Heine–Cantor theorem

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Not to be confused with Cantor's theorem.

In mathematics, the Heine–Cantor theorem, named after Eduard Heine and Georg Cantor, states that if f : MN is a continuous function between two metric spaces, and M is compact, then f is uniformly continuous. An important special case is that every continuous function from a closed interval to the real numbers is uniformly continuous.

Proof[edit]

Suppose that M and N are two metric spaces with metrics dM and dN, respectively. Suppose further that f: M \to N is continuous, and that M is compact. We want to show that f is uniformly continuous, that is, for every \epsilon > 0 there exists  \delta > 0 such that for all points x,y in the domain M,  d_M(x,y) < \delta implies that  d_N(f(x), f(y)) < \epsilon .

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

Let Ux be the open \delta_x/2-neighborhood of x, i.e. the set

 U_x = \left\{ y \mid d_M(x,y) < \frac{1}{2}\delta_x \right\}

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

 U_{x_1}, U_{x_2}, \ldots, U_{x_n}

for some finite set of points  \{x_1, x_2, \ldots, x_n\} \subset M . Each of these open sets has an associated radius  \delta_{x_i}/2 . Let us now define  \delta = \min_{1 \leq i \leq n} \frac{1}{2}\delta_{x_i} , i.e. the minimum radius of these open sets. Since we have a finite number of positive radii, this number  \delta is well-defined. We may now show that this \delta works for the definition of uniform continuity.

Suppose that  d_M(x,y) < \delta 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) < \frac{1}{2}\delta_{x_i} . The Triangle Inequality then implies that

 d_M(x_i, y) \leq d_M(x_i, x) + d_M(x, y) < \frac{1}{2} \delta_{x_i} + \delta \leq \delta_{x_i}

implying that x and y are both at most \delta_{x_i} away from xi. By definition of \delta_{x_i}, this implies that  d_N(f(x_i),f(x)) and  d_N(f(x_i), f(y)) are both less than \epsilon/2. Applying the Triangle Inequality then yields the desired

 d_N(f(x), f(y)) \leq d_N(f(x_i), f(x)) + d_N(f(x_i), f(y)) < \frac{1}{2} \epsilon + \frac{1}{2}\epsilon = \epsilon


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

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