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Let's say we have the closed interval $[0,1]$. Assuming the axiom of choice (if it's necessary), can we enumerate $[0,1]$ with ordinals as $[0,1]= \left\{x_\alpha: \alpha< \mathfrak{c} \right\}$ such that $x_\alpha< x_\beta$ if and only if $\alpha<\beta$?

I'm not a set theorist so forgive me if this question is trivial, but I'm not sure if we can do this or not. I $\textit{think}$ the answer is yes. Intuitively, as I understand it, we can always choose arbitrary ordinals to index our interval as $\left\{ x_{\alpha'}: \alpha'<\mathfrak{c} \right\}$, but these are essentially just labels. I see no obstruction to relabeling them in whatever order we see fit, including the order previously described. Is this the case?

M10687
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  • Just a minor remark, choice is necessary. – Asaf Karagila Dec 13 '16 at 03:10
  • @AsafKaragila Could you possibly briefly explain what you mean by this? The answers below prove that an ordering of the type I described is not possible, so I'm interested in what exactly you are referring to. – M10687 Dec 13 '16 at 03:28
  • The enumeration disagrees with the usual ordering, sure. But its existence follows from the axiom of choice, and we cannot prove the existence of such enumeration without it. – Asaf Karagila Dec 13 '16 at 03:32
  • I have a doubt: when you write "$x_\alpha < x_\beta$" is that the standard total order on $[0,1]$ or an attempt of a definition? Because the anwers below assume the first situation I think. – H. Kissos Dec 13 '16 at 08:04

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Your "enumeration" would be an order-isomorphism $\alpha \mapsto x_{\alpha}$ between $\mathfrak{c} = \{\alpha \mid \alpha < \mathfrak{c}\}$ and the unit interval $[0, 1]$. There are many ways of seeing that no such order-isomorphism can exist: e.g., (a), $\mathfrak{c}$ has no maximal element, but $[0, 1]$ does; (b), $[0, 1]$ is densely ordered but $\mathfrak{c}$ is not; and (c), $\mathfrak{c}$ is well-ordered but $[0, 1]$ is not.

Rob Arthan
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There isn’t even an order-preserving injection $f:\mathfrak{c}\to[0,1]$. If $f:\mathfrak{c}\to\Bbb R$ were an order-preserving injection, no point of its range would be a complete accumulation point of the range from the left. In this answer, however, I showed that if $X$ is any uncountable subset of $\Bbb R$, there is an $x\in X$ that is a complete accumulation point of $X$ from both sides.

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Brian M. Scott
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    Interesting reference and argument. The non-existence of $f$ also follows by arguing that if it did exist there would be uncountably many disjoint open intervals $(f(\alpha), f(\alpha')) \subseteq [0, 1]$ and each one of these intervals would contain a rational number, contradicting the countability of the rationals. (This was actually my first thought about the question, but the approaches I gave in my answer seemed simpler.) – Rob Arthan Dec 13 '16 at 23:19