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Is there an example of a total order with properties

  1. there is a least element and
  2. every element has a (unique) successor

not is not also a well ordering?

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2 Answers 2

up vote 5 down vote accepted

Yes: $\bigl(\{0\}\times\mathbb N\bigr) \cup \bigl(\{1\}\times \mathbb Z\bigr)$ with the lexicographic order.

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If you're unfamiliar with lexicographic ordering, the following set of reals has the same effect (and may be easier to visualize):

$\{1-\frac{1}{n}:n\in\mathbb{N}\}\cup\{1+\frac{1}{n}:n\in\mathbb{N}\}\cup\{3-\frac{1}{n}:n\in\mathbb{N}\}$

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Thanks. You need to exclude 0 as a demoniator. –  Tyson Williams Apr 21 '12 at 17:16
    
@TysonWilliams: No problem. Apologies for the ambiguity. By $\mathbb{N}$, I mean the positive integers. –  Cameron Buie Apr 21 '12 at 18:03

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