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Possible Duplicate:
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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    The well-ordering **requires** (part of) the Axiom of Choice, so we will not be able to exhibit one.2012-10-08
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    Try this out:http://math.stackexchange.com/questions/29237/the-well-ordering-principle2012-10-08
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    Try [Is there a known well-ordering of the reals](http://math.stackexchange.com/questions/6501/is-there-a-known-well-ordering-of-the-reals), or [A way to well-order real line](http://math.stackexchange.com/questions/150992/a-way-to-well-order-real-line).2012-10-08
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    Also: [Explicit well-ordering of $\mathbb{N^N}$](http://math.stackexchange.com/q/23927/16627) and the MathOverflow thread [V=L and a well-ordering of the reals](http://mathoverflow.net/questions/6593/vl-and-a-well-ordering-of-the-reals)2012-10-08
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    Take the any of real numbers, then take any of the rest, again take any of the rest. Continue till the end of time.2012-10-08
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    @Norbert: If time is a continuum then you can only have a countable amount of steps.2012-10-08

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