jaakov
Posts:
11
Registered:
12/3/12


Re: Given a set , is there a disjoint set with an arbitrary cardinality?
Posted:
Dec 3, 2012 5:43 PM


On 03.12.2012 20:08, Butch Malahide wrote: > 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), How?
> 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.
Not quite. You have (T,x) in T <=> T in P(X) and (T,x) not in T. This is not yet a contradiction.
Thank you anyway.
Jaakov.

