Is the space of n-real continous functions defined on open real set complete with uniform distance?

Question

I'm studying the proof of Picard-Lindelöf theorem (Banach fixed-point version), but I come across a doubt. In my notes there is:

Let

$$X:=C^0((t_0-\delta, t_0+\delta), \overline B_{R_0}(x_0)), \space \textrm{ where } \space \overline B_{R_0}(x_0) \subset \mathbb{R}^n,$$

then $(X,d_\infty)$ is complete since is a closed set in $(C^0((t_0-\delta, t_0+\delta), \mathbb{R}^n), d_\infty)$ which is complete.

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Questions:

1. Does $(C^0((t_0-\delta, t_0+\delta), \mathbb{R}^n), d_\infty) $ is complete althought this set contains functions defined on open set?

2. In the case that answer of 1 is "no", is true that $(X,d_\infty)$ is complete? (I think the answer is yes for the fact that $C_b(X,Y)$ is complete when $(Y,d_Y)$ is complete).

Answer 1

The domain does not matter. For any topological space $X$ and any complete metric space $Y$, the set $C_b(X,Y)$ of bounded continuous functions from $X$ to $Y$ is complete. In your example, $X$ is an open interval and $Y$ is a closed ball.

The proof uses standard ingredients from abstract metric spaces:

1. To prove a Cauchy sequence converges, it suffices to prove it has a convergent subsequence.
2. A Cauchy sequence $\{x_n\}$ has a subsequence $\{x_{n_k}\}$ such that $d(x_{n_k}, x_{n_{k+1}}) < 2^{-k}$ for all $k$.

In view of 1 and 2, it suffices to consider a sequence $\{f_n\}$ in $C_b(X,Y)$ such that $d_\infty(f_n, f_{n+1})<2^{-n}$ for all $n$. For each $x\in X$ the sequence $\{f_n(x)\}$ is Cauchy in $Y$, so it has a limit which we call $f(x)$. By the triangle inequality, $$ d_Y(f_n(x),f(x)) \le \sum_{k=n}^\infty d_Y(f_k(x), f_{k+1}(x)) < \sum_{k=n}^\infty 2^{-k} = 2^{1-n}, $$ which shows the convergence $f_n\to f$ is uniform. The boundedness of $f$ is immediate from $d_\infty(f_n,f)\le 2^{1-n}$. The continuity follows by $\epsilon/3$-argument.