# Metric Spaces of Continuous Function

Let X be a metric space defined as such: $$X = f : [0,1] \to \Re : f \,\text{ is continuous}$$ $$d(f,g) = sup_{x\in[0,1]} | f(x) - g(x)|$$

I need to show:
a) The neighborhood, $N_r (0)$, is uncountably infinite.
b) Let E = { $f \in X : f(0) = 0$}. Prove or disprove E is bounded
c)Prove X is not complete (I can't comment, but yes I need to prove it is not complete)

I am not looking for someone to solve the problem for me, Just to point me in the right direction. I think this is saying that X is a metric space of function (?). I don't believe I've run into this before and don't know how to approach this. Any help would be greatly appreciated.

-
$X$ is complete. Surely whoever gave you this problem made a typo. –  David Mitra Jan 14 '13 at 1:01
add comment

## 1 Answer

You need to think abstractly so as not to get confused with the fact that the points in the metric space you consider are themselves functions. You can still think metrically about this situation since the metric defines the distance between such functions and the axioms of a metric space are satisfied.

To solve this problem you need to make sure you understand what each and every term means and then translate it to the problem at hand. So, for instance, the neighborhood $N_r(0)$ means (I'm guessing here) the set $\{f\in X \mid d(f,0) < r \}$. So, for the case at hand, $0$ probably means the constantly $0$ function. Further, $d(f,0)$ is, by definition, $\sup_x|f(x)|$. So, $N_r(0)=\{f\in X\mid \sup_x |f(x)|<r\}$.

For every $0<\epsilon <r$ the function $f(x)=\epsilon$ thus satisfies that $f(x)\in N_r(0)$ and there are uncountably many such functions. Similarly, you proceed with the other items (at least those that are correct).

-
add comment