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 4, 2012 5:19 AM
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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.
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