(Theorem) Continuous functions on compact sets are uniformly continuous

Theorem: Continuity is equals uniform continuity on compact spaces

Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces. If $f:X\to Y$ is continuous and $X$ is compact, then $f$ is uniformly continuous.

Proof from Modern Analysis I.

Given $ϵ>0$, we know for all $x’\in X$, there exists $\delta (x’)>0$ such that

$$d_X(x,x^{\prime })<\delta (x^{\prime })⟹d_Y(f(x),f(x^{\prime }))<ϵ$$

for all $x\in X$. Then $\delta :X\to \mathbb{R}$ is a map.

🔺 Exercise. Show that $\delta :X\to \mathbb{R}$ can be chosen to be continuous.

Since $\delta$ can be chosen to be continuous, by the extreme value theorem, it attains its minimum ${\delta }_0>0$ on the domain $X$. Then, in particular, for this choice of $ϵ>0$ and all $x,x’\in X$, we have

$$d_X(x,x^{\prime })<{\delta }_0⟹d_Y(f(x),f(x^{\prime }))<ϵ$$

as well. Hence, $f$ is uniformly continuous.

❓ Question. Works because choosing the minimum…?

Proof from Topology.

Let $ϵ>0$. Cover $X$ by

$$\mathcal{C}=\{f^{-1}(B(y,ϵ/2){\}}_{y\in Y}.$$

Let $\delta$ be a Lebesgue number for $\mathcal{C}$. If $d(x_1,x_2)<\delta$ (DISTANCE), then

$$\text{diam}\{x_1,x_2\}<\delta ⟹\{x_1,x_2\}\subseteq f^{-1}(B(y,ϵ/2))\text{ for some }y\in Y⟹d_Y(f(x_1),f(x_2))<ϵ.$$