Let us suppose that every subset of $\mathbb R$ that is bounded above has a lowest upper bound. I want to prove as a corollary from this axiom that every subset of $\mathbb R$ that is bounded below has a greatest upper bound. Here is some of the proof I've done:
"Let $S \subset \mathbb R$ be a non-empty set that is bounded below. Let $L \subset \mathbb R$ be the set containing all lower bounds of $S$. Then for all $m \in L, m \leq s$ for all $s \in S$. Thus, by definition, all elements of $S$ are upper bounds of $L$, so $L$ is bounded above. Hence, $L$ has a least upper bound, denoted $\alpha$."
My question is, how do you now show that $\alpha = inf(S)$ if $\alpha = sup(L)$? Thanks in advance.
If $\alpha=\sup L$, must we have$$(\forall s\in S):\alpha\leqslant s?\tag1$$Yes, because otherwise there would be a $s\in S$ such that $\alpha>s$. But then, since $\alpha=\sup L$, there must be some $l\in L$ such that $l>s$. This is impossible, since $L$ is the set of the lower bounds of $S$.
So, $(1)$ holds and therefore $\alpha$ is a lower bound of $S$. Must it be the greatest lower bound? Yes, because if $\beta$ is a lower bound of $S$, then, since $\alpha=\sup L$ and $\beta\in L$, $\beta\leqslant\alpha$.
Therefore, $\alpha$ is the greatest lower bound of $S$, which means that $\alpha=\inf S$.