Supremum of Set of Real Numbers is at least Supremum of Subset/Proof 1

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:

$\sup T \le \sup S$

Proof

The number $\sup S$ is an upper bound for $S$.

Therefore, $\sup S$ is an upper bound for $T$ as $T$ is a non-empty subset of $S$.

Accordingly, $T$ has a supremum by the Continuum Property.

The number $\sup S$ is an upper bound for $T$.

Therefore, $\sup S$ is greater than or equal to $\sup T$ as $\sup T$ is the least upper bound of $T$.

$\blacksquare$