Lines in $\Bbb R^2$ inducing lines in the Torus Viewing the torus as $T^{2} = \mathbb{R}^{2} / \mathbb{Z}^{2} $, let $L$ be a straight line in $\mathbb{R}^{2}$ through the origin. How many path components does $T^{2} \setminus L$ have in terms of the slope of L?
Hint: consider the case where the slope is rational first.
I'm not sure how to solve this. I have read somewhere that the image of $L$ in the torus is dense if and only if the slope is irrational, which seems to make sense, but not entirely sure if this helps solve the question.
 A: If the slope of $L$ is rational then the image of $L$ in $T^2$ is a circle, and the complement of the image is an open annulus hence is connected.
If the slope of $L$ is irrational then the covering map is injective on $L$, and indeed on every line parallel to $L$. Let's consider all covering translates of $L$:
$$L_{(m,n)} = \{(x+m,y+n) \mid (x,y) \in L, (m,n) \in \mathbb Z^2\}
$$
There's only countably many of these covering translates, and each of them has the same injective image in $T^2$. 
What about the uncountably many other lines $L'$ in $\mathbb R^2$ that are parallel to $L$? Well, each of them has countably many covering translates $L'_{(m,n)} = \{(x+m,y+n) \mid (x,y) \in L', (m,n) \in \mathbb Z^2\}$, and each of those translates has the same image in $T^2$. Thus, the uncountable set of lines in $\mathbb R^2$ parallel to $L$ is partitioned into subsets of countably infinite size, each partition element representing lines with the same image in $T^2$, different partition elements representing lines with disjoint image in $T^2$. There are uncountably many partition elements, and hence uncountably many distinct component of $T^2$ minus the image of $L$.
So the upshot is: $T^2$ minus the image of $L$ has either one component (when the slope of $L$ is rational), or an uncountably infinite number of components (when the slope of $L$ is irrational).
