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Topic: question on set theory
Replies: 4   Last Post: Jan 4, 2013 2:17 AM

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 Dan Luecking Posts: 26 Registered: 11/12/08
Re: question on set theory
Posted: Jan 2, 2013 3:30 PM

On Wed, 2 Jan 2013 17:37:20 +0000 (GMT),
"matmzc%hofstra.edu@gtempaccount.com"
<matmzc%hofstra.edu@gtempaccount.com> wrote:

>Happy New Year everybody!
>
> For a set X I will write X < Y to indicate that the cardinality of X is strictly smaller than the cardinality of Y. Let me write P(X) for the power set of X. Consider the proposition:
>
>Prop: If P(X) < P(Y)then X < Y.
>
>Is it possible to prove this in ZFC (without continuum hypothesis)? If not is it perhaps the case that the Prop is equivalent to generalized continuum hypothesis? It is trivial to show that generalized continuum hypothesis implies the proposition, but what about the other direction?
>

This should only require the axiom of choice:

Let X ~ Y mean X and Y are in 1-1 correspondence.
One defines < as follows:
First: X <= Y means X is in 1-1 correspondence with a subset of Y
Then: prove (I think AC is required):
if X <= Y and Y <= X then X ~ Y.
Then define X < Y to mean X <= Y but not X ~ Y.

This yields the tricotomy property: Either X < Y or Y < X or X ~ Y
(exclusive or).

Now if P(X) < P(Y) then there are three cases for X and Y
1. X ~ Y. In this case that 1-1 correspondence of X and Y
induces a 1-1 corespondence between subsets
(elements of P(X)) so P(X) ~ P(Y) and this case is out.
2. Y < X. This means Y is in 1-1 corespondence with a subset
of X. As in case 1, this provides a 1-1 correspondence of
P(Y) with a subset of P(X) so P(Y) < P(X) or P(Y) ~ P(X)
and this case is out.
3. X < Y is the only case remaining.

Dan
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