Question

Prove that any continuous mapping of a compact metric space into a metric space is uniformly continuous. ​

Answer 1

Answer:

The answer is yes, if f is continuous on a compact space then it is uniformly continuous:

Let f:X→Y be continuous, let ε>0 and let X be a compact metric space. Because f is continuous, for every x in X you can find a δx such that f(B(δx,x))⊂B(ε2,f(x)). The balls {B(δx,x)}x∈X form an open cover of X. So do the balls {B(δx2,x)}x∈X. Since X is compact you can find a finite subcover {B(δxi2,xi)}ni=1. (You will see in a second why we are choosing the radii to be half only.)

Now let δ′xi=δxi2.

You want to choose a distance δ such that for any two x,y they lie in the same B(δ′xi,xi) if their distance is less than δ.

How do you do that?

Note that now that you have finitely many δ′xi you can take the minimum over all of them: miniδ′xi. Consider two points x and y. Surely x lies in one of the B(δ′xi,xi) since they cover the whole space and hence x also lies in B(δ′xi,xi) for some i.

Now we want y to also lie in B(δ′xi,xi). And this is where it comes in handy that we chose a subcover with radii divided by two:

If you pick δ:=miniδ′xi (i.e. δ=δxi2 for some i) then y will also lie in B(δxi,xi):

d(xi,y)≤d(xi,x)+d(x,y)<δxi2+minkδxk≤δxi2+δxi2=δxi