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If the the real numbers for instance have a well-ordering, can you not form a bijection from the reals to the natural numbers? Simply map the smallest of the reals to 0, and then iteratively map successors to each other. What is it I'm missing here?

sudoScience
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    Why should it exhaust all reals? Same as: any singleton is well-orderable, but you can't find a bijection with $\mathbb{N}$. – user10354138 Jun 27 '21 at 07:37
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    If in a well-ordered set $X$ you call $0=\min X$ and $x+1:=\min{y\in X,:, x<y}$, then it isn't necessarily true that for all $x\in X\setminus{0}$ there is some $w\in X$ such that $x=w+1$. See for instance $\Bbb N\times\Bbb N$ endowed with the order $$(a,b)\leqslant (c,d)\Leftrightarrow a< c\lor(a=c\land b\le d)$$ The procedure you are sketching gives a bijection between an initial segment of your well-ordering $(\Bbb R,\le)$ and $\Bbb N$, but giving a global bijection is impossible. –  Jun 27 '21 at 08:50
  • @Gae. S. Ah, I see. Since it is pretty much impossible to visualize a well-ordering of R, my intuition was defaulting to a countably infinite list. Your ordering of NxN was a great counter-example that one can actually visualize. – sudoScience Jun 27 '21 at 21:47

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Such a map, from an infinite well ordered set to the natural numbers, might not be defined for all members of the intended domain. In fact it certainly will not if that domain has order type greater than omega.

kinbote
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