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! 
July 14th, 2016, 09:49 AM  #12  
Senior Member Joined: Jun 2015 From: England Posts: 915 Thanks: 271  Quote:
 
July 14th, 2016, 11:58 AM  #13 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,346 Thanks: 988 Math Focus: Wibbly wobbly timeywimey stuff.  
July 14th, 2016, 05:46 PM  #14 
Math Team Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244  
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 xp < $\displaystyle \delta_{p}$ 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 HeineBorel 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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