Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|cite|improve this question
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. – user65985 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

up vote -1 down vote accepted

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.

share|cite|improve this answer

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.