Uniformly Continuous Function is Continuous/Metric Space

Theorem

Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_1, d_1}\right)$ be metric spaces.

Let the mapping $f: M_1 \to M_2$ be uniformly continuous on $M_1$.

Then $f$ is continuous on $M_1$.

Proof

Let $f$ be uniformly continuous on $M_1$.

Let $x \in M_1$.

Let $\epsilon > 0$.

As $f$ is uniformly continuous, $\exists \delta > 0$ such that:

$\forall y \in M_1: d_1 \left({x, y}\right) < \delta: d_2 \left({f \left({x}\right), f \left({y}\right)}\right) < \epsilon$

Thus by definition $f$ is continuous at $x$.

$\blacksquare$