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A well order is a poset such that every non empty subset of this set has a least element. Does "well ordering on $\Bbb R$" mean that every non empty subset of $\Bbb R$ (which is a poset) has a least element?

halrankard
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    Yes. Note that it can't be the usual ordering on $\mathbb{R}$, and indeed it is impossible to construct such an ordering without using some version of the axiom of choice. – user10354138 Aug 03 '20 at 12:23
  • @user10354138 what does the ordering have to do with "having a least element" though? Every subset would be the same no? –  Aug 03 '20 at 12:25
  • I worry that when you say "$\mathbb{R}$ (which is a poset)", you are automatically thinking of $\mathbb{R}$ as a poset with the usual ordering. However, in the claim "$\mathbb{R}$ admits a well-order", the symbol $\mathbb{R}$ only refers to the set of reals, without any further structure. So this claim only says that there is some well-ordering on the set of reals. It says nothing about any kind of compatibility or connection between this well-order and the usual order on $\mathbb{R}$. – halrankard Aug 03 '20 at 13:05
  • It means the same thing as "an enumeration of the rational numbers". – Asaf Karagila Aug 03 '20 at 13:06
  • @halrankard You mean $(ℝ,\preceq)$ and not a specific order like $≤$? I dont get it.. I saw a couple of question on here talking about this and people use for example 0, 1, -1, 2, -2.. to show that the integers are well ordered, but what does this have to do with the well ordering? What is the $\preceq$ in this case? –  Aug 03 '20 at 20:28
  • Yes, the well-ordering on $\mathbb{R}$ will not be the usual order $\leq$. (Of course, a well-ordering of $\mathbb{R}$ is still a "specific" order in the sense that it is a concrete thing; but it's just not going to be the usual ordering $\leq$ of $\mathbb{R}$ since that's not a well-order.) Unlike the integers, I can't explicitly write down a well-ordering of $\mathbb{R}$ because it's existence involves the Axiom of Choice. – halrankard Aug 06 '20 at 12:47

2 Answers2

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A well-ordering on $\Bbb R$ is a binary relation $\preceq$ between elements of $\Bbb R$ such that $(\Bbb R,\preceq)$ is well-ordered. I.e. a relation $\preceq$ which is reflexive, antisymmetric, transitive and such that every non-empty subset $S\subseteq\Bbb R$ has an element $x\in S$ such that, for all $y\in S$, $x\preceq y$.

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It is a theorem of set theory that every set can be endowed with a well-ordering (Zermelo's theorem).

Actually, the axiom of choice, Zorn's lemma and Zermelo's theorem are equivalent.

Bernard
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  • That doesn't quite answer the question, which is what does it mean for the real numbers to be well-ordered. – Asaf Karagila Aug 03 '20 at 13:10
  • I agree, I see it rather as an aside comment. I wanted to insist on the axiom of choice to explain why it is pointless, from my point of view, to explicit the meaning of a well-order on $\mathbf R$. – Bernard Aug 03 '20 at 13:26
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    Well, surely with 152,261 points you can leave a comment under the question... :-) – Asaf Karagila Aug 03 '20 at 13:27