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By well-ordering theorem: "if $A$ is a set, there exists an order relation on $A$ that is a well-ordering."

Consider $A=R$ to be the set of all real numbers. Could you construct an order-relation that is well-ordered on $R$? (As it seems that the usual order is not well-ordered. For example, $(1,2)$ does not have a least element by the usual order.)

  • No you couldn't construct one, you just wave your hands and say the magic word, "Zermelo," and it pops into existence. Strange, no? The link Mark gave you has a good explanation. Ernst Zermelo was the first person to explicitly state the well-ordering theorem and prove it from the Axiom of Choice, in 1904. https://en.wikipedia.org/wiki/Ernst_Zermelo – user4894 Mar 18 '17 at 04:11
  • @user4894 That's not quite true - it is consistent with ZFC that there are definable well-orderings of the reals, just (a) they're pretty complicated nonetheless, and (b) we can't prove in ZFC alone that such a definition works. – Noah Schweber Mar 18 '17 at 05:00
  • @NoahSchweber Do you have a reference that I might be able to understand? That's an interesting comment. – user4894 Mar 18 '17 at 05:09
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    @user4894: In $L$ there is a $\Sigma^1_2$ well-ordering of the reals. This is also provably the lowest we can go as far as the complexity of the ordering. – Asaf Karagila Mar 18 '17 at 05:10
  • @AsafKaragila Minor quibble: I think you mean $\Delta^1_2$. – Noah Schweber Mar 18 '17 at 05:23
  • @Noah: Yes, certainly. – Asaf Karagila Mar 18 '17 at 05:24
  • @AsafKaragila $L$ is the constructible universe, and I know the arithmetic hierarchy tells you how many quantifiers are involved and in what order. So you're saying that it's consistent (true in $L$) that this well-ordering exists; but in other models it may not exist. Have I got that right? – user4894 Mar 18 '17 at 05:24

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