Metrics are Continuous

Last updated Nov 1, 2022

Suppose $(M, d)$ is a Metric Space. Then $d$ is a Continuous Function from $M \times M \to \mathbb{R}$ (where $\mathbb{R}$ is endowed with its standard Distance Function $x,y \mapsto |x - y|$ and $M \times M$ is endowed with the Product Metric Space).

Let $\epsilon > 0$ and let $x,y \in M$. Let $\delta = \frac{\epsilon}{2}$. Suppose $x’,y’ \in M$ so that $\sup\limits {d(x,x’), d(y,y’)} < \delta$. This means $d(x, x’)$ and $d(y,y’)$ are both less than $\delta$. Thus $$2\delta + d(x,y) > d(x’, x) + d(x,y) + d(y,y’) \geq d(x’, y’).$$ Therefore $d(x’,y’) - d(x,y) < 2 \delta$. Likewise $$2\delta + d(x’,y’) > d(x, x’) + d(x’,y’) + d(y’,y) \geq d(x, y).$$ so $d(x,y) - d(x’,y’) < 2 \delta$. Therefore $|d(x,y) - d(x’,y’)| < 2 \delta = \epsilon$. Therefore, by the definition of Function Limit, $\lim\limits_{(x’,y’) \to (x,y)} d(x’,y’) = d(x,y)$. Since $x,y \in M$ were arbitrary, $d$ is a Continuous Function. $\blacksquare$