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 September 20th, 2018, 01:35 PM #1 Newbie   Joined: May 2012 Posts: 5 Thanks: 0 f bounded on A and B, bounded on A U B? Prove that if f is bounded on A and f is also bounded on B, then f is bounded on A U B Proof Since $\displaystyle f$ is bounded on A, there is a real number $\displaystyle m$ such that $\displaystyle \left|f(x) \right| \leq m$, for all x in A. Similarly, since $\displaystyle f$ is bounded on B, there is an $\displaystyle n$ such that $\displaystyle \left|f(x) \right| \leq n$, for all x in B. Let $\displaystyle M = \max(m,n)$ then $\displaystyle \left|f(x) \right| \leq M$, for all x in $\displaystyle A \cup B$. Thus showing that $\displaystyle f$ is bounded on $\displaystyle A \cup B$ Is this proof right? Last edited by skipjack; November 7th, 2018 at 08:47 PM. September 20th, 2018, 02:46 PM #2 Senior Member   Joined: Aug 2017 From: United Kingdom Posts: 313 Thanks: 112 Math Focus: Number Theory, Algebraic Geometry I would like to see a little more detail. In particular, could you explain why it follows that $\displaystyle \left|f(x) \right| \leq M$, for all x in $\displaystyle A \cup B$? Thanks from ag05 September 21st, 2018, 10:26 AM #3 Newbie   Joined: May 2012 Posts: 5 Thanks: 0 Does this work? n,m are positive real numbers. So either \$\displaystyle n=m,\ n
 November 7th, 2018, 08:52 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,969 Thanks: 2219 Both versions look okay to me. Tags bounded Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post samdon123 Calculus 3 November 21st, 2016 09:59 AM WWRtelescoping Advanced Statistics 1 April 13th, 2014 01:40 PM Pintosp Real Analysis 2 April 21st, 2012 01:56 PM wannabe1 Applied Math 1 October 11th, 2009 01:22 PM julien Real Analysis 5 November 1st, 2007 08:35 AM

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