Will S be an empty set if the least upper bound is equal to the greatest lower bound?

Question

Will S be an empty set if the least upper bound is equal to the greatest lower bound?

Also will S be an empty set if the maximum is equal to the minimum?

Answer 1

No.

In the reals empty sets don't have sups or infs or max or mins. In the extended reals $\sup \emptyset = -\infty \ne \infty = \inf \emptyset$. So this is not a property of empty sets.

So if $\sup A = \inf A$ then $A \ne \emptyset$. The real question is, does there exist any set such that $\sup A = \inf A$ or is that an impossibility?

If a set $A$ has $\max A = \min A$ then $\max A \in A$ and $\min A \in A$ and for any $x \in A$, $x \ge \min A = \max A$ and $x \le \min A= \max A$ so by exclusion principal of order $x = \min A = \max A$ so $A$ has exactly one element.

If a set $A$ is such that $\inf A = \sup A$ then means for any $y < \inf A$ there exists an $x_0\in A$ so that $y < x_0 \le \inf A=\sup A$. And for any $w > \sup A $ then there exists a $x_1\in$ so that $\inf A = \sup A \le x_1 < w$.

But if $x_0 < \inf A$ it is not in $A$. So $x_0 = \inf A$. And if $x_1 > \sup A$ is is not in $A$. So $x_1 = \sup A$.

So $A$ will contain only one element.

The only sets with this propertys are the sets with one elements $A = \{x\}$ and $x = \inf A = \sup A = \min A = \max A$.