Uniformly Continuous Function is Continuous

Theorem

Metric Spaces

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 $M_1$.

Real Numbers

Let $I$ be an interval of $\R$.

If a function $f: I \to \R$ is uniformly continuous on $I$, then it is continuous on $I$.

Proof

Proof for Metric Spaces

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

Let us take any $x \in M_1$.

Let $\epsilon > 0$.

By the fact that $f$ 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 $f$ is continuous at $x$.

$\blacksquare$

Proof for Real Numbers

Follows directly from the fact that the real number line is a metric space under the usual metric.

$\blacksquare$

Notes on the proof

The above proof just states that the condition of being continuous is already explicitly included in the condition of being uniformly continuous.