Heine–Cantor theorem

Proof

Suppose that M and N are two metric spaces with metrics d M and d N , respectively. Suppose further that f : M → N is continuous, and that M is compact. We want to show that f is uniformly continuous, that is, for every ε > 0 there exists δ > 0 such that for all points x , y in the domain M , d M ( x , y ) < δ implies that d N ( f ( x ) , f ( y ) ) < ε .

Fix some positive ε > 0 . Then by continuity, for any point x in our domain M , there exists a positive real number δ x > 0 such that d N ( f ( x ) , f ( y ) ) < ε / 2 when y is within δ x of x .

Let U x be the open δ x / 2 -neighborhood of x , i.e. the set

Since each point x is contained in its own U x , we find that the collection { U x ∣ x ∈ M } is an open cover of M . Since M is compact, this cover has a finite subcover. That subcover must be of the form

for some finite set of points { x 1 , x 2 , … , x n } ⊂ M . Each of these open sets has an associated radius δ x i / 2 . Let us now define δ = min 1 ≤ i ≤ n 1 2 δ x i , i.e. the minimum radius of these open sets. Since we have a finite number of positive radii, this number δ is well-defined and positive. We may now show that this δ works for the definition of uniform continuity.

Suppose that d M ( x , y ) < δ for any two x , y in M . Since the sets U x i form an open (sub)cover of our space M , we know that x must lie within one of them, say U x i . Then we have that d M ( x , x i ) < 1 2 δ x i . The Triangle Inequality then implies that

implying that x and y are both at most δ x i away from x i . By definition of δ x i , this implies that d N ( f ( x i ) , f ( x ) ) and d N ( f ( x i ) , f ( y ) ) are both less than ε / 2 . Applying the Triangle Inequality then yields the desired

For an alternative proof in the case of M = [ a , b ] a closed interval, see the article on non-standard calculus.