In message <email@example.com>, MoeBlee <firstname.lastname@example.org> writes >On Aug 24, 3:02 pm, WM <mueck...@rz.fh-augsburg.de> wrote: >> If there were sets that >> could not be well-ordered, then they had no ordinal number. And then >> their cadinal number could not be fixed. > >That is, if we use a certain definition of "card", yes.
There is no associated aleph, yes, but the set still has cardinality and a cardinal can be defined using "Scott's trick".
>> There would exist sets A and >> B for which the cardinalities were not in trichotomy. > >More exactly, it would be the case that we can't prove that for every >set A and B, either card(A) =< B or card(B) <= card(A). > >It would NOT be the case that we could provide examples of sets A and B >that don't satisfy card. > >Most importantly, it would NOT be the case that we could prove: > >There exist sets A and B such that A and B fail trichotomy.
Are you sure? If A cannot be well-ordered but B can, then card(B)<= card(A) is possible, but card(A) <= card(B) is not. However, you can not have card(B) <= card (A) for every well-orderable B, so there will be some B which is incomparable with A. In particular there will be a least aleph incomparable to A. It is called the Hartog's number of A. http://en.wikipedia.org/wiki/Hartogs_number -- David Hartley