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 September 20th, 2018, 02:35 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 09:47 PM.
 September 20th, 2018, 03:46 PM #2 Senior Member   Joined: Aug 2017 From: United Kingdom Posts: 282 Thanks: 85 Math Focus: Algebraic Number Theory, Arithmetic 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, 11:26 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, 09:52 PM #4 Global Moderator   Joined: Dec 2006 Posts: 19,878 Thanks: 1834 Both versions look okay to me.

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