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 October 28th, 2017, 11:24 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 non-empty 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 11:32 AM. October 28th, 2017, 04:08 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 06:39 AM. October 28th, 2017, 06:22 PM   #3
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 Originally Posted by Mathmatizer I have A, B, two non-empty 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!
Suppose $\inf(B) > \sup(A)$. Note that $\inf(B) \leq b$ for every $b\in B$ and also $\sup(A) \geq a$ for every $a\in A$. It follows that for every $a \in A$ and $b \in B$ we have
$a \leq \sup(A) < \inf(B) \leq b$
and we conclude that $b > a$. Hence, $A\bigcap B$ is empty. October 28th, 2017, 11:41 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, 04:26 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 $sup(A)< inf(B)$. Tags empty, intersection, mathematically, show 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 John Travolski Real Analysis 3 September 3rd, 2017 05:36 AM gaussrelatz Applied Math 4 September 26th, 2013 04:37 AM Hamid Behravan New Users 7 September 25th, 2011 08:58 PM turtlejohn Complex Analysis 0 January 20th, 2010 09:49 PM gaussrelatz New Users 1 December 31st, 1969 04:00 PM

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