Heine-Cantor Theorem/Proof 2

Proof
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 and, 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.