TITLE: Does an unbounded open subset of $\mathbb{R}$ contain infinitely many multiples of some number?
QUESTION [25 upvotes]: If $U$ is an unbounded open subset of $\mathbb{R}$, does there exist some $a$ such that infinitely many integer multiples of $a$ belong to U?

REPLY [10 votes]: You can prove this using the Baire category theorem. Let's let $U$ be an unbounded subset of the positive reals, noting that this change does not affect the theorem. Define $U_n = U \cap (n,\infty)$ and define
$$V_n = \{x\in\mathbb R^+ : \text{There exists an integer }k\text{ so that }kx\in U_n\}.$$
This set is open as we could also write
$$V_n=\bigcup_{k=1}^{\infty}\frac{1}kU_n$$
where each of the sets in the union is open.
We note that $V_n$ is dense in $[0,\infty)$ because its complement cannot contain any intervals - note that if an interval $(a,a+\varepsilon)$ was in the complement of $V_n$, then, $U_n$ must be disjoint from every interval of the form $(ka, ka+k\varepsilon)$. However, the union $\bigcup_{k=1}^{\infty}(ka,ka+k\varepsilon)$ has a bounded complement in $\mathbb R^+$ because for any $k$ such that $k\varepsilon > a$, the interval $(ka,ka+k\varepsilon)$ overlaps with $((k+1)a, (k+1)a+(k+1)\varepsilon)$ and thus this union contains all of $(ka,\infty)$. Since $U$ is unbounded, this cannot be.
Note that the set of numbers for which there are infinitely many multiples in $U$ is exactly
$$\mathscr V=\bigcap_{n=1}^{\infty}V_n$$
since if there were only finitely many multiples of $x$ in $U$, then there would be none in some $U_n$. This is a countable intersection of open dense sets in $[0,\infty)$. By the Baire category theorem, $\mathscr V$ is dense in $[0,\infty)$ - and, in particular, it is not empty.