Heine–Cantor theorem

In mathematics, the Heine–Cantor theorem, named after Eduard Heine and Georg Cantor, states that if f : M → N 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 bounded closed interval to the real numbers is uniformly continuous.

Proof

Suppose that M and N are two metric spaces with metrics d_M and d_N, 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 U_x 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 U_x, 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 and positive. 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 x_i. 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.

External links

• Heine–Cantor theorem at PlanetMath.org.
• Proof of Heine–Cantor theorem at PlanetMath.org.