# Showing a metric space is complete.

On the space of continuous functions on $[0,1]$, I have a metric

$$d(f,g) = \sup | \alpha(x) (f(x) -g(x))|,$$

where $\alpha(x)$ is a continuous function and $\alpha(x) \ne 0$.

I'm trying to find whether this space is complete. I think it is complete because the continuous functions on $[0,1]$ are complete and the $\alpha$ doesn't create a problem because it is continuos on a closed interval and so bounded. But this seems a bit too simple?

Thanks

Yes. Since $\alpha(x)$ is continuous with $\alpha(x) \neq 0$, you can write any $f(x) \in \mathcal{C}([0,1])$ as $f(x) = p(x)/\alpha(x)$ for some continuous $p(x)$. Then, $$d(f,g) = d(p/\alpha,q/\alpha) = \sup|\alpha(f-g)| = \sup|\alpha(p/\alpha - q/\alpha)| = \sup|p - q|,$$ which is just the sup norm in $\mathcal{C}([0,1])$. It follows that $(\mathcal{C}([0,1]),d)$ is complete.