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

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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

$S$ 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 $T \subseteq S$ | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \sup T\) | \(\le\) | \(\displaystyle \sup S\) |

$\blacksquare$