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Assuming R is well-ordered relation on a set A. Wikipedia claims that every well ordered relation is also a total one.Even tough in first sight it seems trivial, how can I prove it?

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There’s nothing to prove: it’s part of the definition of well-ordering. –  Brian M. Scott Sep 12 '13 at 6:42
What is a well-ordered relation? Do you mean a well-ordering or a well-founded relation? –  William Sep 12 '13 at 6:45
I meant well ordering. –  Danis Fischer Sep 12 '13 at 6:46
@CoarguAliquis A well-ordering is a total ordering by definition. –  William Sep 12 '13 at 6:46
A well-order may be defined without the restriction that it must be linear as it can be proved as soon as any set has a minimum. Both variations with and without the linearity restrictions are used by mathematicians. –  Keinstein Sep 12 '13 at 11:34

2 Answers 2

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Take an antichain $C⊆A$ in $(A,R)$, that is a set of pairwise non-comparible elements($∀x,y∈C: x\mathrel{\not R} y \text{ and }y\mathrel{\not R}x$). Thus it conains many minimal but no smallest element. On the other hand $C$ contains a smallest element as $R$ is a well-order. This is a contradiction.

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It's simple, you only need to take a pair {a,b} a!=b of a well ordered set, now this pair has a minimum element, so either a < b or b < a

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