# Metric 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 $d_{\infty}: \left({A \times A}\right) \times \left({A \times A}\right) \to \R$ be the metric on $A \times A$ defined by:

- $d_{\infty} \left({\left({x, y}\right), \left({x', y'}\right)}\right) = \max \left\{{d \left({x, x'}\right), d \left({y, y'}\right)}\right\}$

By P-Product Metric Induces Product Topology, $\tau$ is the topology on $A \times A$ induced by $d_{\infty}$.

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 $d_{\infty} \left({\left({x, y}\right), \left({x_0, y_0}\right)}\right) < \dfrac 1 2 \epsilon$.

Then:

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \left\vert{d \left({x, y}\right) - d \left({x_0, y_0}\right)}\right\vert\) | \(\le\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \left\vert{d \left({x, y}\right) - d \left({x_0, y}\right)}\right\vert + \left\vert{d \left({x_0, y}\right) - d \left({x_0, y_0}\right)}\right\vert\) | \(\displaystyle \) | \(\displaystyle \) | Triangle Inequality for Real Numbers | ||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\le\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle d \left({x, x_0}\right) + d \left({y, y_0}\right)\) | \(\displaystyle \) | \(\displaystyle \) | Reverse Triangle Inequality | ||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(<\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \epsilon\) | \(\displaystyle \) | \(\displaystyle \) | by the definition of $d_{\infty}$ |

The result follows from the definition of a continuous mapping.

$\blacksquare$