Set of Strictly Positive Real Numbers has no Smallest Element
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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 $\S 2.10 \ (6)$