My Math Forum  

Go Back   My Math Forum > College Math Forum > Real Analysis

Real Analysis Real Analysis Math Forum


Thanks Tree19Thanks
Reply
 
LinkBack Thread Tools Display Modes
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
Benit13 is offline  
 
July 14th, 2016, 09:49 AM   #12
Senior Member
 
Joined: Jun 2015
From: England

Posts: 915
Thanks: 271

Quote:
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.

Thanks from Benit13 and manus
studiot is offline  
July 14th, 2016, 11:58 AM   #13
Math Team
 
topsquark's Avatar
 
Joined: May 2013
From: The Astral plane

Posts: 2,346
Thanks: 988

Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
Originally Posted by studiot View Post
That is because it was a chose sandwich - y'all chose it.

Wouldn't that be using the axiom of sandwich choice?

-Dan
Thanks from Benit13 and manus
topsquark is offline  
July 14th, 2016, 05:46 PM   #14
Math Team
 
Joined: Nov 2014
From: Australia

Posts: 689
Thanks: 244

Quote:
Originally Posted by zylo View Post
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.
Thanks from topsquark and manus
Azzajazz is offline  
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.
zylo is offline  
Reply

  My Math Forum > College Math Forum > Real Analysis

Tags
bounded, closed, continuous, function, interval



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Uniform Continuity on a Bounded Interval dpsmith Real Analysis 11 September 1st, 2015 05:53 PM
D[f] is bounded, show that f is uniformly continuous king.oslo Real Analysis 1 July 7th, 2014 05:30 AM
Absolutely continuous/bounded variation natt010 Real Analysis 3 May 4th, 2014 03:51 PM
Infimum of an unbouned function on a bounded interval leeSono Real Analysis 2 October 4th, 2013 10:02 AM
cts fnc on a closed interval rose3 Real Analysis 1 December 19th, 2009 02:06 PM





Copyright © 2019 My Math Forum. All rights reserved.