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A few minutes ago I asked a question about a "proof" that $\mathbb{R}$ is enumerable that crossed my mind: What's wrong with this "proof" that $\mathbb{R}$ is enumerable?

I was told to look into ordinal numbers, and that after crossing $\omega$ we stop considering something to be an enumeration.

Why is this the case? Are there negative consequences if we don't put this limitation?

Edit: I always thought of $\mathbb{N}$ as the "counting numbers" - but... when we cross over to ordinals like $\omega$, $\omega+1$, etc, aren't we still effectively counting?

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  • $\begingroup$ It is just a definition. Nothing negative or positive. Just a name to put things into categories. $\endgroup$ – plop Jul 14 '20 at 21:45
  • $\begingroup$ @plop Thank you - so what would we call an algorithm that goes through a well-ordered set that has an ordinal beyond $w$? $\endgroup$ – Novicegrammer Jul 14 '20 at 21:48
  • $\begingroup$ See hypercomputation. $\endgroup$ – plop Jul 14 '20 at 21:52
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    $\begingroup$ Assuming the axiom of choice, every set can be well-ordered, so the concept of enumerability becomes empty. $\endgroup$ – Yuval Filmus Jul 14 '20 at 21:56
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    $\begingroup$ It is a classical theorem in set theory that the axiom of choice is equivalent to the well-ordering principle. $\endgroup$ – Yuval Filmus Jul 14 '20 at 22:00
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A set $S$ is enumerable (or, countable) if we can enumerate it: $$ S = \{s_1,s_2,s_3,\ldots\} $$ In other words, there is a mapping from $\mathbb{N}$ onto $S$.

Cantor showed that $\mathbb{R}$ isn't enumerable.

We can consider more relaxed notions of enumeration. For example, a set $S$ is well-orderable if there is a linear order $<$ on $S$ such that any non-empty subset of $S$ has a minimum. This encompasses your examples, and much more.

The axiom of choice is equivalent to the well-ordering principle, which states that every set can be well-ordered. Hence if you assume the axiom of choice, every set can be enumerated in this sense.

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  • $\begingroup$ I take it in this definition, the enumeration cannot cross $\omega$? $\endgroup$ – Novicegrammer Jul 14 '20 at 22:02
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    $\begingroup$ When you count $1,2,3,\ldots$, do you ever reach $\omega$? $\endgroup$ – Yuval Filmus Jul 14 '20 at 22:03
  • $\begingroup$ Haha! Quite... I see your point. $\endgroup$ – Novicegrammer Jul 14 '20 at 22:04
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    $\begingroup$ @Novicegrammer But when you well-order $\mathbb{R}$ you do cross $\omega_0$ and $\omega_0+1$ and $2\omega_0$, and $\omega_0^{\omega_0}$, ... Assuming the continuum hypothesis you won't need to cross $\omega_1$, but if not, then you may have to cross even more ordinals. $\endgroup$ – plop Jul 14 '20 at 22:27
  • $\begingroup$ @plop Oh thanks, I've been introduced to yet another hypothesis. ...I'll throw this somewhere in the backlog of stuff I have to read. $\endgroup$ – Novicegrammer Jul 14 '20 at 23:48

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