Countable vs Uncountable Intersection I have a specific question regarding the difference between countable and uncountable intersection. Specifically, let $B(x,r)$ be the open ball centered at $x\in \mathbb{C}$ of radius $r>0$. It is clear that
$$\bigcap_{\theta\in \mathbb{R}}B(e^{2\pi i\theta},1.5)=B(0,0.5).$$
However, does
$$\bigcap_{\theta\in \mathbb{Q}}B(e^{2\pi i\theta},1.5)=B(0,0.5)?$$
 A: Consider the point $0.5e^{2\pi i\sqrt{1/2}}$. In the intersection $$\bigcap_{\theta\in \mathbb{R}}B(e^{2\pi i\theta},1.5).$$This point lies outside $B(e^{-2\pi i\sqrt{1/2}})$, and thus is not contained in the intersection. This is exactly as you say, as that point lies on the boundary of the open ball $B(0,0.5)$.
However, which open ball would exclude it from $$\bigcap_{\theta\in \mathbb{Q}}B(e^{2\pi i\theta},1.5)?$$The answer is, there is no such ball.
The end result for the $\Bbb Q$-intersection is, you get $B(0,0.5)$, along with any point on the boundary with irrational-multiple-of-$\pi$ argument (technically any point on the boundary opposite an irrational-multiple-of-$\pi$, which in this case is the same thing, but for other dense sets the distinction might matter).
As a similar example, to illustrate better that this isn't about countability, if you instead intersect over $\theta\in(\Bbb R\setminus\Bbb Q)$, then what you get is $B(0,0.5)$ along with all rational points on the boundary.
