space and $$(E’, d’)$$ is a complete metric space, then the metric space of continuous functions $$(C(E), d_{\infty})$$ is complete.

Proof

For $$C(E)$$ to be complete, every Cauchy sequence in the space must converge in the space. Consider $$f_n: E \rightarrow E’$$, where the sequence $$f_n \in C(E)$$ is such that, given $$\varepsilon > 0$$, there exists some $$N > 0$$ so that

for all $$n,m > N$$ (That is, consider $$f_n$$ Cauchy in $$E’$$.) Now, for any arbitrary $$x_0 \in E$$, note that $$d’(f_n(x_0), f_m(x_0)) < \varepsilon$$ for all $$n,m > N$$ implies that $$f_n(x_0)$$ converges to some point in $$E’$$, call it $$f(x_0)$$, because $$E’$$ is complete. Doing this for each $$x_0 \in E$$ allows for the definition of $$f:E\rightarrow E’$$ as $$f(x) = \lim f_n(x)$$.

We want to show that this limit function $$f$$ is bounded. We can do this with the triangle inequality:

For any $$\varepsilon >0$$ there exists some $$N$$ such that, for all $$n > N$$, $$d_{\infty}(f, f_n) < \varepsilon$$. We also know that there exists $$0 \leq M < \infty$$ such that $$d_{\infty}(f_n,0) < M$$ since $$f_n \in C(E)$$. So $$f$$ is bounded.

Finally, we need to show that $$f_n \rightarrow f$$ in norm, not just pointwise. For this we can just use the triangle inequality again. Since $$f_n$$ is Cauchy, for any $$\varepsilon > 0$$ there exists some $$N’$$ such that, for all $$n,m > N’$$, $$d_{\infty}(f_m, f_n) < \varepsilon$$. Then, by the triangle inequality,

by the same argument as above. Apply the uniform limit theorem to find that $$f \in C(E)$$. QED.

1. We don’t really care what $$E’$$ is, we just care that it’s complete. Most proofs of this choose $$E’ = \mathbb{R}$$, but that isn’t necessary. The only time we even think about $$E’$$ is when we prove that a Cauchy sequence in $$E’$$ actually converges there.
2. We can use the uniform limit theorem because we have indirectly proved that $$f_n \rightarrow f$$ uniformly. How? By using the sup metric. After all, if $$d_{\infty}(f,g) < \varepsilon$$, it means that, at the point in $$E$$ that maximally separates $$f$$ and $$f_n$$ in $$E’$$, they’re only $$\varepsilon$$ apart–so for any other $$x’ \neq x$$, they’re closer.