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I am trying to wrap my head around the fact that $\mathbb{R}$ can be well ordered (under some unknown order relation). Since the well ordering theorem says that every non-empty subset of a set has a least element, and since $\mathbb{R}$ is a non-empty subset of $\mathbb{R}$ it follows that we can identify the least element of $\mathbb{R}$ (under some hitherto unknown order), say $l$ and, assign to it the number 1. Then we can find the least element of $\mathbb{R}-\{l\}$ and assign it the next integer and so on. Clearly this process won't cover the entirety of $\mathbb{R}$ as that would have implied that $\mathbb{R}$ is countable. Do I have this right?

Marcus Junius Brutus
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    Check this: https://math.stackexchange.com/q/6501/42969 – Martin R Dec 30 '24 at 10:32
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    That process wouldn't cover the entirety of $\mathbb R$, essentially because you stopped too early. If you want to "index" a well-ordering of $\mathbb R$, then you would have to use all of the ordinals that appear in $\mathfrak c$. Roughly speaking, to get a well-ordering of $\mathbb R$ you need to pick a real $x_0\in\mathbb R$, a real $x_1\in\mathbb R\setminus{x_0}$, a real $x_2\in\mathbb R\setminus{x_0,x_1}$, ...., and then a real $x_{\omega}\in\mathbb R\setminus{x_n\mid n\in\mathbb N}$, and so on. This goes on for a long time. – Joe Dec 30 '24 at 10:37
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    Yes, you have it right. What you end up with is an initial segment of $\mathbb{R}$ under the well-ordering, isomorhpic to the positive integers. – TonyK Dec 30 '24 at 10:39
  • @Joe I think I get what you're saying but I also think that you formulation is not allowed under recursive definition but only under transfinite recursion which is out of my depth. – Marcus Junius Brutus Dec 30 '24 at 10:53
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    There have been several questions on this very topic, unfortunately, I think they are all deleted (since they were asked by a well-known crank as a platform for his soapboxing), so I can't quite find a duplicate. https://math.stackexchange.com/questions/4024749/is-a-well-ordered-set-always-countable is probably the closest I can find today. – Asaf Karagila Dec 30 '24 at 11:02
  • Yes, you need to use transfinite recursion (among other things) to formalise this argument, and unfortunately the gap between the informal argument I gave and the formalised argument is rather large. Incidentally, I wrote something about this yesterday. – Joe Dec 30 '24 at 12:57

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