Partial order is intersection of linear extensions also if poset is infinite? I cannot understand the statement "every partial order is the intersection of its linear extensions. (See e.g. Davey and
Priestley [DP]" here (page 6). The book Introduction to Lattices and Order on page 32 assumes that $P$ is finite, not infinite. So suppose that the poset $P$ is infinite. 
Is the intersection of linear extensions now the infinite $P$?
 A: This result relies on Szpilrajn extension theorem: any partial order has a linear extension.
Let $\leqslant$ be the partial order on $P$. I claim that ${\leqslant}$ is equal to the intersection $\preccurlyeq$ of all linear extensions of $\leqslant$.
For each pair $(a,b)$ of incomparable elements of $P$, define a new relation $\leqslant_{a,b}$ has follows:
$$r \leqslant_{a,b} s\ \text{ if }
\begin{cases}
r \leqslant s\\ \text{ or}\\
\text{$r \leqslant a$ and $b \leqslant s$}
\end{cases}
$$
I let you verify that $\leqslant_{a,b}$ is also a partial order. 
Let $(a,b) \in P^2$. Then $a \leqslant b$ implies $a \preccurlyeq b$ by construction. Suppose now that $a \preccurlyeq b$ but $a \not\leqslant b$. If $b \leqslant a$, then $b \preccurlyeq a$ and thus $a = b$. If $a$ and $b$ are incomparable for $\leqslant$, then $\leqslant_{b,a}$ is a preorder, which admits a linear extension $\leqslant^*_{b,a}$. Since $a \preccurlyeq b$, one gets $a \leqslant^*_{a,b} b$. But since $b \leqslant_{b,a} a$, one also gets $b \leqslant^*_{b,a} a$. Thus $a = b$, a contradiction. 
