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Is there a known well ordering of the reals?

I am having a hard time wrapping my head around what a well-ordering of $\mathbb{R}$ looks like. I have seen the presentation of a well-ordering of $\mathbb{Z}$ ie $0, -1, 1, -2, 2, \ldots$ but how could you do this type of ordering with $\mathbb{R}$ where numbers are not countable in that way?

Edit: I'm guessing what I'm asking is if $0, -\epsilon , \epsilon, -2\epsilon, 2\epsilon,\ldots$ could be thought of as a well-ordering of $\mathbb{R}$ in a similar fashion.

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marked as duplicate by MJD, Ross Millikan, Hans Lundmark, Norbert, Thomas Oct 8 '12 at 14:08

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The well-ordering requires (part of) the Axiom of Choice, so we will not be able to exhibit one. –  André Nicolas Oct 8 '12 at 2:04
    
Try this out:math.stackexchange.com/questions/29237/… –  Vishesh Oct 8 '12 at 2:05
    
    
Also: Explicit well-ordering of $\mathbb{N^N}$ and the MathOverflow thread V=L and a well-ordering of the reals –  commenter Oct 8 '12 at 2:31
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@Norbert: If time is a continuum then you can only have a countable amount of steps. –  Asaf Karagila Oct 8 '12 at 14:13

2 Answers 2

Nobody can wrap their head around a well-ordering of $\Bbb R$, and nobody knows what one looks like. It is impossible to exhibit one.

$0, -\epsilon , \epsilon, -2\epsilon, 2\epsilon,\ldots$ cannot be a well-ordering of $\Bbb R$, because it is countable. (In particular, it omits $\epsilon\over 2$.) A well-ordering of $\Bbb R$ must contain an uncountable sequence of elements of $\Bbb R$, which means that it is at least as complicated as $\omega_1$, the smallest uncountable ordinal. This means that you would have to supply not only the first $\omega$ elements $0, -\epsilon , \epsilon, -2\epsilon, 2\epsilon,\ldots$, but then a following sequence corresponding to $\omega+1\ldots 2\omega$, and so on for every countable ordinal. Countable ordinals are very complicated.

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If it is impossible to exhibit (or see) such a well-ordering, ie mortals like myself have little intuition about such a thing, why is there so much faith that it is true, since AC=WO? –  tacos_tacos_tacos Oct 8 '12 at 2:08
    
That is a very big topic, and one which has been addressed elsewhere on the site better than I could hope to do. –  MJD Oct 8 '12 at 2:11
    
Possibly relevant to your question: Advantage of accepting the axiom of choice, or Unprovable statements in ZF, in which Asaf left a lot of links to related material. –  MJD Oct 8 '12 at 2:18

We don't know what a well-ordering of the real numbers looks like. Not only because it is a non-constructive object (i.e. its existence is provable, but not describable in ZFC), but also because the length of this well-ordering is undecidable in ZFC.

Namely, suppose that the real numbers could be well-ordered. Take an ordering of minimal length. Is this an order of length $\omega_1$? $\omega_5$? $\omega_{\omega_{\omega_1}}$?

The axioms of ZFC are not sufficient to calculate the exact length of the well-ordering of the reals; and the axioms of ZF are not sufficient to prove the existence of such well-ordering (but I wrote about this enough in the linked posts).

As for the elements? Well, that is impossible to tell if the set is not canonically well-ordered, like the natural numbers. Consider the rationals, those are well-orderable (it is a countable set). What is the least rational in the well-ordering? What is its successor? We can't really tell. We can always choose a well-ordering that its first element is $0$ and the second is $42$; we can describe a few more elements; we can even describe longer pieces. However there is no canonical way to do that.

Similarly even if the real numbers are well-orderable we can't really point out a particular well-ordering because of that. We can always take a permutation of the real numbers to define a new well-ordering.

Either way, however, describing only a countable part of the real numbers is not enough to describe a well-ordering of them all because Cantor's theorem tells us the real numbers are uncountable.

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