The Well-Ordering principle states that for every set $X$, there exists a well-ordering $\leq$ on $X$.

Let us not assume the Axiom of Choice. Let $\alpha$ be an uncountable cardinal and suppose the following is true: For every set $X$ of cardinal $\alpha$, there exists a well-ordering $\leq$ on $X$.

Can one now prove the well-ordering principle on any cardinal?

I ask this because, in a conversation with some people, i heard the claim that the existence of the Vitali set is equivalent to the axiom of choice, and therefore to the well-ordering principle. But now, i highly doubt that is true, since in the construction of the Vitali Set, one only need to assume that the well-ordering principle (or the axiom of choice, depending on how you take the crucial step) is true for sets of cardinal $2^{\aleph_0}$.

Nevertheless, if the restriction of the well-ordering principle to $2^{\aleph_0}$ imply the general well-ordering principle, i can totally see how this would be true.


The answer is negative. And you are correct. To construct a Vitali set one only need choice up to $2^{\aleph_0}$, or that the real numbers can be well ordered (which might happen regardless to how badly choice fails in the universe).

The universe of set theory has sets much much larger than the real numbers, and even more is true: the universe is built in steps, and the real numbers appear fairly soon in the construction, and the axiom of choice could hold for sets up to some step, and then fail incredibly above some point of the construction of the universe.

One last remark would be that cardinal is ambiguous without choice. Specifically, some people use it exclusively to talk about well ordered cardinals, and in that case every set which is equipotent to an ordinal has a well ordering. Just it's true that the real numbers might not have a cardinal. I prefer the broader usage of the term cardinal to talk about cardinals of arbitrary sets, in which case the question has some actual content to it.


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