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.



wellfounded sets and predecessors
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.


