
Re: Given a set , is there a disjoint set with an arbitrary cardinality?
Posted:
Dec 3, 2012 2:08 PM


On Dec 3, 8:21 am, jaakov <removeit_jaakov@deleteit_ro.ru> wrote: > > Given a set X and a cardinal k, is there a set Y such that card(Y)=k and > X is disjoint from Y?
This is equivalent to asking, for a given set X, is there a set Y such that card(Y) = card(X) and X is disjoint from Y?
Namely, given a set X and a cardinal k, we can take a set K with card(K) = k and let X' be the union of X and K. If we can find a set Y' such that card(Y') = card(X') and Y' is disjoint from X', then Y' has a subset Y such that card(Y) = k, and of course Y is disjoint from X.
> Is there a proof of this fact that works without the axiom of regularity > (= axiom of foundation) and does not assume purity of sets?
Let X be a given set. For each set S in P(X), let Y_S = {(S,x): x in X}. Clearly Y_S = X. Assuming that X meets Y_S for each S in P(X), we could define a surjection from X to P(X), in contradiction to Cantor's theorem. Therefore we can choose S in P(X) so that X is disjoint from Y_S.
A more concrete version of this argument, a la Russell: Given a set X, let T = {(S,x): S in P(X), x in X, (S,x) in X, (S,x) not in S} and let Y = {(T,x): x in X}. Clearly Y = X. Assuming X is not disjoint from Y, there is an element x in X such that (T,x) is in X. Now we get the Russell paradox in the form (T,x) is in T <> (T,x) is not in T.

