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Show that if $R$ is a partial ordering on a set $A$, then there exists a linear ordering $R^*$ on $A$ such that $R \subseteq R^*$

GoodDeeds
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RobertM
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    Hint: For the finite case, use induction. For the general case, use the compactness theorem of first order logic. Let $T$ be the theory consisting of the diagram of $(A,R)$ together with the sentence asserting that $R$ is a linear order. Then every finite subset of $T$ has a (finite) model (this is why it's necessary to deal with the finite case first), so by compactness $T$ has a model... – BrianO Apr 15 '16 at 14:57
  • I am torn between closing this as a duplicate, or does the utter lack of substance doesn't deserve the two minutes effort it will take to find that duplicate. – Asaf Karagila Apr 15 '16 at 15:01
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    @AsafKaragila http://math.stackexchange.com/questions/271003/every-partial-order-can-be-extended-to-a-linear-ordering. You answered it :) Note, found it in 1m16s. – BrianO Apr 15 '16 at 15:03
  • @Brian: I know. – Asaf Karagila Apr 15 '16 at 15:05

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This requires the Axiom of Choice: The partial orderings on $A$ are inductively ordered, hence by Zorn's lemma there is a maximal partial ordering $R^\star$ among those containing $R$. Then $R^\star$ is linear (as the lack of this property would quickly contradict its maximality).

Hagen von Eitzen
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