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October 28th, 2017, 10:24 AM  #1 
Newbie Joined: Oct 2017 From: Here Posts: 19 Thanks: 0  How to show mathematically that the intersection is empty
I have A, B, two nonempty subsets of R. A has upper bound only, and B is a bounded set (both upper and lower bounds) Now I need to prove, or disprove, that if inf(B)>sup(A), then the intersection of A and B is an empty set. What I think is that it's true, so I need to prove rather than disprove this claim, but I just don't know how to write it mathematically correct. Right now I do have this general idea of why it's an empty set but can't get it right using all the math on the paper. *For me it seems quite obvious that it's true since the smallest number on B is greater than the largest number on A, but how exactly I write it Thanks! Last edited by Mathmatizer; October 28th, 2017 at 10:32 AM. 
October 28th, 2017, 03:08 PM  #2 
Newbie Joined: May 2017 From: California Posts: 15 Thanks: 1 
I was learning these proofs last week, so take this with a grain of salt... I think you can do this with a contradiction by "Suppose inf(B) > sup(A) and let x is an element of A intersection B"? So x is an element of A and x is an element of B and this would lead to a contradiction. Last edited by skipjack; October 29th, 2017 at 05:39 AM. 
October 28th, 2017, 05:22 PM  #3  
Senior Member Joined: Sep 2016 From: USA Posts: 635 Thanks: 401 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
\[a \leq \sup(A) < \inf(B) \leq b\] and we conclude that $b > a$. Hence, $A\bigcap B$ is empty.  
October 28th, 2017, 10:41 PM  #4 
Newbie Joined: Oct 2017 From: Here Posts: 19 Thanks: 0 
Thanks guys, I was thinking to do something like that, it's just that I wasn't sure saying that if b>a simply means intersection is empty, because it seems too literal rather than math, but I guess it's because it's just a simple question

October 29th, 2017, 03:26 AM  #5 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
Equivalently, proof by contradiction: Suppose there does exist a number, x, in both A and B. Since x is in B, $\displaystyle x\le sup(B)$. Since x is in A, $x\ge inf(B)$. So $inf(B)\le x\le sup(A)$, contradicting $sup(A)< inf(B)$.


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