Heine-Cantor Theorem/Proof 1
Theorem
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $M_1$ be compact.
Let $f: A_1 \to A_2$ be a continuous mapping.
Then $f$ is uniformly continuous.
Proof
Let $\epsilon \in \R_{>0}$ be a strictly positive real number.
For all $x \in A_1$, define:
- $\map \Delta x = \set {\delta \in \R_{>0}: \forall y \in A_1: \map {d_1} {x, y} < 2 \delta \implies \map {d_2} {\map f x, \map f y} < \dfrac \epsilon 2}$
Define:
- $\CC = \set {\map {B_{\delta} } x: x \in A_1, \, \delta \in \map \Delta x}$
where $\map {B_{\delta} } x$ denotes the open $\delta$-ball of $x$ in $M_1$.
From the definition of continuity, it follows that $\CC$ is a cover for $A_1$.
From Open Ball of Metric Space is Open Set, it therefore follows that $\CC$ is an open cover for $A_1$.
By the definition of a compact metric space, there exists a finite subcover $\set {\map {B_{\delta_1} } {x_1}, \map {B_{\delta_2} } {x_2}, \ldots, \map {B_{\delta_n} } {x_n} }$ of $\CC$ for $A_1$.
Define:
- $\delta = \min \set {\delta_1, \delta_2, \ldots, \delta_n}$
Let $x, y \in A_1$ satisfy $\map {d_1} {x, y} < \delta$.
By the definition of a cover, there exists a $k \in \set{1, 2, \ldots, n}$ such that $\map {d_1} {x, x_k} < \delta_k$.
Then:
\(\ds \map {d_1} {y, x_k}\) | \(\le\) | \(\ds \map {d_1} {y, x} + \map {d_1} {x, x_k}\) | Metric Space Axiom $(\text M 2)$: Triangle Inequality | |||||||||||
\(\ds \) | \(<\) | \(\ds \delta + \delta_k\) | Metric Space Axiom $(\text M 3)$ | |||||||||||
\(\ds \) | \(\le\) | \(\ds 2 \delta_k\) |
By the definition of $\map \Delta {x_k}$, it follows that:
- $\map {d_2} {\map f x, \map f {x_k} } < \dfrac \epsilon 2$
- $\map {d_2} {\map f y, \map f {x_k} } < \dfrac \epsilon 2$
Hence:
\(\ds \map {d_2} {\map f x, \map f y}\) | \(\le\) | \(\ds \map {d_2} {\map f x, \map f {x_k} } + \map {d_2} {\map f {x_k}, \map f y}\) | Metric Space Axiom $(\text M 2)$: Triangle Inequality | |||||||||||
\(\ds \) | \(<\) | \(\ds \frac \epsilon 2 + \frac \epsilon 2\) | Metric Space Axiom $(\text M 3)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \epsilon\) |
The result follows from the definition of uniform continuity.
$\blacksquare$
Source of Name
This entry was named for Heinrich Eduard Heine and Georg Cantor.
Sources
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $5$: Compact spaces: $5.8$: Compactness and Uniform Continuity: Proposition $5.8.2$