Compact Space: Locally Continuous $\implies$ Uniformly Continuous

Question

Given metric spaces.
Prove that any locally continuous function on a compact space is uniformly continuous!

Answer 1

By local continuity there is a delta for any point: $$d(x,z)<\delta(z)\implies d(f(x),f(z))<\frac{1}{2}\epsilon$$ By compactness there is finite cover: $$X=\bigcup_{i=1}^N B_{\frac{1}{2}\delta(z_{i_0})}(z_i)$$ Thus any two points close enough: $$d(x,y)<\min_{i=1\ldots N}\frac{1}{2}\delta(z_i)$$ belong to one common ball: $$d(x,z_{i_0})<\frac{1}{2}\delta(z_{i_0})<\delta(z_{i_0})$$ $$d(y,z_{i_0})<d(x,z_{i_0})+d(x,y)<\frac{1}{2}\delta(z_{i_0})+\frac{1}{2}\delta(z_{i_0})=\delta(z_{i_0})$$ and therefore satisfy: $$d(f(x),f(y))<d(f(z),f(z_{i_0}))+d(f(x),f(z_{i_0}))<\frac{1}{2}\epsilon+\frac{1}{2}\epsilon=\epsilon$$

Answer 2

Assuming by "locally continuous" you simply mean "continuous" this is a classical result. You can find the proof in many texts. You can approach it by assuming the function is not uniformly continuous and use that assumption to construct a suitable sequence, and use compactness to deduce a contradiction. Or, you can take a more topological approach and first prove the Lebesgue number lemma. Uniform continuity of continuous functions is an easy consequence.

Answer 3

For every point $x$ find a ball over which the function doesn't change more than by $\epsilon$. The balls of half the radius still cover the set, and since the set is compact there is a finite subcover. The smallest radius $\delta$ of these finitely many smaller balls works for the whole set. Indeed, any two points within $\delta$ of each other will be in one of the larger original balls by the triangle inequality, and values at them can differ by no more than $\epsilon$.