Real Numbers are not Well-Ordered under Conventional Ordering

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\left({\R, \leqslant}\right)$ be the ordered structure consisting of the real numbers under the conventional ordering.

Then $\left({\R, \leqslant}\right)$ is not a well-ordered set.


Proof

Proof by Counterexample

Consider the set:

$A := \left\{ {x \in \R: x > 1}\right\}$

Suppose $a$ were the smallest element of $A$.

Then $a > 1$.

But then:

$1 < \dfrac {a + 1} 2 < a$

and so $a$ has been shown not to be the smallest element of $A$ after all.

Hence $A$ has no smallest element.

Thus there exists a subset of $\R$ which has no smallest element.

Thus $\R$ is not a well-ordered under $\leqslant$.

$\blacksquare$


Sources