Characterizing Property of Supremum of Subset of Real Numbers

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $S \subset \R$ be a non-empty subset of the real numbers.

Let $S$ be bounded above.

Let $\omega \in \R$.


The following are equivalent:

$(1): \quad \omega$ is the supremum of $S$
$(2): \quad \omega$ is an upper bound for $S$
and:
$\forall \epsilon \in \R_{> 0}$ there exists $x \in S$ with $x > \omega - \epsilon$


Proof

$(1)$ implies $(2)$

Let $\omega$ be the supremum of $S$.

Then by definition, $\omega$ is an upper bound for $S$.

Let $\epsilon > 0$.

Because $\omega - \epsilon < \omega$, it is not an upper bound for $S$.

Thus there exists $x\in S$ with $x > \omega - \epsilon$.

$\Box$


$(2)$ implies $(1)$

Let $\omega$ be an upper bound of $S$ such that $\forall \epsilon > 0$ there exists $x \in S$ with $x > \omega - \epsilon$.

Let $d \in \R$ be an upper bound of $S$.

We have to prove that $d \ge \omega$.

Aiming for a contradiction, suppose $d < \omega$.

Let $\epsilon = \omega - d > 0$.

Then there exists $x \in S$ such that $x > \omega - \left({\omega - d}\right) = d$.

But this contradicts our assumption that $d$ is an upper bound of $S$.

$\blacksquare$


Also see