Re: Infinite power sets

David R Tribble wrote:
>> Assuming it is possible to construct set G(A) that contains some,
>> but not all, of the subsets of A, is it necessarily true that:
>> [...]
>
David C. Ullrich wrote:
> Yes - there's simply no reason that card(G(A)) cannot equal card(A),
> whether or not you assume CH. What if G(A) = {{x} : x in A}?

Sorry, my fault for not being clearer.

My intent is to create an infinite set G(A) from some, but not
all, of the possible subsets of A, such that there are indeed
more members in G(A) than in A, and then to determine if
card(G(A)) must then (according to GCH) be required to be the
same as card(P(A)).


Arturo Magidin wrote:
> I suspect you are thinking about the ->Generalized<- Continuum
> Hypothesis: one has to be careful in stating it in the absence of the
> Axiom of Choice. One common way of stating it is:
> GCH: Given a set x, there is no set y such that card(x)<card(y)<card(P(x)).
> Equivalently:
> GCH: For any sets x and y, if card(x) <= card(y) <= card(P(x))
> then card(x)=card(y) or card(y)=card(P(x)).
>
> Note that GCH stated this way (and in a number of other ways) implies
> the Axiom of Choice.
>
David Tribble:
>> So given G(A), a set having more members than A and also (seemingly)
>> having less members than P(A), card(G(A)) should be equal to
>> card(P(A)), right?
>
Arturo Magidin wrote:
> Depends on what you mean by "more members". Do you mean, A set G(A)
> such that G(A) is contained in P(A), and card(A)<card(G(A))? If so,
> then assuming GCH, yes, card(G(A)) would necessarily be equal to
> card(P(A)).

Exactly, as I will try to explain more clearly.

For example, we could build G(A) by adding all the single-member
subsets of A {0}, {1}, {2}, etc., and then use some random process
for choosing n-member subsets (for n>1) out of P(A) and adding
them to G(A), as long as the probability of adding any particular
member of P(A) to G(A) was less than 1, and as long as the resulting
members of G(A) were not countable by the members of A (or however
you say that).

So we end up with a set G(A) where card(G(A)) > card(A), because
there are members in G(A) that can't be mapped to members in A.
But it also *appears* that G(A) has less members than P(A), because
we didn't include all of the members of P(A) when we constructed G(A).
G(A) is thus necessarily a subset of P(A), right?

Then, the question is, must card(G(A)) = card(P(A)) (assuming GCH)?

I think I've answered my own question (=yes), because of the fact
that G(A) is a subset of P(A), and G(A) and P(A) are both infinite
sets, then there must exist a mapping P(A) <-> G(A).

.

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