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.
[Math] Question about supremum $\implies$ infimum
order-theoryordered-fieldsreal-analysissupremum-and-infimum
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"Least upper bound" property is that every nonempty set that is bounded above has a least upper bound; dually for "greatest lower bound", so it is only required that nonempty sets have the property.
(For example, the real numbers have the least upper bound property; if you also required the empty set to have a least upper bound, this would require the reals to have a least element).
Yes, Dedekind completeness is the same thing as the least upper bound property.
For 2: If $S$ is a nonempty set that is bounded below, let $B$ be the set of lower bounds of $S$. Show that $B$ is (i) nonempty; and (ii) bounded above. Conclude that $B$ has a least upper bound. Show that the least upper bound of $B$ is also the greatest lower bound of $S$. The converse is proven dually: the least upper bound of a nonempty set that is bounded above is equal to the greatest lower bound of the set of upper bounds.
The idea is correct, but could be written down more clearly:
Let $A$ be an ordered set that satisfies the lub property. Then $A$ satisfies the glb property:
Let $B$ be a non-empty subset of $A$ that is bounded below, define
$$L(B) = \{x \in A: \forall b \in B: x \le b\}$$ which is the set of lower bounds for $B$, which is by assumption non-empty.
Then for any fixed $b_0 \in B$ ($B$ is non-empty), and any $l \in L(B)$ we have $l \le b_0$. But this says that $L(B)$ is bounded above. So the lub property says that $l_0 = \operatorname{lub}(L(B))$ exists. Claim: $l_0 = \operatorname{glb}(B)$.
For this we need to show two things: $l_0$ is a lower bound for $B$ and there is no greater lower bound for $B$.
So let $b \in B$. Then, as before, $b$ as before is an upper bound for $L(B)$ by the definition of $L(B)$, and as $l_0$ is the least upper bound for $L(B)$ we have $l_0 \le b$. As $b$ was arbitary, $l_0$ is a lower bound for $B$.
Now suppose $m$ is any lower bound for $B$. Then again by definition, $m \in L(B)$. As $l_0$ is the least upper bound for $L(B)$, it is an upperbound for $L(B)$ so $m \le l_0$. So all lower bounds for $B$ are below $l_0$.
Together this says that $l_0 = \operatorname{glb}(B)$ and so $A$ satisfies the glb-property.
The reverse is indeed similar: $\operatorname{lub}(B) = \operatorname{glb}(U(B))$, where $U(B)$ is the set of upperbounds for $B$.
Best Answer
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.