# Completeness of space of continuous functions with added condition

For a topological space $X$, let $C_0(X)$ denote the set of continuous functions $$f:X\to \mathbb{C}$$ such that for all $\epsilon>0$, the set $\{x\in X\mid |f(x)|\geq \epsilon\}$ is compact.

(a) Show that $C_0(X)$ is complete with respect to the norm $\|f\|_\infty:=\sup_{x\in X}|f(x)|$

For (a) I did the standard thing and took $(f_n)$ to be a Cauchy sequence in $C_0(X)$.

For each $x\in X$, $(f_n(x))_{n\in\mathbb{N}}$ is Cauchy and therefore converges to a point, say f(x) (as $\mathbb{C}$ is complete). My claim is that the limit of $(f_n)$ is the pointwise limit of the $f_n(x)$'s.

Fix $\epsilon>0$.

Since $(f_n)$ is a Cauchy, by definition there exists $N\in\mathbb{N}$ such that for all $m,n>N$, $|f_m(x)-f_n(x)|<\epsilon.$ So taking the limit as $n\to\infty$ we get that $|f_m(x)-f(x)|\leq\epsilon.$ Since this is independent of $x$ we actually have $\sup\limits_{x\in X}|f_m(x)-f(x)|=\|f_m-f\|_\infty\leq \epsilon$. Hence $f_m\to f$.

This is where I'm a little unsure. Can I use the fact that uniform limit of continuous function is continuous and how do I know for all $\epsilon>0$, the set $\{x\in X\mid |f(x)|\geq \epsilon\}$ is compact?

For part $(b)$ and $(c)$ I'm having trouble coming up with the isometries.

Any help is appreciated.

• Maybe the two other questions should be asked in a separated thread. Mar 29, 2016 at 21:03

Your proof for the fact that each Cauchy sequence in $C_0(X)$ is convergent is good. Maybe you could write "by definition there exists $N\in\mathbb{N}$ such that for all $m,n>N$ and each $x\in X$, $|f_m(x)-f_n(x)|<\epsilon.$"
Now you have to check that $f$ is an element of $C_0(X)$ and indeed, you have in particular to prove that $f$ is continuous (I assume you know how to do this).
In order to prove that $\{x\in X\mid \left|f(x)\right|\geqslant \varepsilon\}$ is compact for any $\varepsilon$, take $n$ such that $\lVert f-f_n\rVert_{\infty}\lt\varepsilon/2$ and show that $$\left\{x\in X\mid \left|f(x)\right|\geqslant \varepsilon\right\}\subset \left\{x\in X\mid \left|f_n(x)\right|\geqslant \varepsilon/2\right\}.$$ The right hand side is compact and the left hand side is closed.