Can every poset be toplogically sorted?

Question

A partial ordering $\preceq'$ on $S$ is said to be compatible with another partial ordering $\preceq$ if for all $a,b,\in S$,

$$ a \preceq b \Rightarrow a \preceq' b$$

Given a poset $(S, \preceq)$, a toplogical sorting on $S$ is a total order $\preceq'$ that is compatible with $\preceq$. There is a simple constructive proof that every finite poset can be toplogically sorted. Is the same true for infinite sets? Can every poset be toplogically sorted?

In other words, can every partial order be extended to a total order. Yes assuming choice. This was covered in multiple threads before. — Asaf Karagila, May 02 '16 at 18:11

score 5 · Accepted Answer · answered May 02 '16 at 18:13

5

This is the Szpilrajn extension theorem; it’s a result that follows from the axiom of choice. In fact, it follows from but does not imply the weaker compactness theorem for first-order logic or the equivalent Boolean prime ideal theorem. Its proof from the result for finite partial orders and either the compactness theorem or the equivalent ultrafilter lemma is a straightforward application of either.

answered May 02 '16 at 18:13

Brian M. Scott

631,399

Do you have a reference that discusses this? – David M Jul 20 '20 at 04:43
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@DavidM: I don’t, I’m afraid, but you could do worse than to start with this link in Noah Schweber’s answer, especially the answer there by François G. Dorais, and in connection with that the answers to this question by Asaf Karagila and me. – Brian M. Scott Jul 20 '20 at 17:22

score 2 · Answer 2 · edited Apr 13 '17 at 12:58

2

Choice is indeed necessary - for one thing, even the statement "Every set can be linearly ordered" is not provable in ZF alone! See https://mathoverflow.net/questions/37272/are-all-sets-totally-ordered.

edited Apr 13 '17 at 12:58

Community

1

answered May 02 '16 at 18:13

Noah Schweber

260,658

Can every poset be toplogically sorted?

2 Answers2