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Continuous function on a compact metric space is uniformly continuous

How does uniform continuity and continuity coincide in a Compact set ?

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marked as duplicate by lhf, t.b., Benjamin Lim, Rudy the Reindeer, Kannappan Sampath Apr 26 '12 at 12:11

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

See this question: Continuous function on a compact metric space is uniformly continuous. Also, you should tell us in what setting are you working: real functions? functions on metric spaces? functions on uniform spaces? – Martin Sleziak Apr 26 '12 at 10:47
In fact, the question asked in the title is not the same as in the body. In the last one, we have to see whether a continuous function on a compact space is uniformly continuous, whereas in the title it's asked whether continuity implies uniform continuity. In the case of real-valued functions defined on metric spaces, we can show that a continuous function is uniformly continuous if the metric space is compact, and if each continuous function is uniformly continuous, the metric space is compact without isolated points. – Davide Giraudo Apr 26 '12 at 11:30
up vote 1 down vote accepted

Continuity is a local property. On a compact set, you can get global properties by combining a finite number of local properties.

More precisely, continuity means that given $\varepsilon>0$, for each point $x$ there is a $\delta_x>0$ such that points $\delta_x$-near $x$ are sent to points $\varepsilon$-near $f(x)$. In a compact set, you can take a finite number of $\delta_x$ to cover the domain and take $\delta>0$ as the minimum of those and so get a $\delta>0$ that works for all points for the given $\varepsilon$.

PS: I'm using the characterization of compact sets as the ones for which every open cover has a finite subcover.

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The minimum of the finitely many $\delta_{x_1},\ldots,\delta_{x_n}$ is not good enough. If $x$ is arbitrary then $x$ is $\delta$-close to some $x_k$. If $d(x,y) \lt \delta$ then there's no guarantee that $y$ is also close to the same $x_k$, only to some other $x_l$. So you cannot compare the values $f(x)$ and $f(y)$. – t.b. Apr 26 '12 at 10:58
t.b.'s objection can be worked around using the trick in the answer that Martin linked to – Willie Wong Apr 26 '12 at 11:03
@t.b., right, I forgot about the $\delta_x/2$ thing. Thanks. – lhf Apr 26 '12 at 12:28

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