Prove the set $S$ satisfies lub property iff it satisfies glb property

I've been stuck on this and although I have gone to professors for advice, I can't grasp it. Here I tried re-doing my proof and I can bet it is certainly wrong, but I am burnt out. I would like to know not only my mistakes in the proof, as I have looked at it for an hour and can't seem to go anywhere; I would also like a throrough explanation of this in layman terms, because that is exactly what I need right now. What I know is that the infimum of a set bounded from above is the same as the supremum of the set bounded from below. I have no idea on how to construct the argument thoroughly.

Prove that if the ordered set $S, <$ satisfies the greatest lower bound property, then it satisfies the least upper bound property.

Let $B$ be a nonempty subset of $S$ that is bounded below. Let $a \in A$ be the set of all lower bounds of $B$. By definition $A$ is nonempty. There also exists an element $b \in B$ for all $a$ such that $a \leq b$, so $A$ is bounded above (by $b$). Since $S$ satisfies the greatest lower bound property, $inf \hspace{2 pt} B = \alpha$ exists for all $a$ such that $a \leq \alpha \leq b$. If $\alpha < a$ otherwise, then $a$ is not a lower bound of $B$. But if $\beta < \alpha$, then $\beta \not\in B$, since $\alpha$ bounds $B$ from below. Thus $a \leq \alpha$, and so $A$ has a least upper bound in $S$. QED.

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Your proof is mostly OK, except for the fact that it's not a proof of your original statement.

You are trying to prove that any set with greatest lower bounds automatically has least upper bounds. In your proof, you fix an arbitrary subset $B$ of a set $S$ with greatest lower bounds, and then show that a completely different set $A$ has a least upper bound. Instead, you need to show that $B$ has least upper bounds. Let's see how we can alter your proof so that it proves your original statement.

Let $B$ be a nonempty subset of $S$ that is bounded below.

This is (almost) a great place to start. If you're trying to prove something about all objects in a class, it is always good to start by fixing an arbitrary one. But here, you want to show that any non-empty bounded above subset of $S$ has a least upper bound, so you should be fixing $B$ to be a non-empty bounded above subset of $S$.

Let $a\in A$ be the set of all lower bounds of $B$.

First of all, I think you meant to say 'Let $A$ be the set of all lower bounds of $B$'. But even this is not what you want to do. Fixing $A$ to be the set of lower bounds of $B$ can't help you at all, since the only interesting thing you can say about $A$ is that it has a greatest element ($\inf B$), which is equivalent to $B$ having a greatest lower bound. You certainly can't apply the greatest lower bound property to $A$ - you don't even know that $A$ is bounded and $\inf A$, if it exists, is certainly unrelated to $B$.

At some point, you have to apply the greatest lower bound property to some set $C\subset S$, since you can't conclude that $S$ has the least upper bound property if you don't know that it has the greatest lower bound property (e.g., $\mathbb Q$ does not have the greatest lower bound property), but you have to apply it to a set that you know has a greatest lower bound; i.e., a set that is bounded below. Furthermore, it's going to have to be a set of elements that are $\ge$ the elements of $B$; otherwise, it'll have nothing to do with upper bounds of $B$ at all.

The solution, of course, is to fix $C$ to be the set of upper bounds for $B$. Now we're trying to show that $C$ has a least element.

There also exists an element $b\in B$ for all $a$ such that $a\le b$, so $A$ is bounded above (by $b$).

First of all, we aren't interested in saying that $A$ (or $C$) is bounded above, since being bounded above doesn't tell us anything. Being bounded below, of course, tells us that a set has a greatest lower bound. So we want to show that $C$ is bounded below. Secondly, it is ambiguous whether you mean that there is some $b$ that works for all $A$ or whether there is a separate $b(a)$ for each $a$. Both are true, but only the first implies your conclusion.

I'd prefer to write, 'Fix an arbitrary $b\in B$. Then for each $c\in C$, $c$ is an upper bound for $B$ so, in particular, $b\le c$. So $C$ is bounded below by $b$.'

Since $S$ satisfies the greatest lower bound property, $\inf \hspace{2 pt} B = \alpha$ exists for all $a$ such that $a \leq \alpha \leq b$.

This is where you started to go off the rails. Knowing that $B$ has a greatest lower bound can tell you nothing about whether it has a least upper bound. For example, the set $\{3,3.1,3.14,3.141,3.1415,3.14159,\dots\}$ has a greatest lower bound but no least upper bound in $\mathbb Q$.

I think I've given you enough clues for you to finish the proof on your own. For reference, here is my proof of the statement:

Let $B$ be a non-empty subset of $S$ that is bounded above, and let $C$ be the set of all upper bounds for $B$. Fix $b\in B$; then, for each $c\in C$, $c$ is an upper bound for $B$ so, in particular, $b\le c$. So $C$ is bounded below by $b$, which means that it has a greatest lower bound $\gamma$. Since every $b\in B$ is a lower bound for $C$, $\gamma\ge b$ for each $b\in B$. So $\gamma$ is an upper bound for $B$, and is clearly the least upper bound since it is a lower bound for the set of upper bounds. $\Box$

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