bounded continous implies uniformly continuous
I'm trying to show that continuous f : [a, b] > R implies f uniformly continuous.
f continuous if for all e > 0, x in [a, b], there exists d > 0 such that for all y in [a, b], ¦x  y¦ < d implies ¦f(x)  f(y)¦ < e. f uniformly continuous if for all e > 0, there exists d > 0 such that for all x and y in [a, b], ¦x  y¦ < d implies ¦f(x)  f(y)¦ < e. I constructed A(d) = { u in [a, b] : x, y in [a, u], ¦x  y¦ < d implies ¦f(x)  f(y)¦ < e } A = U_{d > 0} A(d) And I think I need to show sup A = b and b is in A, but I'm stuck. 
Re: bounded continous implies uniformly continuous
So you know a characterization of uniform continuity using sequences??
That is: f is uniform continuous iff for all equivalent sequences [itex](x_n)_n[/itex] and [itex](y_n)_n[/itex] holds that [itex](f(x_n))_n[/itex] and [itex](f(y_n))_n[/itex] are also equivalent. 
Re: bounded continous implies uniformly continuous
Was not aware of this characterization until now. My text does not mention this, so there must be another way.

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Re: bounded continous implies uniformly continuous
Isn't it true that on a compact set (which [a,b] is), the sup is equivalent to the maximum (and similarly, the inf is the same as a minimum)?

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[tex]\sup_{x \in K} f(x) = \max_{x \in K} f(x)[/tex] and [tex]\inf_{x \in K} f(x) = \min_{x \in K} f(x)[/tex] How do you propose to use this fact? 
Re: bounded continous implies uniformly continuous
P.S. It would help to know what definition of "compact" you are using. There are several definitions that are equivalent on the real line.
Does your definition involve open covers and finite subcovers, or convergent subsequences, or "closed and bounded", or what? 
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(Disclaimer: I don't really know much math, so people who actually know this stuff should correct me if I'm speaking nonsense!) 
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EDIT: Sorry, I very stupidly wrote sup when I meant inf in my post above, so obviously it made no sense  way too tired to be useful. I'll go back and edit now. 
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Also, I confused Ansatz7 with the original poster (alanlu), who made no mention of compactness of [a,b] and perhaps doesn't have the appropriate machinery (HeineBorel) available. That may have been what he was getting at with this Spivakstyle construction: Quote:

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Re: bounded continous implies uniformly continuous
Right, I know that this only works because [a, b] is compact, as I stated in my first post in the thread. I believe it was you who asked how I would use this fact, which is where everything else came from.
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Re: bounded continous implies uniformly continuous
Ah thanks! Actually, I did arrive at inf { d(x) }, but I wasn't sure how to turn that into something that is guaranteed to be > 0.

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