Greatest Lower Bound Property/Proof 2

Proof
Let $T$ be the set of lower bounds of $S$:


 * $T=\set {x \in \R: x\leq \forall y \in S }$

Since $S$ is bounded below, $T$ is non-empty.

Now, every $x\in T$ and $y\in S$ satisfy $x\leq y$.

That is, $T$ is bounded above by every element of $S$.

By the Continuum Property, $T$ admits a supremum in $\R$.

Let $B=\sup T$.

Since every element of $S$ is an upper bound of $T$ it follows from the definition of least upper bounds that:


 * $\forall y\in S : y\geq B$

Thus $B$ is a lower bound of $S$.

Because $B$ is an upper bound of $T$:


 * $\forall x\in T : x \leq B$

and so $B$ is the greatest lower bound of $S$.