0
$\begingroup$

Following question: Let's assume that W is a wellfounded set, i.e. it has a partial order and every nonempty subset of W has minimal elements with respect to the order.

Now we can easily define a binary relation 'preceeds' with the definition

a.preceeds(b) = b.is_minimal({x: a < x})

I am not able to prove that the fact that an element b has no predecessor (with respect to the preceeds relation) implies that b is minimal in W.

Is it possible in a wellfounded set that an element b has no predecessor but there are elements a below it (i.e. a < b)? If this is possible are there examples?

Thanks for any help.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

Sure. Just take the natural numbers with the usual ordering, and slap on a new maximum element. This is well-founded (it is the ordinal $\omega + 1$), but the maximum element has no predecessor, despite not being minimal either.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thanks. This explains why I am not able to prove this wrong assertion. $\endgroup$
    – Helmut Brandl
    Oct 4, 2012 at 19:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.