# 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:

$\sup T \le \sup S$

## Proof

$S$ 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.

Therefore by Suprema of two Real Sets:

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

We have:

 $\displaystyle \forall \epsilon$ $\in$ $\displaystyle \R_{>0}: 0 < \epsilon$ $\displaystyle \leadsto \ \$ $\displaystyle \forall \epsilon$ $\in$ $\displaystyle \R_{>0}: \forall t \in T: t < t + \epsilon$ $\displaystyle \leadsto \ \$ $\displaystyle \forall \epsilon$ $\in$ $\displaystyle \R_{>0}: \forall t \in T: t < s + \epsilon \land s = t$ $\displaystyle \leadsto \ \$ $\displaystyle \forall \epsilon$ $\in$ $\displaystyle \R_{>0}: \forall t \in T: \exists s \in S: t < s + \epsilon$ as $T \subseteq S$ $\displaystyle \leadsto \ \$ $\displaystyle \sup T$ $\le$ $\displaystyle \sup S$

$\blacksquare$