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I have been reading Munkres' Topology and in Ch. 1 section 10, the Well-Ordering Theorem is introduced as equivalent to AC, there is even a proof of it outlined in the supplementary exercises. However, the main use of this theorem in the book seems to be to show that uncountable well-ordered sets exist. From a bit of research (and even in Munkres' book) you can show that there exist uncountable well-ordered sets without AC.

To my actual question, it is stated all over in various sources that the well-ordering of $\mathbb{R}$, or $\mathbb{R}$ being impossible to well-order is independent of ZF. I can't seem to find a proof or a link to a paper that shows that this is in fact the case. I don't know tons of set theory, just what is given in the first chapter of Munkres' book, so maybe I couldn't even understand the proof, but I would at least like to know it's out there and try to understand it. This is the closest thing I can find to talking about the subject but still doesn't mention the reason why the well-ordering of $\mathbb{R}$ is independent of ZF. Any help is much appreciated.

Jay
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    Doesn't "the axiom of choice is consistent with ZF" already answer your question? –  Apr 19 '18 at 05:19
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    I don't think so; I was looking for a proof that the well-ordered property of the reals was consistent with ZF since other uncountable sets can be well-ordered in just ZF. I am wondering how I know that R isn't one of these sets. – Jay Apr 19 '18 at 05:50
  • It occurred to me that perhaps you meant that in showing that AC is consistent with ZF involves somehow the ordering of the reals. Is this the case? I am very inexperienced in this area, apologies. – Jay Apr 19 '18 at 06:02
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    The axiom of choice proves there is a well-ordering of $\Bbb R$. So if the axiom of choice is consistent with ZF... – Asaf Karagila Apr 19 '18 at 07:41
  • Sorry I thought I changed all the instances of "consistent" to "independent" in editing. I mistakenly asked about consistency originally, but wanted to know about independence. – Jay Apr 19 '18 at 08:50

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The axiom of choice implies that every set can be well-ordered. In particular, it implies that the real numbers can be well-ordered.

So a proof that the axiom of choice is consistent with $\sf ZF$ would be also a proof that the fact that the real numbers can be well-ordered consistently with $\sf ZF$.

The relative consistency of the axiom of choice is normally proved via Gödel's constructible universe, but you need to have a lot more set theory under your belt to fully understand the proof. You can find it in most advanced set theory books (e.g. Jech, or Kunen).

The main difficulty is to prove that no well-ordering of the reals can be explicitly defined, or that a well-ordering of the reals does not imply the full axiom of choice. For these you need even more set theory than before, as the proofs include forcing and symmetric extensions. You can find these proofs also in Jech or Kunen.

Asaf Karagila
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  • Apologies, I think I have asked the wrong question due to my lack of set theory knowledge. I think what I meant to ask was how do we know that "R is well-ordered" is independent of ZF? While I understand that with AC, R is well-ordered, how do we know we need AC to show this? – Jay Apr 19 '18 at 08:24
  • https://math.stackexchange.com/questions/6501/is-there-a-known-well-ordering-of-the-reals – Asaf Karagila Apr 19 '18 at 08:29
  • You should also edit your question... – Asaf Karagila Apr 19 '18 at 08:29
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    @Jay: To help prevent reading too much into that last paragraph, it's worth noting that there is an explicitly defined well-ordering on a set, and it will be independent of ZF whether or not that set is the set of real numbers. –  Apr 19 '18 at 09:40