# Supremum of Set of Real Numbers is at least Supremum of Subset

## 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 1

The number $\sup S$ is an upper bound for $S$.

Therefore, $\sup S$ is an upper bound for $T$ as $T$ is a non-empty subset of $S$.

Accordingly, $T$ has a supremum by the Continuum Property.

The number $\sup S$ is an upper bound for $T$.

Therefore, $\sup S$ is greater than or equal to $\sup T$ as $\sup T$ is the least upper bound of $T$.

$\blacksquare$

## Proof 2

By the Continuum Property, $T$ admits a supremum.

It follows from Supremum of Subset that $\sup T \le \sup S$.

$\blacksquare$

## Proof 3

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

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

$\blacksquare$

## Proof 4

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:

- $\sup T \le \sup S$

$\blacksquare$

## Sources

- 1977: K.G. Binmore:
*Mathematical Analysis: A Straightforward Approach*... (previous) ... (next): $\S 2$: Continuum Property: Exercise $\S 2.13 \ (1)$