Set of Non-Negative Real Numbers is not Well-Ordered by Usual Ordering
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Theorem
The set of non-negative real numbers $\R_{\ge 0}$ is not well-ordered under the usual ordering $\le$.
Proof
Aiming for a contradiction, suppose $\R_{\ge 0}$ is a well-ordered set.
Then by definition, all subsets of $\R_{\ge 0}$ has a smallest element.
Take the subset $\R_{> 0}$:
- $\R_{> 0} = \left\{ {x \in \R_{\ge 0}: x > 0}\right\} = \R_{\ge 0} \setminus \left\{ {0}\right\}$
Suppose $x \in \R_{> 0}$ is a smallest element.
Then $\dfrac x 2 \in \R_{> 0}$.
But $\dfrac x 2 < x$, which contradicts the supposition that $x \in \R_{> 0}$ is a smallest element.
Hence there can be no such smallest element.
So by Proof by Contradiction, $\R_{\ge 0}$ is not well-ordered by $\le$.
$\blacksquare$
Sources
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.1$: Mathematical Induction: Exercise $15 \ \text{(c)}$