Proof that Continuous Functions on Compact Metric Spaces are Uniformly Continuous
Theorem
Let $(X,d_{X})$ be a compact metric space, $(Y,d_{Y})$ a metric space, and $f:X\to Y$ continuous. Then, $f$ is uniformly continuous on $X$.
Explanation
The condition of being compact cannot be omitted.
Proof
Suppose we are given any positive number $\varepsilon >0$. Since $f$ is assumed to be continuous, by definition, for each point $p\in X$, there exists a positive number $\delta_{p}$ satisfying the following equation:
$$ \forall q\in X,\quad d_{X}(p,q)<\delta_{p} \implies d_{Y}(f(p),f(q))<\frac{\varepsilon}{2} $$
Now consider the following set:
$$ N_{p}:= \left\{ q : d_{X}(p,q)<\frac{1}{2}\delta_{p} \right\} $$
Since $p \in N_{p}$, the collection of all $N_{p}$ becomes an open cover of $X$. Since $X$ is assumed to be compact, there exists $p_{1},\cdots,p_{n}$ satisfying the following equation:
$$ \begin{equation} X \subset N_{p_{1}}\cup \cdots \cup N_{p_{n}} \tag{2} \label{eq1} \end{equation} $$
Let us now set $\delta=\frac{1}{2} \min (\delta_{p_{1}},\cdots,\delta_{p_{n}})$. Then, $\delta$ is clearly positive. Now consider two points $p,q \in X$ satisfying $d_{X}(p,q)<\delta$. Then, by $\eqref{eq1}$, there exists $m(1\le m \le n)$ satisfying $p \in N_{p_{m}}$. Therefore, the following holds:
$$ d_{X}(p,p_{m}) \le \frac{1}{2}\delta_{p_{m}} $$
Then, the following equation holds:
$$ d_{X}(q,p_{m}) \le d_{X}(q,p) + d_{X}(p,p_{m}) < \delta + \frac{1}{2}\delta_{p_{m}} \le \delta _{p_{m}} $$
Therefore,
$$ d_{X}(p,q)<\delta \implies d_{Y}(f(p),f(q))\le d_{Y}(f(p),f(p_{m})) + d_{Y}(f(p_{m}),f(q))<\frac{1}{2}\varepsilon+\frac{1}{2}\varepsilon=\varepsilon $$
Thus, $f$ is uniformly continuous.
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