In Exercise 13, section 3 on page 29 of Munkres’ Topology 2E, the problem is stated as follows.
Prove the following:
Theorem. If an order set $A$ has the least upper bound property, then it has the greatest lower bound property.
The following is my attempt to prove the above theorem.
Proof:
Let $A$ be an order set that has the least upper bound property. Suppose $E$ be a nonempty subset of $A$ that does not have the greatest lower bound.
Then, consequently, $ E^c $ the complement of $E$ which is also a nonempty subset of $A$ will not have the least upper bound, contrary to the fact that $A$ has the least upper bound property.
Thus, every nonempty subset of $A$ must have the greatest lower bound. Thus, we conclude that $A$ has the greatest lower bound property. ∎
I’d like to know whether it is a preferable one or not, and if it isn’t, I’d like to request your counter-part if you don’t mind.
If there is already a similar question, I'd like to keep it as a duplicate one.