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 July 14th, 2016, 05:54 AM #11 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,166 Thanks: 738 Math Focus: Physics, mathematical modelling, numerical and computational solutions Let my sandwich be a cheese sandwich After I put my glasses on, it turns out not to be a cheese sandwich. Therefore, contradiction! Gadzooks! The nature of cheesiness of sandwich hath revealed itself unto me! Thanks from topsquark and manus
July 14th, 2016, 09:49 AM   #12
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 Let my sandwich be a cheese sandwich After I put my glasses on, it turns out not to be a cheese sandwich.
That is because it was a chose sandwich - y'all chose it.

July 14th, 2016, 11:58 AM   #13
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Quote:
 Originally Posted by studiot That is because it was a chose sandwich - y'all chose it.
Wouldn't that be using the axiom of sandwich choice?

-Dan

July 14th, 2016, 05:46 PM   #14
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 Originally Posted by zylo I stick by my original contention. If f(x) is continuous on a closed interval, it can never equal $\displaystyle \infty$.
You obviously didn't read my last post carefully enough, as you made exactly the same mistake in this statement as you did before.

 July 27th, 2016, 01:41 PM #15 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 I give up. Can't think of a simple, elementary, proof of OP. So: Since f continuous, for every p of [a,b] |f(x)-f(p)| < $\displaystyle \epsilon$ if |x-p| < $\displaystyle \delta_{p}$ Snce f continuous, f(p) finite, and f(p)- $\displaystyle \epsilon$ < f(x) < f(p) + $\displaystyle \epsilon$ < Mp For every point p there is an interval p-$\displaystyle \delta_{p}$< x < p+$\displaystyle \delta_{p}$ on which f(x) is bounded. These intervals cover [a,b]. By Heine-Borel theorem there is a finite subset of the intervals which cover [a,b]. Then f(x) is bounded by the max value of Mp on these intervals.

 Tags bounded, closed, continuous, function, interval

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