Why $C(0,1)$ is not complete?

Question

I understand that $C([0,1])$ of continuous complex-valued functions on $[0,1]$ is complete wrt sup norm. But why the space $C((0,1))$ of continuous complex-valued functions on $(0,1)$ is not complete?

Answer 1

Notice that not all continuous functions are bounded, so the supremum norm cannot be defined on $C(0,1)$.

If we restrict ourselves to the set of bounded continuous functions on $f:(0,1)\rightarrow M$, where $M$ is a complete metric space then it is true.

Notice that the space of bounded functions with some fixed domain is complete for any domain (see here, the proof only uses that $\mathbb R$ is complete, so it works for $M$ also).

Suppose that $f_1,f_2,\dots$ are continuous functions on $(0,1)$ that converge uniformly to a bounded function $f$.

pick an $x_0\in (0,1)$ and $\epsilon>0$. By hypothesis there is an $N$ such that $|f_N(x)-f(x)|<\epsilon/3$ for all $x\in(0,1)$. There is also a neighbourhood $U$ around $x$ such that $|f_N(x_0)-f_n(x)|<\epsilon/3$ for all $x\in U$.

It follows by the triangle inequality that $|f(x_0)-f(x)|<|f(x_0)-f_N(x_0)|+|f_N(x_0)-f_N(x)|+|f_N(x)-f(x)|<\epsilon$ for all $x\in U$, so $f$ is continuous at $x$, and hence in all $(0,1)$.

So the space of bounded continuous function with domain $(0,1)$ is complete (which is cool since it is the biggest set to which we can assign the supremum norm)