Enumerating an Interval

Question

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?

Answer 1

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.

Answer 2

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.