jaakov
Posts:
11
Registered:
12/3/12


Re: Given a set , is there a disjoint set with an arbitrary cardinality?
Posted:
Dec 4, 2012 5:19 AM


On 04.12.2012 00:28, Butch Malahide wrote: > On Dec 3, 4:43 pm, jaakov<removeit_jaakov@deleteit_ro.ru> wrote: >> 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? > > Consider any x in X. If x is an ordered pair (u,v) with u in P(X), > define f(x) = u; otherwise define f(x) = X. This defines a function > f:X > P(X). If Y_S has nonempty intersection with X, then S is in the > range of f. > >>> 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. > > From the definition of T, namely > > T = {(S,x): blah, blah, (S,x) in X, blah} > > it can be seen that T is a subset of X, that is, T is an element of > P(X). Inasmuch as "T in P(X)" is true, the statement "T in P(X) and > (T,x) not in T" simplifies to "(T,x) not in T". > >> Thank you anyway.
Right. Thank you!
Jaakov.

