Unique infinite subsets of the integers

Question

Edit: Great points on the comments. There is no unique set of unique infinite subsets of the integers. Is this a better question?

What is the largest possible cardinality of a set which is a set of unique infinite subsets of the integers, or to put it another way, a set of infinite subsets of the integers such that the intersection of any two subsets is empty?

Thank you.

Answer 1

I'm not quite clear what you're asking:

• The set of all infinite subsets of $\mathbb{N}$ is the same cardinality as the set of all subsets of $\mathbb{N}$, $2^{\aleph_0}$; this is because the set of finite subsets is countable.

• But since you talk about intersections, it sounds like you mean something else. It's still not clear - for instance, the set of all infinite subsets of $\mathbb{N}$ has the property that its intersection is empty . . .

• but this is true for silly reasons: for instance, $\{$evens$\}\cap\{$odds$\}=\emptyset$, and so this stays true when we add more infinite sets to the collection. So maybe you're trying to ask something different.

• Take 1: maybe you want the intersection of the whole collection to be empty, but not proper subcollection. For instance, let $A_n=\{m: m>n\}$; then $\bigcap A_n=\emptyset$, but any finite intersection is nonempty (in fact, cofinite).

• Take 2: maybe you are asking about having every intersection be empty - that is, you want a family of sets $\{A_i\}$ such that $A_i\cap A_j=\emptyset$ whenever $i\not=j$. Then clearly such a family can be at most countably infinite, since there are only countably many natural numbers in the first place. (Note that if we allow finite intersections, then we switch - there are continuum-sized families!)

Does this help?

Answer 2

If $F$ is any family of pair-wise disjoint subsets of $N,$ then for each $n\in N $ there is at most one $f\in F$ such that $\min f=n$. For $f\in F$ let $G(f)=\min f$ if $f\ne \phi$, and $G(f)=0$ if $f=\phi.$ Then $G$ is a bijection from $F$ to a subset of $N\cup \{0\}$ .So $F$ is countable. Example: For each prime $p$ let $f_p=\{p^n :n\in N\}$ and let $F=\{f_p :p$ prime $\}$.