Distance Function of Metric Space is Continuous

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Theorem

Let $M = \struct {A, d}$ be a metric space.

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

Let $\struct {A \times A, \tau}$ be the product space of $\struct {A, \tau_A}$ and itself.


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


Proof

Let $d_\infty: \paren {A \times A} \times \paren {A \times A} \to \R$ be the metric on $A \times A$ defined by:

$\map {d_\infty} {\tuple {x, y}, \tuple {x', y'} } = \max \set {\map d {x, x'}, \map d {y, y'} }$

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 $\tuple {x_0, y_0} \in A \times A$.

Suppose that $\tuple {x, y} \in A \times A$ and $\map {d_\infty} {\tuple{x, y}, \tuple{x_0, y_0} } < \dfrac 1 2 \epsilon$.

Then:

\(\ds \size {\map d {x, y} - \map d {x_0, y_0} }\) \(\le\) \(\ds \size {\map d {x, y} - \map d {x_0, y} } + \size {\map d {x_0, y} - \map d {x_0, y_0} }\) Triangle Inequality for Real Numbers
\(\ds \) \(\le\) \(\ds \map d {x, x_0} + \map d {y, y_0}\) Reverse Triangle Inequality
\(\ds \) \(<\) \(\ds \epsilon\) by the definition of $d_\infty$

The result follows from the definition of a continuous mapping.

$\blacksquare$