Can we always define a strict total order relation on a set?

Question

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.

Answer 1

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.