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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.

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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.

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

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