# Thread: Compactness and metric space

1. ## Compactness and metric space

Let $\displaystyle (X,d)$ be a compact metric space, which metric is $\displaystyle d$.

Let be p and q two points of $\displaystyle X$.

Can I say that, since $\displaystyle X$ is compact, the distance between p and q is finite and reachable?

Thank you.

2. The distance map $\displaystyle d: X\times X \rightarrow \mathbb{R}^{nonneg}$ is a continuous map from a compact space (compact because X is, and continuity requires some proof if you haven't already proved this).

But the continuous image of a compact set is compact which means closed and bounded in $\displaystyle \mathbb{R}$ so the distance between the two points is definitely finite. I am not sure what you mean by reachable.

3. By reachable I mean that your map d from R*R reaches is maximum and its minimum where the function is defined.

For exemple the function sin(x) defined on [0,pi] has a maximum and a minimum and it reaches its maximum at Pi/2

4. Yeah you are just talking about the extreme value theorem then I think.

Basically here is the deal. The image of the space is compact in $\displaystyle \mathbb{R}$ which is complete. So we know the image is closed an bounded. In particular the inf and sup of this set would be a limit point, but it is closed, so it contains all of its limit points. Thus is achieves both its max and min? Sound good?