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

Theorem

Let $S$ be a set of real numbers.

Let $S$ have a supremum.

Let $T$ be a non-empty subset of $S$.

Then $\sup T$ exists and:

$\sup T \le \sup S$

$S$ 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.

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

We have:

$\ds \forall \epsilon \in \R_{>0}: \, $

$\ds 0$

$<$

$\ds \epsilon$

$\ds \leadsto \ \ $

$\ds \forall \epsilon \in \R_{>0}: \forall t \in T: \, $

$\ds t$

$<$

$\ds t + \epsilon$

$\ds \leadsto \ \ $

$\ds \forall \epsilon \in \R_{>0}: \forall t \in T: \, $

$\ds t$

$<$

$\ds s + \epsilon$

$\, \ds \land \, $

$\ds s$

$=$

$\ds t$

$\ds \leadsto \ \ $

$\ds \forall \epsilon \in \R_{>0}: \forall t \in T: \exists s \in S: \, $

$\ds t$

$<$

$\ds s + \epsilon$

as $T \subseteq S$

$\ds \leadsto \ \ $

$\ds \sup T$

$\le$

$\ds \sup S$

$\blacksquare$