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October 28th, 2017, 11:24 AM   #1
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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.
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October 28th, 2017, 04:08 PM   #2
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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.
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October 28th, 2017, 06:22 PM   #3
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Quote:
Originally Posted by Mathmatizer View Post
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.
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October 28th, 2017, 11:41 PM   #4
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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
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October 29th, 2017, 04:26 AM   #5
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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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