Set of Strictly Positive Real Numbers has no Smallest Element

Theorem

Let $\R_{>0}$ denote the set of strictly positive real numbers.

Then $\R_{>0}$ has no smallest element.

Proof

Aiming for a contradiction, suppose $\R_{>0}$ has a smallest element.

Let $m$ be that smallest element.

Then we have that:

$0 < \dfrac m 2 < m$

But as $0 < \dfrac m 2$ it follows that $\dfrac m 2 \in \R_{>0}$.

This contradicts our assertion that $m$ is the smallest element of $\R_{>0}$.

Hence the result by Proof by Contradiction.

$\blacksquare$

Sources

• 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 2$: Continuum Property: Exercise $\S 2.10 \ (6)$