Supremum of Set of Real Numbers is at least Supremum of Subset/Proof 2
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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 the Continuum Property, $T$ admits a supremum.
It follows from Supremum of Subset that $\sup T \le \sup S$.
$\blacksquare$