Consider $X=\{2,3,4\}$. This set has some interesting properties:


* The number of even numbers in $X$ is 2, an even number
* The number of odd numbers in $X$ is 1, an odd number
* The number of primes in $X$ is 2, a prime
* The number of squares in $X$ is 1, a square


We might wonder if this example can be extended to any number of such similar "properties." Hence:


**Question**:


Suppose $\mathcal{A} = \{A\_1,A\_2,\ldots,A\_k\} \subseteq 2^{\mathbb{N}}$ is a finite family of infinite subsets of nonnegative integers (that is, each $A\_i$ is infinite). Does there exist $X \subseteq \mathbb{N}$ such that $|A\_i \cap X| \in A\_i$ for all $i=1,\ldots,k$?


Note that it is not necessarily the case that such an $X$ exists if the $A\_i$ are allowed to be finite, even if $\mathcal{A}$ satisfies the following obvious necessary condition:


* $\mathrm{min}(A) \leq |A|$ for all $A \in \mathcal{A}$ (where $\mathrm{min}(\emptyset):=\infty$);


consider the example $\mathcal{A}=\{\{1\},\{2,4\},\{1,2,4\}\}$.


Also note that there is not necessarily such an $X$ if $\mathcal{A}$ itself is allowed to be infinite, as in the example $\mathcal{A}=\{A\_1,A\_2,\ldots\}$ where $A\_i =\{i,i+1,i+2,\ldots\}$.



Cardinality of intersection with set belongs to set for all sets in a family
Just take $A\_1=\{2^n : n \in \mathbb N\}$, $A\_2=\{2^n : n \textrm{ even}\}$, and $A\_3=\{2^n : n \textrm{ odd}\}$. If $|A\_2 \cap X| \in A\_2$ and $|A\_3 \cap X| \in A\_3$, then $|A\_1 \cap X|$ is a sum of two distinct powers of $2$, and hence is not in $A\_1$.


