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This question arises from a homework problem asking if $\sup A < \inf B$ means that there is a $c$ such that $a < c < b$ for all $a,b$ in $A,B \subset R$. But that's not that difficult to answer and I was wondering what properties the sets must have for that to be true. Eg. if $A,B \subset U$ what properties must $U$ have such that there must be a $c$.

My inclination is that $U$ must be dense in itself for this to true but I don't really know how to prove this or if this is too strong/weak of a condition.

Thanks for the help

Zaros
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1 Answers1

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Assuming that both $A\subset U$ and $B\subset U$ are non-empty (otherwise defining their infimum and supremum becomes somewhat tricky), and assuming you are looking for a $c\in U$, it is necessary and sufficient that at least one of the following be true:

  1. $\sup A \notin A$ but $\sup A\in U$.
  2. $\inf B \notin B$ but $\sup B\in U$.
  3. $(\sup A,\inf B)\cap U\neq\emptyset$

It is very easy to prove that any of these $3$ conditions by itself is sufficient, simply by taking as $c$ respectively $\sup A$, $\inf B$, or an element of $(\sup A,\inf B)$. It is almost as easy to prove that at least one of them is necessary.

Note that $U$ being dense in (an arbitrary subinterval of) $(\sup A,\inf B)$ is sufficient to guarantee $3$, but it's not necessary (take $A=\{1\}$, $B=\{3\}$ and $U=\{1,2,3\}$).

Anonymous
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  • Thanks for your answer. Is there a way to characterize a space with the third property. Let's say I'm working with arbitrary sets in a non well ordered set so 1 and 2 aren't necessarily true. Eg. what does it mean about the space if an interval (a,b) is empty when $a<b$? – Zaros Jan 22 '17 at 07:04
  • Actually well ordered isn't the term I was looking for but you get the idea. Since I'm working with arbitrary sets not all sets necessarily contain their sup/inf. – Zaros Jan 22 '17 at 07:06
  • I don't think there's a lot more that you can say beyond 3. It's obvious that if $U$ is dense in $(x,y)$, then it's non-empty there, but the opposite is just as obviously not necessarily true. However, if for any choice of $x$ and $y$ in some interval $(t,u)$, $U$ is non-empty in $(x,y)$, then $U$ is (by definition) dense in $(t,u)$. – Anonymous Jan 22 '17 at 07:38