Supremum of Subset of Real Numbers is Arbitrarily Close

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:

$\exists x \in A: b − x < \epsilon$

$\blacksquare$

Also see

• Infimum of Subset of Real Numbers is Arbitrarily Close