This is a follow-up to my question here. Let $A$ and $B$ be sets of real numbers. My question is, under what circumstances is the supremum of $A$ equal to the supremum 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 supremum of $B$ if and only if for any $\epsilon>0$ there exists a $b\in B$ such that $y\geq b>y - \epsilon$. Is there any way to combine these two conditions into a single condition for when the supremum of $A$ is equal to the supremum of $B$? Some kind of inequality involving elements of $A$ and elements of $B$?
Keep in mind that "some kind of inequality" in this answer to the earlier question was actually two inequalities. One inequality was $a\leq b$ and the other inequality was $b \leq a+\epsilon.$
For an answer with just two inequalities (aside from the obligatory $\epsilon > 0$) involving only $\epsilon$ and members of the two sets, try this:
Condition E: For every $\epsilon > 0,$ there exists an $a \in A$ such that $b < a + \epsilon$ for every $b \in B,$ and there exists a $b' \in B$ such that $a' < b' + \epsilon$ for every $a' \in A.$
Rationale:
The mere existence of one pair $(\epsilon, a)$ such that $b < a + \epsilon$ for every $b \in B$ means that $B$ has an upper bound, and therefore $B$ has a supremum. A similar argument shows that $A$ has a supremum.
Now suppose that $\sup A < \sup B.$ Then for $\epsilon = \frac12(\sup B - \sup A),$ there exists $b \in B$ such that $b > \sup B - \epsilon = \sup A + \epsilon,$ and therefore $b > a + \epsilon$ for every $a \in A.$ But that contradicts Condition E, above; therefore Condition E implies that $\sup A \geq \sup B.$ A similar argument (interchanging $A$ and $B$) shows that $\sup B \geq \sup A.$
To express that two sets have the same infimum, we reverse the directions of the two inequalities and the direction in which we take the "epsilon":
For every $\epsilon > 0,$ there exists an $a \in A$ such that $b > a - \epsilon$ for every $b \in B,$ and there exists a $b' \in B$ such that $a' > b' - \epsilon$ for every $a' \in A.$
Looking at Condition E, it seems to involve more numbers than the condition in the answer to the earlier question, since I used the symbols $a'$ and $b'$; but technically, I could have used $a$ and $b$ again, I just chose to write the second set of quantifiers with new symbols for clarity. The earlier question's answer uses the logical conjunction of two separate statements, each of which is quantified using symbols $a$ and $b,$ which technically could just as well (and perhaps more clearly) have been written $a$ and $b$ in the first statement and $a'$ and $b'$ in the second.
In fact the inequality $a \leq b$ actually appears twice in that answer, quantified once for all $a$ and $b$ and quantified the second time for all $\epsilon$ and for some $a$ and $b.$ So that answer actually has three inequalities. For this question we make do with two.