Heine-Cantor Theorem/Proof 1

Theorem
Let $M_1$ and $M_2$ be metric spaces.

Let $f: M_1 \to M_2$ be a continuous mapping.

If $M_1$ is compact, then $f$ is uniformly continuous on $M_1$.

Proof
Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be metric spaces such that $M_1$ is compact.

Let $f: M_1 \to M_2$ be continuous.

Let $\epsilon > 0$.

Then by definition:
 * $\forall x \in M_1: \exists \delta \left({x}\right) > 0: \forall y \in M_1: d_1 \left({x, y}\right) < 2 \delta \left({x}\right) \implies d_2 \left({f \left({x}\right), f \left({y}\right)}\right) < \dfrac \epsilon 2$

The set $\left\{{N_{\delta \left({x}\right)} \left({x}\right): x \in M_1}\right\}$ is an open cover for $M_1$.

As $M_1$ is compact, there is a finite subcover $\left\{{N_{\delta \left({x_1}\right)} \left({x_1}\right), N_{\delta \left({x_2}\right)} \left({x_2}\right), \ldots, N_{\delta \left({x_r}\right)} \left({x_r}\right)}\right\}$.

Now let $\delta = \min \left\{{\delta \left({x_1}\right), \delta \left({x_2}\right), \ldots, \delta \left({x_r}\right)}\right\}$.

Consider any $x, y \in M_1$ which satisfy $d_1 \left({x, y}\right) < \delta$.

There is some $i \in \left\{{1, 2, \ldots, r}\right\}$ such that $d_1 \left({x, x_i}\right) < \delta \left({x_i}\right)$.

Thus:
 * From $d_1 \left({x, x_i}\right) < \delta \left({x_i}\right)$ we have:
 * $d_2 \left({f \left({x}\right), f \left({x_1}\right)}\right) \le \dfrac \epsilon 2$


 * From $d_1 \left({y, x_i}\right) \le 2 \delta\left({x_i}\right)$ we have:
 * $d_2 \left({f \left({y}\right), f \left({x_1}\right)}\right) \le \dfrac \epsilon 2$

So: