most recent 30 from http://mathoverflow.net 2013-05-22T07:21:49Z http://mathoverflow.net/feeds/question/108844 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/108844/wellfounded-sets-and-predecessors Helmut Brandl 2012-10-04T18:48:45Z 2012-10-04T18:57:46Z

<p>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.</p> <p>Now we can easily define a binary relation 'preceeds' with the definition</p> <p>a.preceeds(b) = b.is_minimal({x: a &lt; x})</p> <p>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.</p> <p>Is it possible in a wellfounded set that an element b has no predecessor but there are elements a below it (i.e. a &lt; b)? If this is possible are there examples?</p> <p>Thanks for any help.</p>

http://mathoverflow.net/questions/108844/wellfounded-sets-and-predecessors/108845#108845 Sridhar Ramesh 2012-10-04T18:57:46Z 2012-10-04T18:57:46Z

<p>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.</p>