Does there exist an injection $f:\mathbb{R}\to P(\mathbb{N})$ that satisfies these conditions?

Question

Does there exist an injection $f:\mathbb{R}\to P(\mathbb{N})$ from the reals into the power set of the naturals such that

1. for any $x\in\mathbb{R}$ the set $f(x)$ is infinite, and

2. for any distinct $x,y\in\mathbb{R}$ the intersection $f(x)\cap f(y)$ is finite?

Answer 1

Fix a bijection $\mathbf{N} \leftrightarrow \mathbf{Q}$. For each $x \in \mathbf{R}$ map it to a set of rational numbers $\{x_1,x_2,\dots\}$ corresponding to a sequence that converges to $x$. Call this map $f : \mathbf{R} \to P(\mathbf{Q})$. Then show that $f$ is injective, $f(x)$ is infinite and if $x \ne y$ then $f(x) \cap f(y)$ is finite.

Answer 2

First, get an injection $g : \mathbb{R} \to P(\mathbb{N} \times \mathbb{N})$ satisfying these conditions by setting $g(x) := \{ (n, \lfloor 10^n x \rfloor) : n \in \mathbb{N} \}$. Now, using the fact that $\mathbb{N} \times \mathbb{N}$ is countable, can you see a way to convert $g$ to a function $f$ satisfying the conditions?

Answer 3

HINT:

For each $\alpha \in \mathbb{R}$ consider the infinite strip $$S_{\alpha} \colon \alpha x \le y \le \alpha x + 1$$

Each $S_\alpha$ contains infinitely many points from $\mathbb{Z}^2$ : indeed, it contains at least one point from each vertical. However, for $\alpha \ne \beta$, $S_{\alpha} \cap S_{\beta}$ is compact, so the intersection contains finitely many integral points.