# Continuous mapping on a compact metric space is uniformly continuous

I am struggling with this question:

Prove or give a counterexample: If $f : X \to Y$ is a continuous mapping from a compact metric space $X$, then $f$ is uniformly continuous on $X$.

The answer is yes, if $$f$$ is continuous on a compact space then it is uniformly continuous:

Let $$f: X \to Y$$ be continuous, let $$\varepsilon > 0$$ and let $$X$$ be a compact metric space. Because $$f$$ is continuous, for every $$x$$ in $$X$$ you can find a $$\delta_x$$ such that $$f(B(\delta_x, x)) \subset B({\varepsilon\over 2}, f(x))$$. The balls $$\{B(\delta_x, x)\}_{x \in X}$$ form an open cover of $$X$$. So do the balls $$\left\{B \left(\frac{\delta_x}{2}, x\right)\right\}_{x \in X}$$. Since $$X$$ is compact you can find a finite subcover $$\left\{B \left( \frac{\delta_{x_i}}{2}, x_i \right) \right\}_{i=1}^n$$. (You will see in a second why we are choosing the radii to be half only.)

Now let $$\delta_{x_i}’ = {\delta_{x_i}\over 2}$$.

You want to choose a distance $$\delta$$ such that for any two $$x,y$$ they lie in the same $$B(\delta_{x_i}’, x_i)$$ if their distance is less than $$\delta$$.

How do you do that?

Note that now that you have finitely many $$\delta_{x_i}’$$ you can take the minimum over all of them: $$\min_i \delta_{x_i}’$$. Consider two points $$x$$ and $$y$$. Surely $$x$$ lies in one of the $$B(\delta_{x_i}’, x_i)$$ since they cover the whole space and hence $$x$$ also lies in $$B(\delta_{x_i}’, x_i)$$ for some $$i$$.

Now we want $$y$$ to also lie in $$B(\delta_{x_i}’, x_i)$$. And this is where it comes in handy that we chose a subcover with radii divided by two:

If you pick $$\delta : = \min_i \delta_{x_i}’$$ (i.e. $$\delta = \frac{\delta_{x_i}}{2}$$ for some $$i$$) then $$y$$ will also lie in $$B(\delta_{x_i}, x_i)$$:

$$d(x_i, y) \leq d(x_i, x) + d(x,y) < \frac{\delta_{x_i}}{2} + \min_k \delta_{x_k} \leq \frac{\delta_{x_i}}{2} + \frac{\delta_{x_i}}{2} = \delta_{x_i}$$.

Hope this helps.