Cardinality of subsets with finite intersections

Question

Let $\ F_0 $ be a family of disjoint subsets of $ C$. $\ |C|= \aleph_0$.

Prove that $\ (*) |F_0|\leq\aleph_0 $.

This part was relatively simple, in the presence of choice an injection can be defined from each set to one of its elements in C. By Cantor the conclusion is simple.

I'd like to show that $\ |F_1|\leq\aleph_0$ where $\ F_1 $ is a family of subsets of $C$ in which for any two distinct $\ A,B\in F_1$, $A\cap B \leq 1 $.

From here I'd like to inductivly prove $\ |F_n|\leq\aleph_0$ where $\ F_n $ is a family of subsets of $C$ in which for any two distinct $\ A,B\in F_n$, $A\cap B \leq n $.

I'd like to isolate the intersections and show each family as a disjoint family and use $ (*) $ from above. My problem is properly defining an injection.

Answer 1

You don’t need induction.

HINT: For each $a\subseteq C$ of cardinality $n+1$ let $F_n(a)=\{f\in F_n:a\subseteq f\}$, and show that $|F_n(a)|\le 1$.

Answer 2

Hint (for both the $n=1$ case and the inductive step): Suppose $F_n$ is uncountable, and let $F_n(x)=\{A\in F_n:x\in A\}$ for each $x\in C$. Show that $F_n(x)$ must be uncountable for some $x\in C$, and then apply the induction hypothesis to $\{A\setminus\{x\}:A\in F_n(x)\}$.