Zorn's Lemma in Kelley's book: What is wrong in my reasoning of this seemingly "counterexample"?

Question

In this post I have stated how Zorn's Lemma is stated in Kelley's book. Different formulas of Zorn's Lemma However, consider the set $A=\{a_1, a_2, a_3,b_1,b_2,b_3\}$, with the partial ordering $<$(by Kelley's definition): for $a_i$, $a_i<a_j$ iff $i\le j$, $a_i<b_j$ iff $i\le j$, and for $b_i$, $b_i<a_j$ iff $i\le j$, $b_i<b_j$ iff $i\le j$. Then it seems $A$ satisfies the hypothesis of Zorn's Lemma on Kelley's book (Since every chain has upper bound $a_3$), but $A$ has no maximal elements since if we assume it has, then it can only be $a_3$ or $b_3$, but neither of them is. What is going wrong in my reasoning?

Answer 1

You are taking Kelley's definition of "partial order" (namely, "transitive relation"), but you are not taking Kelley's definition of "maximal element". Kelly writes (pp. 33)

A maximal element of the ordered set $A$ is an element $x$ such that $x$ follows each comparable element of $A$; that is, if $y\in A$, then either $y$ precedes $x$ or $x$ does not precede $y$.

He defines "precedes" on pp 13, where he says "If $\lt$ is an ordering and $x\lt y$, then it is customary to say that $x$ precedes $y$ or $x$ is less than $y$. So his definition is:

A maximal element of the ordered set $A$ is an element $x$ such that for every $y\in A$, either $y\lt x$ or else $x\not\lt y$.

With this definition, your claim that neither $b_3$ nor $a_3$ are maximal is incorrect. In fact, they are both maximal.

Indeed, take $a_3$, and let $y$ be any other element of $A$. Then $y\lt a_3$: for either $y=a_i$ with $i\leq 3$, so $y\lt a_3$; or else $y=b_j$ with $j\leq 3$, so $y\lt a_3$. So $a_3$ satisfies the definition. Likewise, since $x\lt b_3$ for every $x\in A$, it follows that $x$ is maximal under Kelley's definition. Thus, the set does have maximal elements.

Answer 2

Taking $i=j=1$ in: "$a_i<b_j$ iff. $i\le j$" and $i=j=1$ in "$b_i<a_j$ iff. $i\le j$" then you have: $$a_1<b_1<a_1$$So $(A,<)$ is not a poset as the antireflexivity (?) axiom is violated. If you wanted this to be a poset, you'd have to at least identify $b_1$ with $a_1$, or drop the $=$ in $i\le j$, just look at $i<j$.

But if we do that, then both $a_3$ and $b_3$ are maximal elements and Zorn's lemma is not shown to be false.

If you insist on a frankly bad definition of partial order, i.e. one where $a_1<b_1<a_1$ is permitted (a preorder e.g.) then the Zorn's lemma still holds so long as you interpret "maximal" as:

$m$ is maximal in the preorder (which is what $(A,<)$ is, it is not a poset) $(X,<)$ if for all $x\in X$ which are comparable to $m$, it is true that $x\le m$.

This statement leaves open the possibility that $x<m<x$. In this sense - and note your linked question didn't ask about "maximal" when it should have - your example is still not a counterexample as $a_3,b_3$ can be considered maximal. In the "separative quotient" (see your linked question) $a_3\sim b_3$ and $[a_3]$ would be maximal.