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Topic: Given a set , is there a disjoint set with an arbitrary cardinality?
Replies: 28   Last Post: Dec 4, 2012 5:50 PM

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 Butch Malahide Posts: 894 Registered: 6/29/05
Re: Given a set , is there a disjoint set with an arbitrary cardinality?
Posted: Dec 3, 2012 2:08 PM
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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.

Date Subject Author
12/3/12 jaakov
12/3/12 forbisgaryg@gmail.com
12/3/12 Aatu Koskensilta
12/3/12 jaakov
12/3/12 Carsten Schultz
12/3/12 jaakov
12/3/12 Aatu Koskensilta
12/3/12 jaakov
12/3/12 Aatu Koskensilta
12/3/12 jaakov
12/3/12 Carsten Schultz
12/3/12 jaakov
12/3/12 Aatu Koskensilta
12/3/12 Butch Malahide
12/3/12 jaakov
12/3/12 Butch Malahide
12/4/12 jaakov
12/4/12 forbisgaryg@gmail.com
12/4/12 William Elliot
12/4/12 forbisgaryg@gmail.com
12/4/12 William Elliot
12/4/12 William Elliot
12/4/12 jaakov
12/4/12 William Elliot
12/4/12 jaakov
12/4/12 Shmuel (Seymour J.) Metz
12/4/12 Spammer
12/4/12 jaakov

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