Characterizing Property of Supremum of Subset of Real Numbers

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

Let $S$ be bounded above.

Let $\omega \in \R$.


 * 1) $\omega$ is the supremum of $S$
 * 2) * $\omega$ is an upper bound for $S$ and
 * 3) *$\forall \epsilon>0$ there exists $x \in S$ with $x > \omega - \epsilon$
 * 1) *$\forall \epsilon>0$ there exists $x \in S$ with $x > \omega - \epsilon$

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 $x > \omega - \epsilon$.

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 $d \geq \omega$.

Suppose $d < \omega$.

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

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

But then $d$ is not an upper bound of $S$.

Also see

 * Characterizing Property of Infimum of Subset of Real Numbers