Show the existence of family of sets

Question

Show that there exists a family with cardinality of $c$, of subsets of $\mathbb{N}$, such that an intersection of any three elements of the family is an infinite (countable) set and the intersection of any four elements of the family is a finite set.

Other than the fact that set of infinite subsets of $\mathbb{N}$ is infinite, I couldn't come up with a way to start. I suppose that such family should be constructed, possibly using set $\mathbb{Q}$, which would then imply that the statement is true for $\mathbb{N}$. Any ideas on how to solve this are welcome.

Edit
Malady suggested in the comments that i try constructing four sets such that the given statement is true for them, which is what I did.
So first I'll define four sets $A=\mathbb{N}$, $B=\{\frac{1}{n}:n\in\mathbb{N}\}$, $C=\{\frac{2}{n}:n\in\mathbb{N},\gcd(2,n)\}$ and $D=\{\frac{3}{n}:n\in\mathbb{N},\gcd(3,n)\}$. Four sets satisfying the conditions from the statement are: $$A\cup B\cup C$$ $$A\cup B\cup D$$ $$A\cup D\cup C$$ $$D\cup B\cup C$$ This suggests that we can, possibly, construct such family by defining some countable subsets of a countable set and then constructing the elements of the family as some union of those subsets. But I'm not sure how to generalize this so that the family is uncountable.

Answer 1

I will define a family of infinite subsets of the countably infinite set $M=\bigcup_{n\in\mathbb N}M_n$ where $M_n$ is the set of all $3\times n$ matrices of $0$s and $1$s; namely, the family $\{X_a:a\in\{0,1\}^\mathbb N\}$ where $$X_a=\bigcup_{n\in\mathbb N}\{A\in M_n:\text{ some row of }A\text{ is an initial segment of }a\}\subseteq M.$$ If $a,b,c,d$ are four distinct infinite sequences of $0$s and $1$s, then $X_a\cap X_b\cap X_c$ is an infinite subset of $M$ but $X_a\cap X_b\cap X_c\cap X_d$ is finite.

$X_a\cap X_b\cap X_c$ is infinite because, for each $n\in\mathbb N$, there is a $3\times n$ matrix whose first row is an initial segment of $a$, whose second row is an initial segment of $b$, and whose third row is an initial segment of $c$.

For some $N\in\mathbb N$ the $N^\text{th}$ initial segments of $a$, $b$, $c$, and $d$ are all different. Hence $X_a\cap X_b\cap X_c\cap X_d\subseteq\bigcup_{n\lt N}M_n$ which is finite.

More generally, for any positive integer $k$ we can construct a family $\mathcal A\subseteq\mathcal P(\mathbb N)$ with $|\mathcal A|=\mathfrak c$ such that the intersection of any $k$ elements of $\mathcal A$ is infinite while the intersection of any $k+1$ elements of $\mathcal A$ is finite.

Answer 2

I have a more geometrical answer inspired by this answer, which is longer but maybe more motivated.

First, note that hyperplanes have a property very similar to what we are looking for: $n$ general $n$ dimensional hyperplanes of an $n+1$ dimensional space will have a $1$ dimensional intersection- which contains infinitely many points. But intersect $n+1$ general hyperplanes and you get a $0$ dimensional space, which is finite.

To turn this into countable sets, consider the set $\mathbb{Z}^{n+1}$, and, for any given $n$ dimensional subspace, take the subset of points that are a distance $< 1$ away from it (using the max norm for convenience). Any distinct subspaces will give different sets. (Pick a vector in $S_1$ but not $S_2$ and scale it up until it's a long distance from $S_2$. There will be an integer point within $1$ of this vetor that can't be near $S_2$.)

If we intersect the sets generated by $n$ general subspaces, our intersection includes, in particular, the points that are a distance $1$ away from the $1$ dimensional space the subspaces intersect at. Which means it contains infinitely many points, as every point on the line is within $1$ of an integer point.

Now intersect $n+1$ general subspaces, and we get $\{0\}$. Intuitively, no point in our intersection can get too far from $0$, as our subspaces get further and further apart from each other; explicitly, we can say this: Consider all the vectors $x$ in the unit sphere, and compute $f(x)$ the distance to the furthest of the $n+1$ subspaces. This is a $\max$ of continuous functions, so it is continuous, and the sphere is compact, so $f$ attains a minimum, $m>0$. So if we scale up any point $x$ in the sphere by $\frac1m$ it will be a distance of at least $1$ from the furthest plane, so any point of size $\ge \frac1m$ has at least one of our subsets of $\mathbb{Z}^{n+1}$ that it isn't part of. Which means it's not in our intersection. So all the points in our intersection of subsets of $\mathbb{Z}^{n+1}$ are contained in a sphere of radius $\frac1m$, which means there are finitely many.

But can we choose a $\mathfrak{c}$-sized set of hyperplanes through $0$ such that any $n+1$ of them are general (i.e. their normal vectors are linearly independent)? There are many ways to do this, one being to look at the contour $(1, x, x^2, \dots, x^n)$ for $x>0$ and pick $\mathfrak{c}$-many normal vectors from this contour. Any $n+1$ of these will then form a Vandermonde matrix, which we know has nonzero determinant, so they must be linearly independent.

We have our sets and the proof is complete.