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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 $P$?

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  • $\begingroup$ Sorry, my bad... $\endgroup$ – Asaf Karagila Mar 13 '16 at 17:26
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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.

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