I want to find an example of a continuos function on an open interval S⊂R, which is not uniform continuos. I use f(x)=1/x on the open interval from (0,1).
But I also want to determine if such examples are possible when S is compact?
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2Could you clarify your question? You want a continuous function on a compact set $S$ but... [fill in what you want here to happen (outside of $S$?)]. – Alex R. Jun 01 '16 at 18:23
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Oh guys sorry, i forgot to write uniform continuos. Edited the post now. – Biggiez Jun 01 '16 at 18:26
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$\sin(1/x)$ will do the trick too on $(0,1)$ – reuns Jun 01 '16 at 18:30
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and no, a function continuous on $[a,b]$ is also uniformly continuous on $[a,b]$ (same with compacts) – reuns Jun 01 '16 at 18:32
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@AlexR. So I want to determine if there exists continuos functions on an open interval S⊂R, which is not uniform continuos. But this time S is compact. – Biggiez Jun 01 '16 at 18:32
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@user1952009 Dont you mean the reverse ? Uniform continuity implies continuity? – Biggiez Jun 01 '16 at 18:33
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1no I mean the wiki/Heine–Cantor_theorem that if $f$ is continuous on a compact then it is uniformly continuous on it – reuns Jun 01 '16 at 18:34
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Okay so if f is continuous on a compact set then this implies that f is uniformly continuous? But why is this true? Im not sure if I understand the idea of a compact set to 100%. – Biggiez Jun 01 '16 at 18:38
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Did you read the proof? Yes, the idea of a compact set is a little slippery, but it is hugely important if you are going to get anywhere in analysis class. – Doug M Jun 01 '16 at 18:40
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Yeah I read the proof, but they use some different notations that im not so used to. – Biggiez Jun 01 '16 at 18:42
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Any continuous function on a compact set is uniformly continuous on this set. – Jun 01 '16 at 18:54
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@Masacroso Is there a logical interpretation on why it makes sense? Or visualize it in a way? – Biggiez Jun 01 '16 at 19:00
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See here or here. – Jun 01 '16 at 19:03
2 Answers
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For example, $\;f(x)=\cfrac1x\;$ is continuous in $\;(0,1)\;$ but not uniformly continuous, because for example
$$\left\{\,\frac1n\,\right\}_{n\ge2}\,,\,\,\left\{\,\frac1{n+1}\,\right\}_{n\ge2}\subset (0,1)\;\;\;\text{and}\;\;\;\left|\frac1n-\frac1{n+1}\right|=\frac1{n(n+1)}\xrightarrow[n\to\infty]{}0$$
yet
$$\left|f\left(\frac1n\right)-f\left(\frac1{n+1}\right)\right|=1\rlap{\;\;\;\;/}\xrightarrow[n\to\infty]{}0$$
DonAntonio
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Yeah, this made sense. But is this related to my question on wheter there exists continuos functions on an open interval S⊂R, which is not uniform continuos. But this time S is compact? – Biggiez Jun 01 '16 at 18:35
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For a compact set in a metric space, continuity is uniform continuity. This follows from alternate definition of compactness (every open cover has a finite subcover). – DBS Jun 01 '16 at 18:38
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@Biggiez First, you were already answered in the comments, as the comment above mentions. Second, that is a theorem: a continuous real function on a closed, bounded (=compact) interval is u.c. there. – DonAntonio Jun 01 '16 at 18:39
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Uniform continuity is a tighter definition than regular continuity. So, if f is uniformly continuous then f is continuous.
However, if f is a continuous function on a compact domain the f is uniformly continuous on the domain.
Proof:
If $f$ is continuous, then the for every open set in the image of $f$, the pre-image is an open set.
That is, for every epsilon ball in the image ($|f(x) - f(y)|<\epsilon$) There is a corresponding delta ball in the pre-image ($|x-y| < \delta$).
We cover image of $f$, in open balls, each of radius $\frac\epsilon2$. Each of those balls has its corresponding delta ball. The union of these delta balls forms an open cover of the domain.
The domain is compact! Every open cover has a finite sub-cover. Choose finitely many many of these delta balls.
If there are finitely many delta balls then there is a minimum radius $\frac\delta2$.
For every $(x,y)$ in the domain within a ball of radius $\frac\delta2, |x-y|<\delta$ everything inside that delta ball maps into a corresponding ball of radius $\frac \epsilon2, |f(x) - f(y)| < \epsilon$.
Doug M
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If f(x) and f(y) are within a ball of radius $\epsilon/2$, the distance between f(x) and f(y) is less than epsilon. – Doug M Jun 01 '16 at 20:00