Extending a partial order to a total/linear order using Zorn's Lemma Here is the question. Given a partially ordered set $(A, \preccurlyeq)$ show that there is a total/linear order $\leq$ on $A$ such that $a \preccurlyeq b$ implies $a \leq b$ for all $a$ and $b$ in $A$, i.e. the total order extends that partial order.
First, I know there are several other questions about this here, namely here (referred to later as Question A), and here. I am posting a new question because the answers for those do not address specifically where I am getting stuck. Well, Question A does but I believe the posted answer is incorrect or at least incomplete, see my comment on the answer there.
I am trying to use the approach in the previous questions, in particular Question A, though I have changed the notation a little to avoid some ambiguity present there. The approach is to define the set
$$
L = \{(B, \leq_B) \,|\, \text{$B \subseteq A$ and $\leq_B$ is a linear order on $B$ such} \\
\text{that $x \preccurlyeq y$ implies $x \leq_B y$ for all $x$ and $y$ in $B$}\} \,.
$$
We can then define a relation $\trianglelefteq$ on $L$ as follows:
$$
\text{$(B, \leq_B) \trianglelefteq (C, \leq_C)$ if and only if $B \subseteq C$ and $\leq_B \subseteq \leq_C$}
$$
for any $(B, \leq_B)$ and $(C, \leq_C)$ in $L$.
It is easy to show that $\trianglelefteq$ is a partial order on $L$ and that each chain of $L$ has an upper bound, thereby meeting the conditions of Zorn's Lemma.
Therefore $L$ has a maximal element $(\bar{A}, \leq_{\bar{A}})$. The problem I am having is in actually showing that $\bar{A} = A$ so that $\leq_{\bar{A}}$ is then a linear order on $A$ as discussed in Question A, again whose answer I think is invalid because it seems to have mixed up the greatest element with the maximal element. If $(\bar{A}, \leq_{\bar{A}})$ were the greatest element then showing the result would be no sweat. Since it is merely a maximal element though, it seems that the approach should be  a proof by contradiction as is so often the case when we are dealing with maximal elements.
We clearly have that $\bar{A} \subseteq A$ by the definition of $L$ (and the fact that $(\bar{A}, \leq_{\bar{A}}) \in L$) but we have to show that $A \subseteq \bar{A}$. So suppose to the contrary that this is not the case so that there is an $a \in A$ where $a \notin \bar{A}$. What we would like to do is construct an element of $(B, \leq_B)$ of $L$ where $(\bar{A}, \leq_{\bar{A}}) \trianglelefteq (B, \leq_B)$, which would be a contradiction since $(\bar{A}, \leq_{\bar{A}})$ is a maximal element.
Clearly the set $B$ should probably be $\bar{A} \cup \{a\}$ since we need to have that $\bar{A} \subseteq B$, but the difficulty is in constructing the ordering $\leq_B$ of $B$ such that the conditions of $L$ are met so that $(B, \leq_B) \in L$. Unless I am missing a simple way to do this, things get messy really fast when trying to ensure transitivity and the fact that it is an extension of $\preccurlyeq$.
Am I missing something simple or is there a better way to prove that $\bar{A} = A$?
 A: The following answer was posted here by the user bof but for some reason it was removed.
I worked through all of the details of his/her answer and as far as I could tell it is correct and I was able to prove this question.
Here is that answer:
For a partially ordered set $(A, \preccurlyeq)$ define the set
$$
P = \{R \subseteq A \times A \,|\, \text{$R$ is a partial order on $A$ and $\preccurlyeq \subseteq R$}\} \,.
$$
Then $\subseteq$ partially orders $P$ and it is easy to show that every $\subseteq$-chain of $P$ has an upper bound.
Thus the conditions of Zorn's Lemma are satisfied so that there is a $\subseteq$-maximal element $\leq$ of $P$.
To show that $\leq$ is a linear ordering, we assume that it is not so that there are $a$ and $b$ in $A$ such that $(a,b) \notin \leq$ and $(b,a) \notin \leq$.
We then define the relation
$$
R = \leq \cup \{(x,y) \in A \times A \,|\, x \leq a \text{ and } b \leq y\} \,.
$$
It is a little tedious but not difficult to then show that $R \in P$ but that $\leq \subset R$, which contradicts the fact that $\leq$ is a maximal element of $P$.
So it must be that $\leq$ is in fact linear.
Again, I am not sure why the answer was removed.
It could be the case that there is an error in the argument that I did not find, so if anyone discovers this please add a comment.
A: Your proof is incorrect, the partial orders all extend a given order on $A$, they are all partial orders on $A$. There are no $B$ and $C$.
The point is that given two points which are incomparable, you can arbitrarily decide that $a<b$, and take the transitive closure of that relation. Then show it is a partial order extending the alleged maximal, therefore there were no incomparable elements.
