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Suppose that and are two metric spaces with metrics and , respectively. Suppose further that is continuous, and that is compact. We want to show that is uniformly continuous, that is, for every there exists such that for all points in the domain, implies that .
Fix some . By continuity, for any point in the domain , there exists some such that when is within of .
Since each point is contained in its own , we find that the collection is an open cover of . Since is compact, this cover has a finite subcover where . Each of these open sets has an associated radius . Let us now define , i.e. the minimum radius of these open sets. Since we have a finite number of positive radii, this minimum is well-defined and positive. We now show that this works for the definition of uniform continuity.
Suppose that for any two in . Since the sets form an open (sub)cover of our space , we know that must lie within one of them, say . Then we have that . The triangle inequality then implies that
implying that and are both at most away from . By definition of , this implies that and are both less than . Applying the triangle inequality then yields the desired
For an alternative proof in the case of , a closed interval, see the article Non-standard calculus.