
Re: Given a set , is there a disjoint set with an arbitrary cardinality?
Posted:
Dec 3, 2012 11:33 AM


jaakov <removeit_jaakov@deleteit_ro.ru> writes:
> Thank you. To finish this proof, we have to prove that your class, let > us call it Z, contains an initial segment of cardinality k. However, > informally speaking, X may hypothetically contain too many ordinals, > still being a set, such that too few ordinals remain in Z. Could you > please expand on why Z contains an initial segment of cardinality k?
Assuming choice, the cardinality k of X is an (initial) ordinal, so take the set X' = {alpha  alpha < k} and let Y be the set {x + lambda  x in X'}, where lambda is an ordinal > any ordinal in X.
 Aatu Koskensilta (aatu.koskensilta@uta.fi)
"Wovon man nicht sprechen kann, darüber muss man schweigen"  Ludwig Wittgenstein, Tractatus LogicoPhilosophicus

