My Math Forum  

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

Real Analysis Real Analysis Math Forum

Thanks Tree1Thanks
  • 1 Post By SDK
LinkBack Thread Tools Display Modes
February 13th, 2017, 11:24 AM   #1
Senior Member
Joined: Jan 2016
From: Blackpool

Posts: 103
Thanks: 2

Proving a continuous function is bounded

How would i prove that the function F:[0,1]->Reals is bounded?
Jaket1 is offline  
February 13th, 2017, 02:40 PM   #2
Senior Member
Joined: Sep 2016
From: USA

Posts: 578
Thanks: 345

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.
Thanks from Jaket1
SDK is offline  

  My Math Forum > College Math Forum > Real Analysis

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 08:19 AM

Copyright © 2019 My Math Forum. All rights reserved.