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By well-ordering theorem (from axiom of choice), any set can be well-ordered. Well-ordering says a set can be counted from bottom up, but how can such a set be uncountable?

Andrés E. Caicedo
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Rui Liu
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    The definition of well ordering is not that a set can be counted from bottom up. The definition of well ordering is that any non-empty subset has a least element. – Riley Mar 05 '18 at 21:16
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    @Riley I do know the definition of well ordering. However, intuitively well ordering does mean that we can grab the least element, then the second least, then the third least... right? This feels that a well ordering set should be able to be counted. Is there a satisfying explanation why such a set can exist? – Rui Liu Mar 05 '18 at 21:19
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    You are correct that well-ordering implies you can grab the $n$-th least element for any natural number $n$. But the converse is not necessarily true. And this doesn't mean you can get every element this way. It just means you can get a countable list this way. – Riley Mar 05 '18 at 21:25
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    "This feels that [...]" It may feel that way, but that's not true. The counter-examples of uncountable well-ordered sets show it -- your question basically boils down to "my intuition tells me something [which is not true], how can this be"? Well, intuition is often not a good guide in mathematics. – Clement C. Mar 05 '18 at 21:26
  • So you could have a countable list of $n$th elements, but that list doesn't necessarily include every element of the original set (or of whatever infinite subset you applied this to). – David K Mar 05 '18 at 21:27
  • If I wrote an answer like the one at https://matheducators.stackexchange.com/questions/689/how-can-i-familiarize-elementary-school-students-with-infinities-larger-than-a/14983#14983 except that I wrote something like "There are 2 definitions of an ordinal number, a Von-Neumann ordinal which can be defined in ZF and an ordinal defined in type theory as anything that can be defined in terms of a proof system of pure number theory of arbitrary high level, which includes only ordinal numbers smaller than the Church-Kleene ordinal. There is a well-ordered set of order type $\omega_1$ but using the – Timothy Feb 06 '19 at 00:54
  • number theory definition of an ordinal number, that is a well-ordered set but there is no ordinal number that it corresponds to. It has an order type of a Von-Neumann ordinal which is a totally different type of object than an ordinal number itself using the number theory definition of an ordinal number," but written in the right way, would that answer your question? – Timothy Feb 06 '19 at 00:57

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Consider the set $\mathbb{N}\cup\{\omega\}$, which is well-ordered by the order defined by $x\le y$ if $y=\omega$ or $x,y\in\mathbb{N}$ and $x\le y$.

This is a well-ordered set, but you cannot use the order on it to inductively enumerate its elements. Indeed, by taking the least element and continuing, you will only be able to get $\{1,2,3,\ldots\}=\mathbb{N}$.

I'm not claiming that this set is uncountable (because it clearly isn't). I only hope this gives some intuition on how an uncountable well-ordered set could possibly exist if one accepts the axiom of choice.

Edit: As pointed out by Hanul Jeon in the comments, the existence of an uncountable well-ordered set does not require the axiom of choice. (See here: A well-order on a uncountable set). However it is required to well-order ANY set (via the well-ordering theorem).

John Griffin
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There's a wrong assumption there. Well ordering doesn't mean that “a set can be counted from bottom up”. It means that every non-empty subset has a least element.

José Carlos Santos
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Notice also that there are uncountable ordinals (see First uncountable ordinal; see also Simple example of uncountable ordinal).
Every ordinal is a well-ordered set (by the relation $\in$).
If there are uncountable ordinals, then clearly there are uncountable well-ordered sets.

amrsa
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