Heine-Cantor Theorem
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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 1
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:
| \(\ds \map {d_1} {y, x_k}\) | \(\le\) | \(\ds \map {d_1} {y, x} + \map {d_1} {x, x_k}\) | Definition of 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:
| \(\ds \map {d_2} {\map f x, \map f y}\) | \(\le\) | \(\ds \map {d_2} {\map f x, \map f {x_k} } + \map {d_2} {\map f {x_k}, \map f y}\) | Definition of 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.
$\blacksquare$
Proof 2
Let $A_1 \times A_1$ and $A_2 \times A_2$ be considered with the product topology.
Let $F: A_1 \times A_1 \to A_2 \times A_2$ be the mapping defined as:
- $\map F {x, y} = \tuple {\map f x, \map f y}$
By Projection from Product Topology is Continuous, we have that the (first and second) projections on $A_1 \times A_1$ are continuous.
By Composite of Continuous Mappings is Continuous and Continuous Mapping to Product Space, it follows that $F$ is continuous.
By Distance Function of Metric Space is Continuous and Composite of Continuous Mappings is Continuous, it follows that $d_2 \circ F: A_1 \times A_1 \to \R$ is continuous.
Let $\epsilon \in \R_{>0}$ be a strictly positive real number.
By Continuity Defined from Closed Sets, the set:
- $C = \paren {d_2 \circ F}^{-1} \sqbrk {\hointr \epsilon \to} = \set {\tuple {x, y} \in A_1 \times A_1: \map {d_2} {\map f x, \map f y} \ge \epsilon}$
is closed in $A_1 \times A_1$.
By Topological Product of Compact Spaces, $A_1 \times A_1$ is compact.
By Closed Subspace of Compact Space is Compact, $C$ is compact.
By Distance Function of Metric Space is Continuous and Continuous Image of Compact Space is Compact, $d_1 \sqbrk C$ is compact.
Therefore, $d_1 \sqbrk C$ has a smallest element $\delta$.
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By Metric Space Axiom $\text M 1$ and Metric Space Axiom $\text M 4$, we have that $\delta > 0$.
By construction, it follows that:
- $\forall x, y \in A_1: \map {d_1} {x, y} < \delta \implies \map {d_2} {\map f x, \map f y} < \epsilon$
Hence, $f$ is uniformly continuous.
$\blacksquare$
Also see
Source of Name
This entry was named for Heinrich Eduard Heine and Georg Cantor.
Sources
- 1953: Walter Rudin: Principles of Mathematical Analysis ... (previous): $4.19$