# Heine-Cantor Theorem

## Theorem

Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ 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:

- $\Delta \left({x}\right) = \left\{{\delta \in \R_{>0}: \forall y \in A_1: d_1 \left({x, y}\right) < 2 \delta \implies d_2 \left({f \left({x}\right), f \left({y}\right)}\right) < \dfrac \epsilon 2}\right\}$

Define:

- $\mathcal C = \left\{{B_{\delta} \left({x}\right): x \in A_1, \, \delta \in \Delta \left({x}\right)}\right\}$

where $B_{\delta} \left({x}\right)$ denotes the open $\delta$-ball of $x$ in $M_1$.

From the definition of continuity, it follows that $\mathcal C$ is a cover for $A_1$.

From Open Ball is Open Set, it therefore follows that $\mathcal C$ is an open cover for $A_1$.

By the definition of a compact metric space, there exists a finite subcover $\left\{{B_{\delta_1} \left({x_1}\right), B_{\delta_2} \left({x_2}\right), \ldots, B_{\delta_n} \left({x_n}\right)}\right\}$ of $\mathcal C$ for $A_1$.

Define:

- $\delta = \min \left\{{\delta_1, \delta_2, \ldots, \delta_n}\right\}$

Let $x, y \in A_1$ satisfy $d_1 \left({x, y}\right) < \delta$.

By the definition of a cover, there exists a $k \in \left\{{1, 2, \ldots, n}\right\}$ such that $d_1 \left({x, x_k}\right) < \delta_k$.

Then:

\(\displaystyle d_1 \left({y, x_k}\right)\) | \(\le\) | \(\displaystyle d_1 \left({y, x}\right) + d_1 \left({x, x_k}\right)\) | Triangle Inequality | ||||||||||

\(\displaystyle \) | \(<\) | \(\displaystyle \delta + \delta_k\) | Metric Space Axiom $\left({M3}\right)$ | ||||||||||

\(\displaystyle \) | \(\le\) | \(\displaystyle 2 \delta_k\) |

By the definition of $\Delta \left({x_k}\right)$, it follows that:

- $d_2 \left({f \left({x}\right), f \left({x_k}\right)}\right) < \dfrac \epsilon 2$

- $d_2 \left({f \left({y}\right), f \left({x_k}\right)}\right) < \dfrac \epsilon 2$

Hence:

\(\displaystyle d_2 \left({f \left({x}\right), f \left({y}\right)}\right)\) | \(\le\) | \(\displaystyle d_2 \left({f \left({x}\right), f \left({x_k}\right)}\right) + d_2 \left({f \left({x_k}\right), f \left({y}\right)}\right)\) | Triangle Inequality | ||||||||||

\(\displaystyle \) | \(<\) | \(\displaystyle \frac \epsilon 2 + \frac \epsilon 2\) | Metric Space Axiom $\left({M3}\right)$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \epsilon\) |

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:

- $F \left({x, y}\right) = \left({f \left({x}\right), f \left({y}\right)}\right)$

By Projection from Product Topology is Continuous, we have that the (first and second) projections on $A_1 \times A_1$ are continuous.

By Continuity of Composite Mapping and Continuous Mapping to Topological Product, it follows that $F$ is continuous.

By Metric is Continuous and Continuity of Composite Mapping, 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:

- $C = \left({d_2 \circ F}\right)^{-1} \left({\left[{\epsilon \,.\,.\, {+\infty}}\right)}\right) = \left\{{\left({x, y}\right) \in A_1 \times A_1: d_2 \left({f \left({x}\right), f \left({y}\right)}\right) \ge \epsilon}\right\}$

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 Metric is Continuous and Continuous Image of Compact Space is Compact, $d_1 \left({C}\right)$ is compact.

Therefore, $d_1 \left({C}\right)$ has a smallest element $\delta$.

By metric space axioms $(M1)$ and $(M4)$, we have that $\delta > 0$.

By construction, it follows that:

- $\forall x, y \in A_1: d_1 \left({x, y}\right) < \delta \implies d_2 \left({f \left({x}\right), f \left({y}\right)}\right) < \epsilon$

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): $4.19$