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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 $\displaystyle \leftf(x) \right \leq m$, for all x in A. Similarly, since $\displaystyle f$ is bounded on B, there is an $\displaystyle n$ such that $\displaystyle \leftf(x) \right \leq n$, for all x in B. Let $\displaystyle M = \max(m,n)$ then $\displaystyle \leftf(x) \right \leq M$, 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 $\displaystyle \leftf(x) \right \leq M$, for all x in $\displaystyle A \cup B$?

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<m, $ or $\displaystyle m<n$. If $\displaystyle n=m$ then $\displaystyle f(x) \leq n=m $ For all x in A $\displaystyle f(x) \leq n=m $ For all x in b Thus for either set $\displaystyle n=m$ is a bound, and so, $\displaystyle f$ is bounded on $\displaystyle A \cup B$. If $\displaystyle n < m$ then $\displaystyle f(x) \leq m $ For all x in A $\displaystyle f(x) \leq n<m $ For all x in B Thus for either set $\displaystyle m$ is a bound, and so, $\displaystyle f$ is bounded on $\displaystyle A \cup B$ If $\displaystyle m < n$ then $\displaystyle f(x) \leq m<n $ For all x in A $\displaystyle f(x) \leq n $ For all x in B Thus for either set $\displaystyle n$ is a bound, and so, $\displaystyle f$ is bounded on $\displaystyle A \cup B$ So if $\displaystyle f$ is bounded on sets A and B, then $\displaystyle f$ is bounded on there union $\displaystyle A \cup B$, which is what I needed to show. I hope I got something right. Thanks for responding to post. If I missed something or everything, let me know. Last edited by skipjack; November 7th, 2018 at 09:49 PM. 
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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