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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.

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  • $\begingroup$ "How come there is no loss of generality here?" Because there's a proof? What you asking for the proof? You seem to know it, so what exactly do you want? Edit: Regarding the second part of the question, you start with a complete ordered field $\mathbb F$ and you take a subset $\mathbb K$ of $\mathbb F$ with the lup property. Then you want to prove that any subset $\mathbb L$ of $\mathbb F$ has the lup property, is that it? $\endgroup$ – Git Gud Dec 5 '14 at 15:12
  • $\begingroup$ @user130018, I am afraid that you would have to be more specific. $\endgroup$ – Manolito Pérez Dec 5 '14 at 15:13
  • $\begingroup$ I see that the proof shows the infimum of one set is the supremum of another set, but I don't understand why this implies that every set with a lower bound necessarily has an infimum $\endgroup$ – mr eyeglasses Dec 5 '14 at 15:15
  • $\begingroup$ @user130018 Would a proof of what's mentioned in the comment above be an answer to your question? $\endgroup$ – Git Gud Dec 5 '14 at 15:16
  • $\begingroup$ @GitGud We start with a complete ordered field $\mathbb{F}$ and we take a subset $\mathbb{K}$ of $\mathbb{F}$ with the least upper bound property. Then we want to prove that any subset $\mathbb{L}$ of $\mathbb{F}$ has the greatest lower bound property. If this is what you meant, then yes $\endgroup$ – mr eyeglasses Dec 5 '14 at 15:18
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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.

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  • $\begingroup$ What about partially ordered sets with the lub property, such as finite lattices? Wouldn't this proof still hold? $\endgroup$ – Arthur Dec 5 '14 at 16:12
  • $\begingroup$ @Arthur The claim certainly holds (but I'd use the fact that any finite lattice is complete instead). The proof also holds for any poset $P$, it doesn't need to be a chain. At first I thought it was necessary that $P$ was a chain because of the proof I envisioned, but then I changed my proof and your comment made me realise that this hypothesis can be relaxed. $\endgroup$ – Git Gud Dec 5 '14 at 16:23

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