Question about supremum $\implies$ infimum

Question

For all the proofs I've seen that if a set has the least upper bound property, then that set also has the greatest lower bound property, they assert something like if a set has lower bounds, then the elements of that set are upper bounds for the set of lower bounds and proceed the proof from there. How come there is no loss of generality here? I wanted to prove that given a subset in a complete ordered field that has the least upper bound property, any other arbitrary subset in that field with a lower bound has the greatest lower bound property, not just show the specific case that the supremum of one subset is the infimum of another subset.

Answer 1

I'm interpreting the question as asking to prove that given an ordered field $\mathbb F$, if $\mathbb F$ has the lub property, then $\mathbb F$ has the glb property. Or equivalentely that if all non-empty bounded above subsets of $\mathbb F$ have a supremum, then all non-empty and bounded below subsets of $\mathbb F$ have an infimum.

Ordered fields are irrelevant here, this can be generalized to any poset.

Let $P$ be a poset such that each of its non-empty and bounded above subsets has a supremum.

The goal is to prove that any non-empty and bounded below subset $A$ of $P$ has an infimum.

Being a universal statement, one way to prove it is to start by taking an arbitrary non-empty and bounded below subset $A$ of $P$.

Now consider the set $\downarrow A$, where $\downarrow A:=\{p\in P\colon \forall a\in A(p\leq a)\}$, that is, consider the set the lower bounds of $A$.

The set $\downarrow A$ isn't empty because $A$ is bounded below. It is also bounded above by any element of $A$.

Hence it's possible to use the hypothesis that any non-empty and bounded above subset of $P$ has a supremum particularized to $\downarrow A$.

Let $s:=\sup\left(\downarrow A\right)$.

Claim: $s=\inf(A)$.

Proof: It suffices to prove that $s$ is a lower bound of $A$ and that $\forall p\in \downarrow A(p\leq s)$. The latter part follows immediately from the fact that $s$ is an upper bound of $\downarrow A$. The first part follows from the fact that any element of $A$ is an upperbound of $\downarrow A$ and hence greater than the supremum of $\downarrow A$ which is $s$.

Since $A$ was an arbitrary bounded below and non-empty subset of $P$, it wasproved that $P$ has the glp property.