Heine-Cantor Theorem/Proof 1

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
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 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:

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:

The result follows from the definition of uniform continuity.