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Does there exist an uncountable $P \subset \mathcal{P}(\mathbb{N}) $ with the property that for any distinct $x,y \in P$, $|x \cap y|$ is prime?

A more general, but likely harder, question: is it possible to characterize the set $\mathcal{A}$ of all subsets $S \subset \mathbb{N}$ with the following property: there is some uncountable $P \subset \mathcal{P}(\mathbb{N})$ such that for all distinct $x,y \in P$, $|x \cap y| \in S$.

It's easy to show no $S$ finite belongs to $\mathcal{A}$. It's also not hard to show every $\text{mod} \ m$ equivalence class, belongs to $\mathcal{A}$.

One reasonable sounding idea is that this is related to the asymptotic density of $S$ in $\mathbb{N}$.

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  • $\begingroup$ Just to be sure. By "$x\cap y\in S$" you mean $x\cap y=\{z\}$ for some $z\in S$, right? $\endgroup$ – Maurizio Moreschi Oct 16 '18 at 16:49
  • $\begingroup$ What do you mean by $x \cap y$ prime? Do you mean that it's a subset of the set of primes? $\endgroup$ – user44191 Oct 16 '18 at 16:50
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    $\begingroup$ Sorry. That was a typo. I should have taken its cardinality. I also renamed the title to make it more appropriate. $\endgroup$ – Arsh Jhaj Oct 16 '18 at 16:51
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    $\begingroup$ @YaakovBaruch That works for $S=\mathbb{N}$, but for arbitrary infinite $S$ I think it's actually much harder to modify that than it is to work with binary trees. $\endgroup$ – Noah Schweber Oct 16 '18 at 17:28
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    $\begingroup$ @user44191 In a general commutative ring, a subset (usually an ideal) $P$ can be called prime iff $ab\in P\implies a\in P\vee b\in P$; the prime integers generate exactly the prime ideals in the integers in this sense, but it also makes sense in any commutative (semi)ring. I thought this was what the OP meant and found the question fascinating, asking for intersections to be exactly the ideals generated by primes in $\mathbb{N}$. $\endgroup$ – Alec Rhea Oct 16 '18 at 19:12
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I guess this is a variant of Noah's construction:

Let $S$ be an infinite subset of $\mathbf{N}=\{0,1,2,\dots\}$. Define a leafless rooted tree $V_S$, starting from a root at level 0, such that given a vertex $v$ of level $n\ge 0$, $v$ has exactly 1 successor if $n\notin S$ and exactly 2 successors if $n\in S$.

Let $\mathcal{W}\subset\mathcal{P}(V_S)$ be the set geodesic rays not containing the root, based at a vertex of level 1; since $S$ is infinite, $\mathcal{W}$ has cardinal $2^{\aleph_0}$. Then for any distinct $U,V\in\mathcal{W}$, $U\cap V$ is either empty, or is the geodesic segment from the some vertex of level 1 to the branching point $v$ at which $U,V$ fork; this has cardinal $n$, the level of $v$; also in case $0\notin S$, $U\cap V$ cannot be empty. Conversely every such geodesic segment has the form $U\cap V$ for suitable $U,V$, and so is the empty set if $0\in S$. So the set of $|U\cap V|$ when $U,V$ range over distinct elements of $\mathcal{W}$, is exactly $S$.

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  • $\begingroup$ Note that this is choice-free. $\endgroup$ – YCor Oct 16 '18 at 18:18
  • $\begingroup$ My construction is also choice free. (We can always just choose the lexicographically least extension.) $\endgroup$ – Noah Schweber Oct 16 '18 at 18:52
  • $\begingroup$ @NoahSchweber I also meant I don't made any choice (not only in the sense of not using the axiom of choice). $\endgroup$ – YCor Oct 16 '18 at 19:30
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    $\begingroup$ Thanks for the upvotes and for acceptation of the answer, but I should emphasize that I did my answer after Noam's was written. $\endgroup$ – YCor Oct 20 '18 at 13:17
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I claim that any infinite set has this property.

Specifically, I'll construct recursively a map $F$ from finite binary strings to finite binary strings with the following properties:

  • $\sigma\preccurlyeq\tau$ iff $F(\sigma)\preccurlyeq F(\tau)$.

  • For each $\sigma$, if $f, g$ are two infinite binary strings whose maximal common initial segment is $\sigma$, then $Set(f)\cap Set(g)=\{n<\vert F(\sigma)\vert: F(\sigma)(n)=1\}$.

    • Here "$F(h)$" is shorthand for the infinite binary sequence $\bigcup_{\sigma\prec h}f(\sigma)$, and "$Set(h)$" is the set whose characteristic function is $h$.
  • For each $\sigma$, $\vert F(\sigma)^{-1}(\{1\})\vert\in S$.

Letting $\mathfrak{S}_F=F(f): f\in 2^\mathbb{N}$, we get an uncountable - in fact, size continuum - family of sets whose pairwise intersections always have cardinality lying in $S$.


So now all we need to do is build $F$. But this is a standard finite extension argument, using the fact that $S$ is infinite to say that we always have "enough freedom" to extend the function as desired.

Specifically, the crucial lemma is the following:

  • Suppose I have a finite set of finite binary strings $\{\tau_i: 1\le i\le n\}$. Then I can find sets $\{\alpha_i: 1\le i\le n\},\{\beta_i: 1\le i\le n\}$ such that:

    • For each $i$, $\alpha_i,\beta_i$ are incomparable proper extensions of $\tau_i$ and $\{n: \alpha_i(n)=\beta_i(n)=1\}$ has cardinality in $S$.

    • We get no new intersections in incomparable strings: for $i\not=j$ and $\eta,\theta\in\{\alpha,\beta\}$, we have $\{n: \alpha_i(n)=\beta_j(n)=1\}=\{n:\tau_i(n)=\tau_j(n)=1\}$.

We then apply this, over and over, to build $F$: e.g. think of $F(010)$ as being build via a $\sigma$-move off the empty string, and then a $\tau$-move off the resulting string, and then a $\sigma$-move off that string.

Note that this is really just an elaboration of the classical argument: the usual construction of a size-continuum almost disjoint family is just the set of sets of finite binary strings, which in the notation above is $\mathfrak{S}_{id}$. Taking $F=id$ works since we only have to keep getting incomparable extensions. Once we add the "$S$-requirement," though, we need to mix in some additional work, but this requirement can be satisfied by imposing very mild restraints. This type of argument can also be used to construct a perfect set of linearly (or otherwise) independent reals without choice, and - in computability theory - to show that there is a perfect set of Turing-incomparable reals.


And I think this can be easily tweaked to get the set of cardinalities of intersections to be exactly $S$; see my comment below.

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  • $\begingroup$ Can you arrange that the set of cardinal of intersections is $S$ (and not only a subset of $S$)? $\endgroup$ – YCor Oct 16 '18 at 17:15
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    $\begingroup$ @YCor I think so: given any $n$ and a tree like the above, we can build a single new set which intersects one of the paths in a set of exactly size $n$ and whose intersection with any other set stays in $S$ (by going "node-by-node" up the tree, so we only have countably many requirements to meet). Now we do that for each $n\in S$, noting that it doesn't get any harder if instead of a tree like the above we have a tree like the above and finitely many additional sets. $\endgroup$ – Noah Schweber Oct 16 '18 at 17:19
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The following argument uses less set-theory than the earlier ones!

For each positive real number $\alpha$ we can write the binary expansion $\alpha=\sum_{i=0}^{\infty} a_i 2^{N-i}$ with $a_0=1$ and $a_i\in\{0,1\}$ for $i\geq 1$. To avoid ambiguity (from an endless sequence of 1's) we also assume that for all $j$ there is a $k>j$ so that $a_k=0$.

We take $S(\alpha)=\{ \alpha_n : n>0 \}$ where $\alpha_n=\sum_{i=0}^n a_i 2^{N-i}$. Then $S(\alpha)\subset\mathbb{Q}$, so we have an uncountable collection of subsets of $\mathbb{Q}$ which is in bijection with $\mathbb{N}$. Moreover, $S(\alpha)\cap S(\beta)$ is finite when $\alpha\neq\beta$. (It consists of those truncated binary expansions of $\alpha$ and $\beta$ which are equal.)

Now take the subset of reals which consists of those $\alpha$ for which $a_i=1$ unless $i$ is prime. It is not too difficult to see that these $\alpha$ are also uncountable. For example you can map reals to such numbers by "filling in 1's" in the binary expansion as required so that the old $a_i$'s become the new $a_{\pi(i)}$ where $\pi(n)$ is the $n$-th prime.

Moreover, it follows that $S(\alpha)\cap S(\beta)$ is a prime for every such $\alpha$ and $\beta$.

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