Heine–Cantor theorem
In mathematics, the Heine–Cantor theorem, named after Eduard Heine and Georg Cantor, states that if [math]\displaystyle{ f \colon M \to N }[/math] is a continuous function between two metric spaces [math]\displaystyle{ M }[/math] and [math]\displaystyle{ N }[/math], and [math]\displaystyle{ M }[/math] is compact, then [math]\displaystyle{ 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]\displaystyle{ M }[/math] and [math]\displaystyle{ N }[/math] are two metric spaces with metrics [math]\displaystyle{ d_M }[/math] and [math]\displaystyle{ d_N }[/math], respectively. Suppose further that a function [math]\displaystyle{ f: M \to N }[/math] is continuous and [math]\displaystyle{ M }[/math] is compact. We want to show that [math]\displaystyle{ f }[/math] is uniformly continuous, that is, for every positive real number [math]\displaystyle{ \varepsilon \gt 0 }[/math] there exists a positive real number [math]\displaystyle{ \delta \gt 0 }[/math] such that for all points [math]\displaystyle{ x, y }[/math] in the function domain [math]\displaystyle{ M }[/math], [math]\displaystyle{ d_M(x,y) \lt \delta }[/math] implies that [math]\displaystyle{ d_N(f(x), f(y)) \lt \varepsilon }[/math].
Consider some positive real number [math]\displaystyle{ \varepsilon \gt 0 }[/math]. By continuity, for any point [math]\displaystyle{ x }[/math] in the domain [math]\displaystyle{ M }[/math], there exists some positive real number [math]\displaystyle{ \delta_x \gt 0 }[/math] such that [math]\displaystyle{ d_N(f(x),f(y)) \lt \varepsilon/2 }[/math] when [math]\displaystyle{ d_M(x,y) \lt \delta _x }[/math], i.e., a fact that [math]\displaystyle{ y }[/math] is within [math]\displaystyle{ \delta_x }[/math] of [math]\displaystyle{ x }[/math] implies that [math]\displaystyle{ f(y) }[/math] is within [math]\displaystyle{ \varepsilon / 2 }[/math] of [math]\displaystyle{ f(x) }[/math].
Let [math]\displaystyle{ U_x }[/math] be the open [math]\displaystyle{ \delta_x/2 }[/math]-neighborhood of [math]\displaystyle{ x }[/math], i.e. the set
- [math]\displaystyle{ U_x = \left\{ y \mid d_M(x,y) \lt \frac 1 2 \delta_x \right\}. }[/math]
Since each point [math]\displaystyle{ x }[/math] is contained in its own [math]\displaystyle{ U_x }[/math], we find that the collection [math]\displaystyle{ \{U_x \mid x \in M\} }[/math] is an open cover of [math]\displaystyle{ M }[/math]. Since [math]\displaystyle{ M }[/math] is compact, this cover has a finite subcover [math]\displaystyle{ \{U_{x_1}, U_{x_2}, \ldots, U_{x_n}\} }[/math] where [math]\displaystyle{ x_1, x_2, \ldots, x_n \in M }[/math]. Each of these open sets has an associated radius [math]\displaystyle{ \delta_{x_i}/2 }[/math]. Let us now define [math]\displaystyle{ \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]\displaystyle{ \delta }[/math] is well-defined and positive. We now show that this [math]\displaystyle{ \delta }[/math] works for the definition of uniform continuity.
Suppose that [math]\displaystyle{ d_M(x,y) \lt \delta }[/math] for any two [math]\displaystyle{ x, y }[/math] in [math]\displaystyle{ M }[/math]. Since the sets [math]\displaystyle{ U_{x_i} }[/math] form an open (sub)cover of our space [math]\displaystyle{ M }[/math], we know that [math]\displaystyle{ x }[/math] must lie within one of them, say [math]\displaystyle{ U_{x_i} }[/math]. Then we have that [math]\displaystyle{ d_M(x, x_i) \lt \frac{1}{2}\delta_{x_i} }[/math]. The triangle inequality then implies that
- [math]\displaystyle{ d_M(x_i, y) \leq d_M(x_i, x) + d_M(x, y) \lt \frac{1}{2} \delta_{x_i} + \delta \leq \delta_{x_i}, }[/math]
implying that [math]\displaystyle{ x }[/math] and [math]\displaystyle{ y }[/math] are both at most [math]\displaystyle{ \delta_{x_i} }[/math] away from [math]\displaystyle{ x_i }[/math]. By definition of [math]\displaystyle{ \delta_{x_i} }[/math], this implies that [math]\displaystyle{ d_N(f(x_i),f(x)) }[/math] and [math]\displaystyle{ d_N(f(x_i), f(y)) }[/math] are both less than [math]\displaystyle{ \varepsilon/2 }[/math]. Applying the triangle inequality then yields the desired
- [math]\displaystyle{ d_N(f(x), f(y)) \leq d_N(f(x_i), f(x)) + d_N(f(x_i), f(y)) \lt \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. }[/math]
For an alternative proof in the case of [math]\displaystyle{ M = [a, b] }[/math], a closed interval, see the article Non-standard calculus.
See also
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