Proof verification: If $A$ is an ordered set with the least upper bound property, then $A$ also has the greatest lower bound property.

Question

Let $A$ be an ordered set with the least upper bound property. Let $B \subset A$ be a nonempty set that is bounded below, and define $$L = \{ x \in A \mid \forall b \in B, x \le b \}.$$ Let $l \in L$, and so $l$ is a lower bound of $B$, which implies that $l \le \inf B$, so $L$ is bounded above by $\inf B$. Because $A$ has the least upper bound property, it follows that $\sup L \le \inf B$. Let $\beta \in A$ be a lower bound of $B$, then $\beta \in L$ by construction, and so $\beta \le \sup L$, but this is true for any lower bound of $B$, including the greatest lower bound.

Answer 1

I hope you've solved this but if you haven't......

So, you need to do an explicit construction here. You've correctly defined the set of all lower bounds of your non-empty set $B$: $$L = \{x \in A| \forall b \in B: x \leq b \}$$

Now, instead of bounding $L$ by $\inf(B)$ (whose existence you're trying to prove), just take any element of $B$, say $b$. Then: $$\forall x \in L: x \leq b$$

This follows by definition of $L$. But now, notice that $L$ is non-empty (because $B$ is bounded below) and that it is bounded above. By the Least Upper Bound Property, it follows that $\sup(L)$ exists. Now, we claim that $\sup(L)$ is the greatest lower bound. I'll leave you to prove that this is, indeed, the case.