
Real Analysis Real Analysis Math Forum 
 LinkBack  Thread Tools  Display Modes 
February 13th, 2017, 12:24 PM  #1 
Member Joined: Jan 2016 From: Blackpool Posts: 40 Thanks: 0  Proving a continuous function is bounded
How would i prove that the function F:[0,1]>Reals is bounded?

February 13th, 2017, 03:40 PM  #2 
Senior Member Joined: Sep 2016 From: USA Posts: 114 Thanks: 44 Math Focus: Dynamical systems, analytic function theory, numerics 
1. Recall the definition of continuity. $F$ is continuous means that for any $\epsilon > 0$, for every $x$, there exists a $\delta$ (depending on $x$) such that $F(y)  F(x) < \epsilon$ whenever $y \in B_{\delta}(x)$. 2. Convince yourself of the following. Let $\delta(x)$ denote the appropriate $\delta$ for a specified $x$. Then $$ \bigcup_{x \in [0,1]} B_{\delta(x)}(x)$$ is an open cover of $[0,1]$. 3. Recall that $[0,1]$ is compact. What does this say about the open cover above? 4. Profit. 

Tags 
bounded, continuous, function, proving 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Continuous function on closed interval is bounded.  zylo  Real Analysis  14  July 27th, 2016 12:41 PM 
Problem with proving that a function is continuous  Gil  Topology  1  August 18th, 2014 09:36 AM 
Absolutely continuous/bounded variation  natt010  Real Analysis  3  May 4th, 2014 02:51 PM 
Proving continuous function on z  William_33  Complex Analysis  2  April 16th, 2013 11:59 AM 
Proving that a space is continuous, continuous at 0, and bdd  thedoctor818  Real Analysis  17  November 9th, 2010 09:19 AM 