Zorn's Lemma Implies Zermelo's Well-Ordering Theorem/Mistake

Source Work

 * Prologue
 * $2$. Orderings: (P.3)
 * $2$. Orderings: (P.3)

Mistake

 * Every nonempty set $X$ can be well-ordered.


 * Consider the collection $\WW$ of well orderings of subsets of $X$.
 * Such well orderings may be regarded as subsets of $X \times X$, so $\WW$ is partially ordered by inclusion.
 * It is easy to verify that the hypotheses of Zorn's lemma are satisfied, so $\WW$ has a maximal element.

Correction
It is insufficient for the well-orderings to be ordered by inclusion $\subseteq$.

We additionally need some kind of stability of the initial segment through a chain, for example as established in the main proof.

Proceeding with just $\subseteq$ as partial ordering of $\WW$, consider the sequence of well-ordered sets:


 * $W_n = \set{ -i: 0 < i \le n} \subseteq \Z$

endowed with the standard ordering.

Then $\sequence{ W_n }_{n \in \N}$ is a chain in $\WW$, but there is no upper bound, for:


 * $\ds \bigcup_{n \in \N} W_n = \set{ -i: i \in \N_{>0} } = \Z_{<0}$

has no smallest element with respect to $<$.

Hence the hypotheses of Zorn's Lemma are not satisfied.