Compact Space: Locally Continuous $\implies$ Uniformly Continuous

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

• what does "locally continuous" mean? – Ittay Weiss Jun 9 '14 at 22:33
• just the ordinary continuity ;) see en.wikipedia.org/wiki/… – C-Star-W-Star Jun 9 '14 at 22:36
• then don't call it by a name nobody uses ;) – Ittay Weiss Jun 9 '14 at 22:38
• Yes I know it just looked better in the title ;) – C-Star-W-Star Jun 9 '14 at 23:09

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.

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 $\epsilon$.

• I think you need some $\delta/2$ work in this proof *as in Freeze_S's answer. As it stands, you seem to assume that any two points that are within $\delta$ of each other are together in one of the sets in your finite subcover. – Andreas Blass Jun 10 '14 at 16:05
• Good catch, I fixed it without halving explicitly. – Conifold Jun 10 '14 at 21:41

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: $$d(f(x),f(y))<d(f(z),f(z_{i_0}))+d(f(x),f(z_{i_0}))<\frac{1}{2}\epsilon+\frac{1}{2}\epsilon=\epsilon$$

• @Downvoter: Could u plz explain! – C-Star-W-Star Jun 9 '14 at 23:10
• I didn't downvote, but I conjecture that the downvoter believed Conifold's answer (or something similar) and therefore thought that yours was unnecessarily complicated. – Andreas Blass Jun 10 '14 at 16:07
• Complicated but necessary, isn't it? – C-Star-W-Star Jun 10 '14 at 16:24
• I didn't check that every $\frac12$ in your answer is necessary, but some of them are surely necessary, as I indicated in a comment on Conifold's answer. – Andreas Blass Jun 10 '14 at 16:28