Uniformly Continuous Function is Continuous

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 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 $$x$$.

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

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.