Re: Every set x equinumerous with a set y disjoint from x?


David C. Ullrich wrote:
On 27 Aug 2006 15:59:21 -0700, "MoeBlee" <jazzmobe@xxxxxxxxxxx> wrote:

David C. Ullrich wrote:
Say S is a set. Some elements of S may be ordered
pairs; let T be the set of y such that there exists
x such that (x,y) is in S. Now card(T) < card(P(S));
choose y in P(S) not equal to any element of T,
and consider S x {y}.

Then T = range(S), which is okay (S has a range even if S is not a
relation).

But how do justify card(T) < card(P(S))?

I can justify it this way, but it uses the axiom of choice:

Az card(range(z)) =< card(z) < card(Pz).

I used the axiom of choice to derive Az card(range(z)) =< card(z).

So how did you get card(T) < card(P(S)) without the axiom of choice?

Come now.

Without AC I'm not sure what card even means. So forget that.
But come on now - all we need here is that there exists y such
that y is not in T; the fact that card(T) < card(P(S)) was just
supposed to be a convenient way to see there is such a thing,
but in fact P(S) is irrelevant.


If we have regularity then we can define the cardinal of a set to be
the set of all sets equipollent to that set of least possible rank. But
in any event we can just interpret card(T) < card(P(S)) to mean: there
is an injection from T to P(S), and there is no bijection from T to
P(S). Your argument is fine.

Theorem: For every T there exists y such that y is not an element of
T.

Proof: Surely you've seen the proof of this? QED.

MoeBlee


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David C. Ullrich

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