Supremum of Set of Real Numbers is at least Supremum of Subset/Proof 3

Proof
$S$ is bounded above as $S$ has a supremum.

Therefore, $T$ is bounded above as $T$ is a subset of $S$.

Accordingly, $T$ admits a supremum by the Continuum Property as $T$ is non-empty.

We know that $\sup T$ and $\sup S$ exist.

Therefore:


 * $\forall \epsilon \in \R_{>0}: \forall t \in T: \exists s \in S: t < s + \epsilon \iff \sup T \le \sup S$ by Suprema of two Real Sets

We have: