Supremum of Set of Real Numbers is at least Supremum of Subset/Proof 4

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

By definition $\sup S$ is an upper bound for $S$.

Thus:

$\forall x \in S: x \le \sup S$

As $T \subseteq S$ we have by definition of subset that:

$\forall x \in T: x \in S$

Hence:

$\forall x \in T: x \le \sup S$

So by definition $\sup S$ is an upper bound for $T$.

So $\sup S$ is at least as big as the smallest upper bound for $T$

Thus by definition of supremum:

$\sup T \le \sup S$

$\blacksquare$