jaakov
Posts:
11
Registered:
12/3/12
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Re: Given a set , is there a disjoint set with an arbitrary cardinality?
Posted:
Dec 3, 2012 5:43 PM
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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.
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