Mapping from Standard Discrete Metric on Real Number Line is Continuous

Theorem
Let $\R$ be the real number line.

Let $\left({\R, d_1}\right)$ be the metric space such that $d_1$ be the Euclidean metric on $\R$.

Let $\left({\R, d_2}\right)$ be the metric space such that $d_2$ be the discrete metric on $\R$.

Let $f: \left({\R, d_2}\right) \to \left({\R, d_1}\right)$ be a real function.

Then $f$ is $\left({d_2, d_1}\right)$-continuous on $\R$.

Proof
Let $\epsilon \in \R: \epsilon > 0$.

Let $\delta = 1$.

Let $x \in \R$.

Let $y \in \R$ such that $d_2 \left({x, y}\right) < \delta$.

That is, $d_2 \left({x, y}\right) < 1$.

By the definition of the discrete metric on $\R$, that would mean that $d_2 \left({x, y}\right) = 0$ and so $x = y$.

Thus $f \left({x}\right) = f \left({y}\right)$.

By definition of a metric, that means:
 * $d_1 \left({f \left({x}\right), f \left({y}\right)}\right) = 0 < \epsilon$

Thus the conditions for $\left({d_2, d_1}\right)$-continuity at a point are fulfilled.

This is true for all $x \in \R$.

So by definition $f$ is $\left({d_2, d_1}\right)$-continuous on $\R$.