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A relation '$ < $' defined on a set $ A $ is called a strict total order (or a linear order, according to Munkres) relation if

  1. for any $ a\ne b $ from $ A $, either $ a<b $, or $ b<a $;
  2. for no $ a\in A $, $ a<a $;
  3. for any $ a $, $ b $, $ c $ from $ A $, $ a<b $ and $ b<c $ together imply $ a<c $.

I want to know whether it is always possible to define a strict total order relation on any given set. I know that if $ A $ is the given set, $ X $ is a strict totally ordered set, and $ f\mathpunct{:} A\to X $ is an injection, then such a relation can be induced in the following way $$ a_{1}<a_{2} \text{ if and only if }f(a_{1})<f(a_{2}). $$ However, in this case, my question reduces to: Given a set $ A $, can we always find a strict totally ordered set $ X $ to support such a map $ f $?

I must mention here that my primary objective is to have the answer to the question posed in the title. The above discussion is just one possible way to get the answer. If there are other trivial (or non-trivial) ways, please let me know about that.

Subhajit Paul
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1 Answers1

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Yes, every set can be linearly ordered.

Famously, the axiom of choice is equivalent to the well-ordering theorem, which states that every set can be well-ordered. A well-order on a set $S$ is a linear order on $S$ with the additional property that every nonempty $K \subseteq S$ has a least element.

Furthermore, the fact that all sets can be linearly ordered cannot be proved if we throw away the axiom of choice entirely. If we abandon the axiom of choice, it is consistent that there is some set which cannot be linearly ordered. In fact, this set can be $P(\mathbb{R})$.

Nevertheless, the fact that all sets can be linearly ordered is a statement far weaker than the full axiom of choice. Using the compactness theorem for propositional logic, we can start by proving any finite set can be linearly ordered and extend this to show all sets can be linearly ordered. The compactness theorem is significantly weaker than the axiom of choice. It is not that difficult to extend the proof to show that any partial order can be extended to a linear order.

Mark Saving
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    The compactness theorem implies indeed a little more than the current question: every partial ordering can be extended to a total ordering. On the other hand, in The axiom of Choice, Jech gives the example of a model of ZF where all sets can be totally ordered but there exists some partial ordering that cannot be extended to a total order (chapter VII). Thus (modulo ZF) the compactness theorem is strictly stronger than “every set admits a total ordering”. – jp boucheron Feb 21 '23 at 21:38
  • Thank you for the clarity. Can you please suggest some references which further elaborates your answer? – Subhajit Paul Feb 22 '23 at 04:13
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    For a proof that the axiom of choice implies all sets can be well-ordered, you would probably want to read an introductory book on set theory (unfortunately, I don’t know a good one). For the compactness theorem, I think a decent introductory text to first-order logic is Alex Kruckman’s model theory notes, though they don’t have that many exercises. Learning either of these fields is a fair bit of work. Proving the compactness theorem doesn’t imply choice and that we can, in the absence of choice, have a set which isn’t linearly ordered requires knowledge of forcing, a pretty advanced field. – Mark Saving Feb 22 '23 at 04:59