# Set Theory: Help with a proof

• Nov 11th 2010, 02:51 AM
somebd
Set Theory: Help with a proof
Prove that exist X ⊂ P(N)

P(N) is a power set of natural numbers
1 cardinality of X is continuum
2 for each A (element of X) in X, A is infinite
3 for each A and B in X, A is not equal to B, A intersection with B is finite

Please, I have no idea how to solve this :o
• Nov 11th 2010, 07:37 AM
MoeBlee
First, I'd state it more clearly:

Let N be the set of natural numbers.
Let P be the powerset operation.
Let R be the set of real numbers.

Show that there exists an X such that:

X is a subset of PN
card(X) = card(R)
Every A in X is infinite
If A and B are distinct members of X then the intersection of A and B is finite
• Nov 11th 2010, 07:41 AM
somebd
thank you very much for the ideas!
• Nov 11th 2010, 09:53 AM
MoeBlee
I haven't had time to think about solving, but a fact might help.

R is 1-1 with the set of infinite subsets of N.

So maybe we can cut that set down to a set of members that have pairwise only finite intersections, while also keeping enough to have a 1-1 with R.

And is the axiom of choice used? (If so, let W be a well ordering of R, then maybe use transfinite recursion on R to define a function f such that for each x in R we have f(x) is an infinite subset of N that has only a finite intersection with any f(y) for y such <y x> in W?)

By the way, what book is this from?
• Nov 11th 2010, 09:56 AM
somebd
Our professor gave us this task at the last lecture (I study in Estonia) so I don't know if it is from any particular book.
And thank you once again :)
• Nov 11th 2010, 01:59 PM
emakarov
Hint: consider a complete infinite binary tree.

Apparently, this is a theorem by W. Sierpinski. For example, this article by Erdős et al. says the following. Two sets $\displaystyle A_1$ and $\displaystyle A_2$ are called almost disjoint if $\displaystyle |A_1\cap A_2|<A_i\quad(i=1,2)$. "An old and well known theorem of W. Sierpinski is that an infinite set of power m contains more than m subsets of power m which are pairwise almost disjoint." One can take $\displaystyle m=\aleph_0$. I am not sure whether continuum hypothesis is needed to bridge the gap between "more than m subsets" and "continuum subsets."

I also found this fact in Google Books, but the pages with the hint are missing.