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From what I understand, the well-ordering theorem assumes ANY set can be made well-ordered depending on how you define the ordering.

For example, $\mathbb{Z}$ is not well-ordered according to the conventional ordering of (<). But, if we define a new ordering, say ($\prec$), s.t. for any $a,b \in \mathbb{Z}$, we have $a \prec b$ whenever $|a| \le |b|$, then we can order the set of integers as $\{0, 1 ,-1, 2,-2,3, -3,...\}$ and thus make the set well-ordered. Under this new ordering the least element is $0$, and every nonempty subset will also have a least element.

If the well-ordering axiom is true, then there exists an ordering s.t. the $\mathbb{R}$ are also well-ordered, but how is this possible given the $\mathbb{R}$ is uncountable?

In other words, if there was such an ordering, say ordering $(@)$, then that ordering would effectively be a function makes the reals listable, would it not? Under $@$, there would be some number, say $n_1$, that would be the least element of $\mathbb{R}$... and another number, $n_2$, that would be the least element of $\mathbb{R}-\{n_1\}$... and another number, $n_3$, that would be the least element of $\mathbb{R}-\{n_1,n_2\}$, and so on and so forth... but we know the reals cannot be placed into a one-to-one correspondence with $\mathbb{N}$ as $n_1,n_2,n_3$,... This seems like a contradiction to me.

Can anyone resolve this for me without getting too steeped in jargon and notation? Thanks.

Andrés E. Caicedo
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RyRy the Fly Guy
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  • An order isn't literally a list, it's just a relation that is total (law of trichotomy). So, uncountable sets can be well ordered. – Rushabh Mehta Oct 05 '19 at 05:45
  • The ordering may not be a list, but couldn't you use the ordering to create a list? For example, by recursively removing the least element of the $\mathbb{R}$ starting with the $\mathbb{R}$? – RyRy the Fly Guy Oct 05 '19 at 05:49
  • Who said doing that will cover all of $\mathbb R$? See this – Rushabh Mehta Oct 05 '19 at 05:50
  • Here's a simple example of why your approach fails. What if you have the set ${2^n;|;n\in\mathbb N}\cup{3^n;|;n\in\mathbb N}$, and the well ordering $2^n<2^m$ if $n<m$, $3^n<3^m$ if $n<m$, and $2^n<3^m$ for all $n,m$. Try your approach on this countable set. – Rushabh Mehta Oct 05 '19 at 05:53
  • If there was an ordering that suddenly gave every nonempty subset of the $\mathbb{R}$ a least element, then it seems apparent to me that you COULD generate such a list. Albeit, it would be an infinite list. Can you show that it would not cover all the real numbers? – RyRy the Fly Guy Oct 05 '19 at 05:54
  • Try your approach on my constructed example. – Rushabh Mehta Oct 05 '19 at 05:54
  • Did you see the issue? – Rushabh Mehta Oct 05 '19 at 05:57
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    Same example as Don's, perhaps phrased more transparently: Order the natural numbers like $0,2,4,6,\ldots, 1,3,5,7,\ldots.$ This is a well ordering. Now take the least number ($0$), the next least ($2$), and so on. This produces a list of the even numbers, not the whole set. – spaceisdarkgreen Oct 05 '19 at 07:04
  • And your challenge to show why it would not cover the real numbers misses the point entirely. As you've noted, the fact that the real numbers are uncountable means this would produce a contradiction. The onus is on you to show that it does cover the whole thing and thus confirm that there is a contradiction. But anyway, the example I just gave shows that there is no reason to expect that it would, even if there were no contradictions due to cardinality. – spaceisdarkgreen Oct 05 '19 at 07:12
  • Whether it's possible or not is a proposition that someone can prove or disprove with depending on thier choice of proof strategy. I don't see how the onus is on me to show something that I dont fully understand, nor do I see how you demanding I prove something I dont fully understand is constructive. I was not posing a challenge but hoping that if someone showed how it did not cover the reals then the error or gap in my understanding would be exposed. – RyRy the Fly Guy Oct 05 '19 at 07:33
  • Fair enough. I phrased that unnecessarily confrontationally. I just wanted to set the logic straight and point out that the reason we know it can't cover the whole thing was already discovered by you. Did the above example help you see why we shouldn't have expected it needed to in the first place? – spaceisdarkgreen Oct 05 '19 at 07:34
  • Yeah. It's starting to become more clear. Thanks – RyRy the Fly Guy Oct 05 '19 at 07:40
  • Sure, i think the key sentence there to unpack is 'this is a well ordering'. An ordering can have limit points with an infinite amount of stuff below and still satisfy the well ordering property. – spaceisdarkgreen Oct 05 '19 at 08:04
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    The axiom of choice is a red herring. There are uncountable well-ordered sets, even if the real numbers cannot be well-ordered (and the axiom of choice fails significantly). – Asaf Karagila Oct 05 '19 at 09:07
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    I want to amplify Asaf's comment, because it took a weirdly long time for someone to clearly point out that there are uncountable well-orderings. Being well-ordered is entirely unrelated to countability. – Malice Vidrine Oct 05 '19 at 17:27
  • @MaliceVidrine Well, OP's question was essentially asking what's wrong with their argument that every well-ordering is countable (in fact of type $\omega$). Their question was premised on a certain uncountable well ordering purportedly existing and asking how could that be consistent. It's definitely worth pointing out that choice is actually not required for the existence of uncountable well-orderings, and they could have asked the same question about e.g. $\omega_1,$ but I don't think it's the central issue. – spaceisdarkgreen Oct 05 '19 at 18:07
  • Doh, my bad. Skimmed much too quickly again... – Malice Vidrine Oct 05 '19 at 18:58

2 Answers2

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First, your list of elements of $\mathbb Z$ can be regarded as well-ordered, but not using the relation you have stated, since that makes both $n\prec -n$ and $-n\prec n$.

Secondly a well-ordered set corresponds to an ordinal number. A countable set is one which has a bijection to the ordinal $\omega$ (which can be regarded as the ordered set $\{0,1,2,3,4, \dots\}$).

To illustrate the difference, consider the integers with the ordering $0, 1, 2, 3 \dots -1, -2, -3, \dots$. This has order type $2\omega$ and is a well-ordering (you should check this). Using your strategy for the reals you would list the positive integers, but would never get to the negative integers.

Countably infinite sets can have many different well-orderings of different order types (uncountably many in fact). But they all have a well-ordering of type $\omega$.

Your strategy for the reals fails to show they are countable because it doesn't list every real number. Cantor's diagonal argument shows that no strategy can work.

Perhaps others will inject a little more technical precision here. The fact that all sets can be well-ordered does not follow from the "other" axioms of set theory. It is sometimes posed as an axiom itself, and is sometimes regarded as a consequence of the Axiom of Choice.

Mark Bennet
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  • i appreciate you pointing out the shortcomings of the ordering i defined on the integers. However, I am NOT trying to make a point that it's possible to list the real numbers. I'm aware of Cantor's argument and that is the very reason i feel the well-ordering theorem or "axiom" poses a contradiction – RyRy the Fly Guy Oct 05 '19 at 06:09
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    @RyRytheFlyGuy It doesn't pose a contradiction precisely because your procedure to enumerate $\mathbb{Z}$, given the well-ordering in the answer, does not in fact enumerate $\mathbb{Z}$. If your "contradiction" were indeed a contradiction, that procedure would have to enumerate $\mathbb{Z}$. – Patrick Stevens Oct 05 '19 at 06:39
  • i'm not trying to order the integers. that was just an example. the point is i'm talking about the real numbers – RyRy the Fly Guy Oct 05 '19 at 06:41
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    @RyRytheFlyGuy I used the example of an alternative well-ordering of the integers to show that even if you begin with a set you well know to be countable, the procedure you want to adopt for the real numbers need not produce a complete enumeration. Indeed every infinite ordinal has an initial segment order isomorphic to $\omega$ and your enumeration technique only counts that initial segment. – Mark Bennet Oct 05 '19 at 16:51
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It is indeed possible to generate an infinite list-more precisely, a list of order type $\omega$-by recursively choosing the least element of any well ordered set. This list will usually not contain the whole set, and indeed cannot when the set is uncountable. You haven't given any justification for why you intuit that this process should exhaust the set, so it's difficult to say more.

Kevin Carlson
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  • can you explain why this method would not exhaust the whole set if every subset of the reals had a least element? – RyRy the Fly Guy Oct 05 '19 at 06:39
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    @RyRytheFlyGuy Well, it simply doesn't. Can you explain why it would? If you understand Cantor's argument, you understand there are different scales of infinity. "Going on forever" doesn't give any reason to exhaust a set. – Kevin Carlson Oct 05 '19 at 06:47
  • maybe the issue is I'm lacking what it means for something to be of order type $\omega$. – RyRy the Fly Guy Oct 05 '19 at 06:52
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    @RyRytheFlyGuy Order-isomorphic to the naturals in their standard ordering. In other words, ordered like this: $x,x,x,x,x\ldots,$ as opposed to like this: $x,x,x,x\ldots, x,x,x,x,\ldots$ as in the counterexamples given above. (Or as opposed to any of the many other patterns the order of a set can take.) – spaceisdarkgreen Oct 05 '19 at 07:30