Mapping from Standard Discrete Metric on Real Number Line is Continuous

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

Let $\struct {\R, d_1}$ be the metric space such that $d_1$ be the Euclidean metric on $\R$.

Let $\struct {\R, d_2}$ be the metric space such that $d_2$ be the standard discrete metric on $\R$.

Let $f: \tuple {\R, d_2} \to \tuple {\R, d_1}$ be a real function.

Then $f$ is $\tuple {d_2, d_1}$-continuous on $\R$.

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

Let $\delta = 1$.

Let $x \in \R$.

Let $y \in \R$ such that $\map {d_2} {x, y} < \delta$.

That is, $\map {d_2} {x, y} < 1$.

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

Thus $\map f x = \map f y$.

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

Thus the conditions for $\tuple {d_2, d_1}$-continuity at a point are fulfilled.

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

So by definition $f$ is $\tuple {d_2, d_1}$-continuous on $\R$.