Uniformly Continuous Function is Continuous/Metric Space
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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 $x$.
$\blacksquare$