Set of partial orderings and Zorn's lemma

Question

Let $X$ be a set and let $P$ be the set of all partial orderings of $X$. ⪯⪯$(P, \leq_p)$ where $\leq_p$ is the ordering such that $\leq\leq_p \leq '$ if $(x\leq y)\Rightarrow (x\leq ' y) \forall x,y\in X$ is a poset. I want to show, using Zorn's lemma, that given any partial ordering $\leq$ there exists a total ordering $\leq '$ such that $\leq\leq_p\leq '$; to accomplish this I tried to apply Zorn's lemma to the set $C:=\{\leq '':\leq\leq_p\leq ''\}$ I tried to show that every totally ordered subset $D$ of $C$ an upper bound (defined in the ⪯text as an element $x\in D$ such that $y\leq_p x \forall y\in D$) but I haven't been able to do so ( I tried by defining an ordering $\leq_m$ such that $x\leq_m y \Leftrightarrow \exists \leq \in D$ such that $x\leq y$ which should be an upper bound but I haven't been able to show that such an ordering belongs to D i.e. that it is an upper bound and not a strict upper bound).

(Note: I know that another user asked a similar question in an almost three years old post and tried a similar approach in proving this fact but he didn't explain why the ordering $\leq_m$ is an upper bound for $D$, that's why I'm asking this question)

So I'd appreciate any help in finishing this proof.

Best regards,

lorenzo.

Answer 1

In Zorn's lemma, an upper bound of a set $D$ does not need to be an element of $D$. That is, an upper bound of $D$ is defined to be just an element $c\in C$ such that $d\leq c$ for all $d\in D$. So you don't need to show that $\leq_m$ is an element of $D$; you've already shown that you can apply Zorn's lemma to find a maximal element of $C$.

Answer 2

As Eric points out, an upper bound need not be an element of the set that it is bounding.

In $\Bbb R$, $1$ is an upper bound of $(0,1)$ but still $1\notin(0,1)$. It is true that $1$ is not an upper bound if we only consider the partial order $(0,1)$.

To recap, an upper bound of $D$ in $P$, is an element $p\in P$, such that for all $d\in D$, $d\leq p$. But there is no requirement that $p\in D$.

In the case of partial orders, the increasing union of partial orders is indeed a partial order. Of course, it has no reason to be an element of the chain, but it is a partial order. So the collection of partial orders (which extend some given $\leq$, if you want), satisfy the conditions for Zorn's lemma, and therefore there is a maximal element there. It is not hard to check that this maximal element is a linear order.