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 $\delta$.


By metric space axioms $(\text M 1)$ and $(\text M 4)$, 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