Families of subsets with "small" intersections

Question

A classic problem in real analysis asserts the existence of $\mathfrak{c}$-sized $X \subset \mathcal{P} (\mathbb{N})$ with $|x \cap y|$ finite for any $x,y \in X$. The solution is fairly simple: swap $\mathbb{N}$ with $\mathbb{Q}$ without loss of generality, and for each $x \in \mathbb{R}$ let $p_x \subset \mathbb{Q}$ be some strictly increasing sequence converging to $x$. Then $X=\{p_x\}_{x \in \mathbb{R}}$ does the trick.

I'm wondering if this problem admits the following generalization.

For any infinite set $S$, there exists an $X \subset \mathcal{P}(S)$ which satisfies the following two conditions: $|X| = |\mathcal{P}(S)|$ and $|x \cap y|< |S|$ for any $x,y \in X$.

A variation of this question would be positing that the size of $|x \cap y|$ have a strict upper bound independent of $|S|$.

That is to say:

There exists a cardinal $\kappa$ such that for any infinite set $S$, there exists an $X \subset \mathcal{P}(S)$ with $|X|=|\mathcal{P}(S)|$ and $|x \cap y|<\kappa$ for any $x,y \in X$.

Answer 1

What you describe in the first part is what is called an almost disjoint family of subsets of $\Bbb N$. We call $x,y\in\mathcal P(\Bbb N)$ almost disjoint when $x\cap y$ is finite, and thus an almost disjoint family is a family in which all members are mutually almost disjoint.

We can generalise this definition of almost disjointness to any infinite cardinality $\kappa$, by saying sets $x$ and $y$ of size $\kappa$ are $\lambda$-almost disjoint, if $|x\cap y|<\lambda$. Suppose that $|S|=\kappa$, then your generalisation is the statement that there exists a family $X\subset\mathcal P(S)$ of size $2^\kappa$, such that $X$ is $\lambda$-almost disjoint.

This turns out to be quite an interesting statement, and certainly not one that is generally provable, even if we take $\lambda$ as large as $\kappa$ itself (which I will assume in the remainder, and thus write "almost disjoint" instead of "$\kappa$-almost disjoint").

It is a consistent statement, since it is a consequence of $\mathsf{ZFC}+\mathsf{GCH}$:

For each $f:\kappa\to \{0,1\}$, define a set $x_f=\{f\restriction \alpha\mid\alpha<\kappa\}$. If $f,f':\kappa\to\{0,1\}$ are distinct functions, then it's easy to see that $|x_f\cap x_{f'}|<\kappa$. The set $X=\{x_f\mid f:\kappa\to\{0,1\}\}$ is therefore almost disjoint, and has size $2^\kappa$.

We use $\mathsf{GCH}$ to show that $X\subset\mathcal P(S)$ for some $S$ with $|S|=\kappa$. This $S$ is of course the set of functions from any $\alpha<\kappa$ to $\{0,1\}$. Therefore $|S|=2^{<\kappa}$, which is equal to $\kappa$ by $\mathsf{GCH}$.

However, it is also consistent with $\mathsf{ZFC}$ that the statement fails for certain $\kappa$. See for example the last part of this answer on MathOverflow in which a model is produced in which no almost disjoint subset of $\omega_1$ exists of size $2^{\omega_1}$.

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If $\kappa$ is regular, we can however always find an almost disjoint family of size $\kappa^+$:

Let $X\subset\mathcal{P}(S)$ and $|X|=\kappa$, then we can find an $y\subset S$ such that $y$ is almost disjoint from all sets in $X$. Let $\langle x_\alpha\mid\alpha<\kappa\rangle$ enumerate $X$. Note that we can assume without loss of generality that $|x_\alpha|=\kappa$ for each $\alpha$.

As $|x_\alpha\cap x_\beta|<\kappa$ for each $\beta<\alpha$, we see that $x_\alpha\setminus \bigcup_{\beta<\alpha} x_\beta$ is nonempty, since the $\alpha$ many sets $\{x_\alpha\cap x_\beta\mid \beta<\alpha\}$ cannot cover all of $x_\alpha$ by $|x_\alpha|=\kappa$ and $\kappa$ being regular. Therefore, let $\gamma_\alpha$ be any any element in $x_\alpha\setminus\bigcup_{\beta<\alpha}x_\beta$, and let $y=\{\gamma_\alpha\mid \alpha<\kappa\}$. Then $y\cap x_\alpha\subset \{\gamma_\beta\mid \beta\leq \alpha\}$, and thus $|y\cap x_\alpha|<\kappa$.