I'm meant to find an example where every condition needed for the Arzelá-Ascoli Theorem holds except for compactness. What would be a good example here - simply an open interval, like (0,1), or might it be easier to prove if I had some bound which depended on n itself?
Proving the Arzelá-Ascoli Theorem fails without compactness
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analysis
compactness
arzela-ascoli
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0One very useful thing that comes from compactness in the proof is the space being totally bounded.(i.e. for each $\epsilon>0$, the space can be covered by a finite number of open balls of radius $\epsilon$.) $(0,1)$ is totally bounded, so it may help if you look in a space that is not. If you need more help, let me know. I have an example in mind that isn't hard to understand, but it isn't on $(0,1)$. – 2017-02-05
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0Hint: Take $\mathbb R$. Then, cook up a sequence, in which $f_1$ is a simple geometric form, and for which each succesive member is a suitable translate. – 2017-02-05
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0Possible duplicate of http://math.stackexchange.com/questions/567252/the-arzel%C3%A0-ascoli-theorem-fails-on-a-half-open-interval?rq=1 – 2017-02-05
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0@SeanEnglish I did start to consider ℝ, but each example that I come up with has something wrong. I considered fn(x) = n, and fn(x) = 0 for n even, and 1 for n odd, but neither work. Is there a simple example similar to what I've been coming up with that works, or am I on the wrong track? – 2017-02-05
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0$\mathbb{R}$ is suitable. You want your functions to be bounded, so stick between 0 and 1. It might help to observe that a family of continuous piece-wise linear functions are equicontinuous if they only attain finitely many slopes. – 2017-02-05
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0Did you figure it out? – 2017-02-10
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0@SeanEnglish I did figure it out, thank you! Sorry to not get back to you sooner! – 2017-02-11