Any Set of Ordinals is Well-Ordered

Question

Is this a theorem of ZF or a theorem of ZFC?

My suspicion is that it is a theorem of ZFC (the proof I've seen in P. L. Clark's online notes on countable ordinals requires selecting an element from each set of a countable collection of sets).

Does the situation change if the ordinals in the collection are themselves countable?

Answer 1

This theorem, in fact, is provable in ZF. To see this, let $\alpha_i,i\in I$ be a set of ordinals. If we take union $\bigcup_{i\in I}(\alpha_i+1)$, we will get an ordinal $\beta$ greater than all of $\alpha_i$. But now $\{\alpha_i\}_{i\in I}$ is a subset of $\beta$, and $\beta$ is by definition well-ordered. Finally, every subset of a well-ordered set is well-ordered, and we can finish here.

Every step in this proof is valid without choice.

Answer 2

It is not hard to show directly from the definitions:

1. If $(A,<)$ is a well-ordered set, then for every $B\subseteq A$, $(B,<\restriction_B)$ is also well-ordered.

2. Every set of ordinals is a subset of an ordinal.

So without using the axiom of choice we have that every set of ordinals is well-ordered.