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:
\(\ds \forall \epsilon \in \R_{>0}: \, \) | \(\ds 0\) | \(<\) | \(\ds \epsilon\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \forall \epsilon \in \R_{>0}: \forall t \in T: \, \) | \(\ds t\) | \(<\) | \(\ds t + \epsilon\) | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \forall \epsilon \in \R_{>0}: \forall t \in T: \, \) | \(\ds t\) | \(<\) | \(\ds s + \epsilon\) | ||||||||||
\(\, \ds \land \, \) | \(\ds s\) | \(=\) | \(\ds t\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \forall \epsilon \in \R_{>0}: \forall t \in T: \exists s \in S: \, \) | \(\ds t\) | \(<\) | \(\ds s + \epsilon\) | as $T \subseteq S$ | |||||||||
\(\ds \leadsto \ \ \) | \(\ds \sup T\) | \(\le\) | \(\ds \sup S\) |
$\blacksquare$