Date: Dec 3, 2012 6:28 PM
Author: Butch Malahide
Subject: Re: Given a set , is there a disjoint set with an arbitrary cardinality?

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.