Supremum of Subset of Real Numbers is Arbitrarily Close
Jump to navigation
Jump to search
Theorem
Let $A \subseteq \R$ be a subset of the real numbers.
Let $b$ be a supremum of $A$.
Let $\epsilon \in \R_{>0}$.
Then:
- $\exists x \in A: b − x < \epsilon$
Proof
Note that $A$ is non-empty as the empty set does not admit a supremum (in $\R$).
Suppose $\epsilon \in \R_{>0}$ such that:
- $\forall x \in A: b − x \ge \epsilon$
Then:
- $\forall x \in A: b − \epsilon \ge x$
and so $b − \epsilon$ would be an upper bound of $A$ which is less than $b$.
But since $b$ is a supremum of $A$ there can be no such $b − \epsilon$.
From that contradiction it follows that:
- $\exists x \in A: b − x < \epsilon$
$\blacksquare$