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

Proof
By the Continuum Property, $T$ admits a supremum.

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

By Suprema of two Real Sets, $\sup T \le \sup S$ :


 * $\forall \epsilon \in \R_{>0}: \forall t \in T: \exists s \in S: t < s + \epsilon$

So let $\epsilon \in \R_{>0}$ and $t \in T$ be arbitrary.

Then since $T \subseteq S$, also $t \in S$, and:


 * $t < t + \epsilon$

as desired.

The result follows.