Distance Function of Metric Space is Continuous

Theorem
Let $M = \left({A, d}\right)$ be a metric space.

Let $\tau_A$ be the topology on $A$ induced by $d$.

Let $\left({A \times A, \tau}\right)$ be the topological product of $\left({A, \tau_A}\right)$ and itself.

Then the metric $d: A \times A \to \R$ is a continuous mapping.

Proof
Let $d_1: \left({A \times A}\right) \times \left({A \times A}\right) \to \R$ be the metric on $A \times A$ defined by:
 * $d_1 \left({\left({x, y}\right), \left({x', y'}\right)}\right) = d \left({x, x'}\right) + d \left({y, y'}\right)$

By Product Space Metric Induces Product Topology, $\tau$ is the topology on $A \times A$ induced by $d_1$.

Let $\epsilon \in \R_{>0}$ be a strictly positive real number.

Let $\left({x_0, y_0}\right) \in A \times A$.

Suppose that $\left({x, y}\right) \in A \times A$ and $d_1 \left({\left({x, y}\right), \left({x_0, y_0}\right)}\right) < \epsilon$.

Then:

The result follows from the definition of a continuous mapping.