# Uniform continuity on bounded open intervals

If $f$ is uniformly continuous on a bounded open interval $(a,b)$, then $\lim_{x \to a^+} f(x)$ and $\lim_{x \to b^-} f(x)$ exist and are finite.

I was reading this book: http://classicalrealanalysis.info/documents/T-CalculusIntegral-AllChapters-Portrait.pdf

the above is one direction of theorem 1.12, and is proven in exercise 65. I was having some trouble understanding the proof, and was hoping that someone might be able to explain it more in-depth.

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What parts you don't understand? Try to be specific. – Pedro Tamaroff Sep 16 '13 at 1:42
I suppose that I don't quite understand the point of the oscillation of a function, and its use in the proof. – user95072 Sep 16 '13 at 1:55
Can you at least see, intuitively, why the oscillation is zero if and only if the function is continuous there, say? – Pedro Tamaroff Sep 16 '13 at 1:58
Yes, that makes sense – user95072 Sep 16 '13 at 2:06

## 1 Answer

Extension of uniformly continuous functions is discussed in detail in $\S$ 10.11 of these notes.

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(I guess this is the way to go, the other notes are strange.) – Pedro Tamaroff Sep 16 '13 at 2:17
@Peter: For some reason this basic result is hard to find in most elementary analysis courses. I remember thinking that when I taught undergrad analysis at McGill almost ten years ago and making a point of putting it in the notes. IIRC though, the course text used at the time (by Russell Gordon; it is really quite a good book) did treat this material, and I took my discussion more or less from there. – Pete L. Clark Sep 16 '13 at 2:23