Let $A$ and $B$ be sets of real numbers. My question is, under what circumstances is the supremum of $A$ equal to the infimum of $B$?

Now $x$ is the supremum of $A$ if and only if for any $\epsilon>0$ there exists an $a\in A$ such that $x\geq a>x-\epsilon$. And $y$ is the infimum of $B$ if and only if there exists a $b\in B$ such that $y\leq b<y+\epsilon$. Is there any way to combine these two conditions into a single condition for when the supremum of $A$ is equal to the infimum of $B$? Some kind of inequality involving elements of $A$ and elements of $B$?


1 Answer 1


You might try to ask for something like $$\forall a\in A\forall b\in B(a\leq b)\wedge\forall \epsilon\exists a\in A\exists b\in B (a\leq b\leq a+\epsilon),$$ which actually means that $A$ is below $B$ as a set and thee two sets have elements that are arbitrarily close. But I am not sure if this exactly what you are looking for.

Since every element of $A$ is below $B$, the supremum of $A$ $x$ cannot be above the infimum of $B$ $y$, i.e. $x\leq y$. But if we had $x<y$, then say using $\epsilon=|x-y|/2$, we cannot find any element of $B$ at distance less than $\epsilon$ from $x$ (and so from every element of $A$), contradicting the formula above. This proves that the formula above implies that $x=y$.

On the other hand, if $x=y$, given any $\epsilon$, we can find $a$ and $b$ with distance from $x$ less than $\epsilon/2$. In particular, $a\leq b\leq a+\epsilon$. Moreover, clearly every element of $A$ is not above $x$ and every element of $B$ is not below it, so the formula follows.

  • $\begingroup$ This is indeed the kind of thing I'm looking for. But is it correct? $\endgroup$ Oct 25, 2018 at 23:06
  • $\begingroup$ It is. I will edit my answer to make it clearer. $\endgroup$
    – Leo163
    Oct 25, 2018 at 23:08
  • $\begingroup$ So is it a necessary and sufficient condition? $\endgroup$ Oct 25, 2018 at 23:11
  • $\begingroup$ Yes it is. I added the proof. $\endgroup$
    – Leo163
    Oct 25, 2018 at 23:15
  • $\begingroup$ OK thanks for your answer. $\endgroup$ Oct 25, 2018 at 23:17

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