Prove the existence of $X \subset \mathcal{P}(\mathbb{N})$ with continuum cardinality, such that each intersection of two of its elements is finite.

Question

Prove that there exists a set $X \subset \mathcal{P}(\mathbb{N})$ so that $|X| = \mathfrak{c}$ and for every $A,B\in X$, the set $A \cap B$ is finite.

Since this statement feels unintuitive, it's difficult to come up with ideas on how to prove it. Surely if the set X has continuum cardinality, some intersections of it's elements are bound to be infinite as well?

My thought then went to assembling $X$ so that all elements in $X$ contain the number 1 and no other common elements. However, that seems quite difficult, when we have to make a continuum cardinality of these sets where the only common element is 1.

So basically, I'm stuck and I'd like help.

Answer 1

For each $b:\mathbb{N}\rightarrow \{0,1\}$, we construct an element $X_b$ of $X$ in the following way:

All numbers written in binary:

$X_b=\{ 1, 1b_1, 1b_1b_2, 1b_1b_2b_3, \ldots\}.$

Then $X$ satisfies the desired properties.

Answer 2

There are various ways of constructing such $X$. One is given by the previous answer, here is my favourite (together with the one in i707107's answer) :

For $x\in \mathbb{R}$, pick a sequence $(q_n^x)$ of rationals that converges to $x$.

Then $X=\{\{q_n^x, n\in \mathbb{N}\}, x\in \mathbb{R}\}$ is a subset of $\mathcal{P}(\mathbb{Q})$ of size $\mathfrak{c}$, by unicity of limits, and if $q_n \to x, r_n \to y$ and infinitely many $q_n $ are also $r_m$'s, then one can extract a subsequence of $q_n$ that tends to $y$, so that $x=y$ (unicity of limits), so distinct elements of $X$ have a finite intersection. It is easy to transfer it to $\mathcal{P}(\mathbb{N})$ since $\mathbb{Q}$ is countable.