Poset to total order via Zorn's lemma?

Question

This is a question from Burris/Sankappanavar's "Universal Algebra":

If < is a partial order on A, show that there is a total order <* on A such that if $a \in A$ and $ b \in A,$ then $ a < b \implies a<^* b. $ (Hint: Use Zorn's lemma.)

I'm struggling with this since Zorn's lemma requires that every chain have an upper bound before concluding that the poset must have a maximal element. Any ideas?

Answer 1

As the hint suggests this is done using Zorn's lemma, but not by applying it to the poset $(A, <)$. Let $P$ be the poset of all partial orders on $A$ extending $<$. That is, the elements of $P$ are partial orders $<'$ on $A$ such that $a < b$ implies $a <' b$ for all $a,b \in A$. We order the elements of $P$ by extension. That is, given partial orders $<_1$ and $<_2$ on $A$ we say that $<_1$ is less than (or equal to) $<_2$ in $P$ if $a <_1 b$ implies $a <_2 b$ for all $a, b \in A$.

Now it is straightforward to check that $P$ is indeed a poset and that every chain has an upper bound. Also $P$ is clearly not empty as $<$ is in $P$. So we can apply Zorn's lemma and obtain a maximal element $<^*$, we claim that $<^*$ is a total order. From here on I will work with $\leq^*$ instead of $<^*$ as that is more convenient. If $\leq^*$ weren't a total order then there are $a,b \in A$ such that $a \not \leq^* b$ and $b \not \leq^* a$. we can define a partial order $\leq^{\#}$ on $A$ as follows: $c \leq^{\#} d$ iff either

1. $c \leq^* d$ or,
2. $c \leq^* a$ and $b \leq^* d$.

What this essentially does is force in the relation $a \leq^{\#} b$. We verify that $\leq^{\#}$ is a partial order on $A$.

• Reflexivity is clearly still true, as we only added relations compared to $\leq^*$.
• For transitivity we let $c_1 \leq^{\#} c_2 \leq^{\#} c_3$ and make a case distinction.
  • Both the first and second $\leq^{\#}$ are due to point 1 in the definition. So $c_1 \leq^* c_2 \leq^* c_3$, which means that $c_1 \leq^* c_3$ and hence $c_1 \leq^* c_3$.
  • One of the $\leq^{\#}$ relations is due to point 1 and the other is due to point 2. Say $c_1 \leq^* c_2$ and $c_2 \leq^* a$ and $b \leq^* c_3$. Then also $c_1 \leq^* a$ and hence $c_1 \leq^{\#} c_3$. The other case is analogous.
  • Both $\leq^{\#}$ relations are due to piont 2. This is in fact impossible, as that would imply that $b \leq^* c_2 \leq^* a$ and hence $b \leq^* a$.
• For antisymmetry we let $c \leq^{\#} d$ and $d \leq^{\#} c$. Again we split in cases.
  • If both $\leq^{\#}$ relations are due to point 1. Then $c = d$ follows from antisymmetry of $\leq^*$.
  • If one of the $\leq^{\#}$ relations is due to point 1 and the other due to point 2, say $c \leq^* d$ and $d \leq^* a$ and $b \leq^* c$, then we have $b \leq^* a$, which is impossible.
  • If both $\leq^{\#}$ relations are due to point 2 then we have in particular that $b \leq^* c \leq^* a$, which is again impossible.

So we conclude that $\leq^{\#}$ is indeed a partial order on $A$ and due to point 1 in its definition it extends $\leq^*$. At the same time $\leq^{\#}$ is not the same as $\leq^*$ because it will satisfy $a \leq^{\#} b$. This contradicts maximality of $\leq^*$, so $\leq^*$ must have been total.