linear extension of a partial order

**(linear extension)**

Given a set $X$ equipped with a partial ordering $\leq$, then a *linear extension* is a linear order $\leq_{lin}$ on the same set $S$, such that the identity function $id_S$ is order-preserving

$(S,\leq) \overset{ id_S }{\longrightarrow} \left(S,\leq_{lin}\right)
\,.$

**(existence of linear extensions)**

For finite sets linear extensions (Def. ) always exist. For non-finite sets this is still the case using the axiom of choice.

A proof under AC was first published in (Marczewski 30). The proposition actually follows from the weaker choice principle called the ultrafilter principle, by appeal to the compactness theorem, as follows.

It is a very simple matter to show linear extensions exist in the finite case: one may proceed by induction. Any finite $n$-element poset has a minimal element $x$ (meaning $y \leq x$ implies $y = x$). By induction the restricted partial order on $X \setminus \{x\}$ admits a linear extension, and then one may simply prepend $x$ to that linear order to complete the inductive step.

The rest is a routine application of compactness for propositional theories. Let $(X, \leq)$ be a partially ordered set, and introduce a signature consisting of constants $c_x$, one for each $x \in X$, and a binary relation $L$. Introduce axioms $\neg(c_x = c_y)$ whenever $x \neq y$ in $X$, and $L(c_x, c_y)$ whenever $x \leq y$ in the poset $X$, and axioms stating that $L$ is a linear order. By the previous paragraph, the resulting theory is finitely satisfiable upon interpreting each $c_x$ as $x$. Hence the theory is satisfiable. Taking any model $M$, and interpreting the constants in $M$, and restricting $L$ to them, we obtain a linear extension on $X$.

- Edward Marczewski,
*Sur l’extension de l’ordre partiel*, Fundamenta Mathematicae, 16: 386–389 (1930) (pdf)

See also

- Wikipedia,
*Linear extension*

Last revised on January 27, 2019 at 20:24:07. See the history of this page for a list of all contributions to it.