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Suppose $X \subseteq \mathbb{R}$ is well-ordered by $<$, the normal ordering on $\mathbb{R}$. Is it then countable? If so, how to prove this. I've tried to set up a bijection as follows: $$ \phi : \mathbb{N} \to X : n \mapsto \ <\textrm{-least element of } X \backslash \phi(\mathbb{N} \upharpoonright n) $$ I.e, $0$ maps to the least element of $X$, 1 maps to the least element of $X \backslash \{x_1\}$, $2$ maps to the least element of $X \backslash \{x_1,x_2\}$ etc. This is injective by construction, but I'm not sure if it is surjective.

Andrés E. Caicedo
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Ordinalz
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  • Since you don't use that $X \subset \mathbb{R}$, this can't be right (not any well-ordered set $X$ is countable). – Olivier Roche Nov 13 '19 at 14:30
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    This follows from the fact that any uncountable subset of $\mathbb{R}$ has (and contains one of) a limit point. See this: https://math.stackexchange.com/questions/310113/accumulation-points-of-uncountable-sets – freakish Nov 13 '19 at 14:30
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    Try to find a rational number strictly between each two elements of X. If you've ever seen the theorem that a monotonic function can have at most countably many discontinuities, the proof uses a similar idea. – Matthew Towers Nov 13 '19 at 14:30
  • As a curiosity thing, using this reasoning you can prove that in $\mathbb{R}^n$, there is at most a countable number of disjoint open balls. This is very surprising in my opinion. As a challenge, try to prove this. – Cezar98 Nov 13 '19 at 14:40

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Assume W is an uncountable, well ordered subset of R.
For each x in W, Let x' be the successor of x in the well order.
{ (x,x') : x in W } is a uncountable collection of pairwise disjoint open intervals,
That is impossible because in each of those intervals there is a different rational and there are only countablely many rationals but uncountable many intervals.

William Elliot
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