Английская Википедия:Heine–Cantor theorem
Шаблон:DistinguishШаблон:No footnotes In mathematics, the Heine–Cantor theorem, named after Eduard Heine and Georg Cantor, states that if <math>f \colon M \to N</math> is a continuous function between two metric spaces <math>M</math> and <math>N</math>, and <math>M</math> is compact, then <math>f</math> is uniformly continuous. An important special case is that every continuous function from a closed bounded interval to the real numbers is uniformly continuous.
Proof
Suppose that <math>M</math> and <math>N</math> are two metric spaces with metrics <math>d_M</math> and <math>d_N</math>, respectively. Suppose further that a function <math>f: M \to N</math> is continuous and <math> M </math> is compact. We want to show that <math>f</math> is uniformly continuous, that is, for every positive real number <math>\varepsilon > 0</math> there exists a positive real number <math>\delta > 0</math> such that for all points <math>x, y</math> in the function domain <math>M</math>, <math>d_M(x,y) < \delta</math> implies that <math>d_N(f(x), f(y)) < \varepsilon</math>.
Consider some positive real number <math>\varepsilon > 0</math>. By continuity, for any point <math>x</math> in the domain <math>M</math>, there exists some positive real number <math>\delta_x > 0</math> such that <math>d_N(f(x),f(y)) < \varepsilon/2</math> when <math>d_M(x,y) < \delta _x</math>, i.e., a fact that <math>y</math> is within <math>\delta_x</math> of <math>x</math> implies that <math>f(y)</math> is within <math>\varepsilon / 2</math> of <math>f(x)</math>.
Let <math>U_x</math> be the open <math>\delta_x/2</math>-neighborhood of <math>x</math>, i.e. the set
- <math>U_x = \left\{ y \mid d_M(x,y) < \frac 1 2 \delta_x \right\}.</math>
Since each point <math>x</math> is contained in its own <math>U_x</math>, we find that the collection <math>\{U_x \mid x \in M\}</math> is an open cover of <math>M</math>. Since <math>M</math> is compact, this cover has a finite subcover <math>\{U_{x_1}, U_{x_2}, \ldots, U_{x_n}\}</math> where <math>x_1, x_2, \ldots, x_n \in M</math>. Each of these open sets has an associated radius <math>\delta_{x_i}/2</math>. Let us now define <math>\delta = \min_{1 \leq i \leq n} \delta_{x_i}/2</math>, i.e. the minimum radius of these open sets. Since we have a finite number of positive radii, this minimum <math>\delta</math> is well-defined and positive. We now show that this <math>\delta</math> works for the definition of uniform continuity.
Suppose that <math>d_M(x,y) < \delta</math> for any two <math>x, y</math> in <math>M</math>. Since the sets <math>U_{x_i}</math> form an open (sub)cover of our space <math>M</math>, we know that <math>x</math> must lie within one of them, say <math>U_{x_i}</math>. Then we have that <math>d_M(x, x_i) < \frac{1}{2}\delta_{x_i}</math>. The triangle inequality then implies that
- <math>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},</math>
implying that <math>x</math> and <math>y</math> are both at most <math>\delta_{x_i}</math> away from <math>x_i</math>. By definition of <math>\delta_{x_i}</math>, this implies that <math>d_N(f(x_i),f(x))</math> and <math>d_N(f(x_i), f(y))</math> are both less than <math>\varepsilon/2</math>. Applying the triangle inequality then yields the desired
- <math>d_N(f(x), f(y)) \leq d_N(f(x_i), f(x)) + d_N(f(x_i), f(y)) < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.</math>
For an alternative proof in the case of <math>M = [a, b]</math>, a closed interval, see the article Non-standard calculus.
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