# 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.

There are two key points to remember about this proof:

- 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.
- 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.

**N.B.**: Thank you to Robin Vogel for pointing out an error in an earlier version of this
proof.