Are there uncountably many $A\_\alpha $ of subsets of $\mathbb{N}$ with the following two properties:


1. Each $A\_\alpha$ has positive upper natural density
2. $A\_\alpha \cap A\_\beta$ is a finite set for $\alpha \neq \beta$


If the answer is no then the next question:


Are there uncountably many $A\_\alpha $ of subsets of $\mathbb{N}$ with the following two properties:


1. Each $A\_\alpha$ has positive upper natural density
2. $A\_\alpha \cap A\_\beta$ has zero natural density for $\alpha \neq \beta$


**Note:** As indicated [at this MathSE post](https://math.stackexchange.com/questions/162387/), there are uncountably many infinite subsets of $\mathbb{N}$ with pairwise finite intersection. However,  

I could not modify the method indicated in the discussion of that post to get an answer to my question.



Uncountably many subsets of the natural numbers with certain natural density condition
Amazingly, the answer to the main question is yes. For each $n$ let $I\_n = [2^{2^n}, 2^{2^{n+1}})$. Then let $\mathcal{C}$ be an uncountable family of infinite subsets of $\mathbb{N}$ any two of which have finite intersection. For each $B\in \mathcal{C}$ let $B' = \bigcup \{I\_n: n \in B\}$. Then $\{B': B \in \mathcal{C}\}$ is the desired family. The point is that if $B$ is any infinite subset of $\mathbb{N}$ then $B'$ has upper density 1.


Edit: or just take $I\_n = [n!, (n+1)!)$, so that $\frac{|I\_n|}{(n+1)!} = \frac{n}{n+1}$.


