Suppose that $\alpha$ is irrational. Let $T^2$ be the 2-torus, i.e. $T^2 = \mathbb{R}^2/\mathbb{Z}^2$, and $f: T^2 \to T^2$ given by $f(x,y) = (x+\alpha, x+y)$. Show that every non-empty, open, $f$-invariant set is dense.
Let $U$ be such a set. $U$ is non-empty and open, so there are $u \in U$ and $\epsilon > 0$ such that $B_{\epsilon}(u)\subseteq U$. The orbit of $B_{\epsilon}(u)$, $B := \{f^n(x): x \in B_{\epsilon}(u), n \in \mathbb{N}_0\}$ satisfies the criteria, so it suffices to show that $B$ is dense ($f$ is a homeomorphism, so an open mapping, hence $B$ is open).
By the pigeon-hole principle, we know that the orbit of any point is dense in the $x$-dimension. Suppose $(a,b) \in T^2$ and $\delta > 0$. Since $B$ must be dense, we have $B \cap B_{\delta}(a,b) \neq \emptyset$. Since you are dense in the first direction, you infinitely often are $\delta$ close to $a$ in the $x$-dimension ($\delta > 0$). So I thought maybe by applying the pigeon-hole principle again, to the set of these kind of visits (being $\delta$ close to $a$ in one dimension), you have to be close to $(a,b)$ at least once. However, I am not able to convert this idea into a rigorous proof (if that's even possible). Any ideas?
Edit: I think I found a solution; we can choose $N$ such that $f^N(x,y)$ is close to $a$ in the first dimension. We can choose $N$ large enough to then shift our first coordinate only a tiny amount to $x'$ to make the second dimension of $f^N(x', y)$ exactly equal to $b$. Then $f^N(x',y)$ is close to $(a,b)$ and $(x',y)$ is still in $B$.