Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 4 / 13E / b

For the torus quotient, a point has the same image as a point of $B(u,r)$ precisely when they differ by an integer lattice vector. ThereforeThis union is open in ${\mathbb{R}}^2$, so $U$ is open by the quotient topology. Distinct integer lattice vectors are at least distance one apart, whereas $2r<1$. Thus these open balls are disjoint, with a positive gap between any two.

Each ball is connected, since it is convex. It is both open and closed relative to the displayed union: its complement there is the union of all the other open balls. No connected component can meet two such relatively separated pieces. Consequently the components are exactly the infinitely many balls $B(u+m,r)$. Translation by $-m$ is a homeomorphism from each component onto $B(u,r)$. The statement uses the intended positive radius $0<r<1/2$.

Use the elementary irrational-rotation density fact: if $\alpha$ is irrational, $\{n\alpha \mathrm{mod}1:n\in \mathbb{Z}\}$ is dense in the circle. One proof uses the pigeonhole principle on fractional parts of $0,\alpha ,\dots ,N\alpha$. Two lie within $1/N$; taking their difference and, if needed, its negative yields an integer multiple of $\alpha$ with fractional part $\delta \in (0,1/N)$. The multiples $0,\delta ,\dots ,\lfloor 1/\delta \rfloor \delta$ form a $1/N$-net of the circle and are themselves integer multiples of $\alpha$ modulo one. Taking $N$ arbitrarily large proves density.

For any target coordinates $(u,v)$ modulo one, points of the line with parameter $t=u+n$ have first coordinate exactly $u$ modulo one and second coordinate $\alpha u+n\alpha$ modulo one. The density fact approximates $v$ arbitrarily well. Since products of small coordinate intervals form a neighbourhood basis of the torus,Nevertheless the point $\pi (0,\alpha /2)$ does not lie in $\pi (L)$. Such an equality would require a line parameter $t=n\in \mathbb{Z}$ and $\alpha (n-1/2)\in \mathbb{Z}$, contradicting irrationality. HenceThis is the irrational winding of the torus: dense need not mean surjective.