Greatest Lower Bound Property/Proof 2

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


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

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 supremum that:


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

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

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


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

and so $B$ is the infimum of $S$.