Proving that every forward orbit is dense for a homeomorphism on the torus.

Question

Consider the map $f:\mathbb{T}^2\to \mathbb{T}^2$, given by $f(x,y)=(x+\alpha\mod 1,y+x\mod 1)$, where $\alpha$ is irrational. I have been struggling on and off for the last week to prove that every forward orbit of $f$ is dense in the torus. It is fairly simple to show via induction that, for any $n\in\mathbb N$, $$f^n(x,y)=(x+n\alpha\mod 1,y+nx+\frac{n(n-1)}{2}\alpha\mod 1).$$ Using this I was able to show that if the forward orbit of $(x_0,y_0)$ is dense in the torus, then so too is the forward orbit of $(x_0,y)$ for any $y\in S^1$. We also know that if $A$ is an $f$ invariant open subset of the torus then it must be dense in the torus (I got help here proving this. This function is giving me nightmares.)

Now I'm trying to use these facts to prove that the forward orbit of any point is dense in the torus. I have a feeling that I need to use the Baire Category Theorem somewhere, but I am not sure where. The only potentially useful application of the BCT I have come up with so far is to use it to prove that the complement of any orbit is dense in the torus. Now I was thinking that maybe the compactness of the Torus, along with the previously proven facts would enable us to draw the desired conclusion somehow, but I have yet to find that somehow. I have also tried constructing the forward orbit of a point as the countable intersection of countable open sets directly, but without any success.

Any ideas would be greatly appreciated, as this problem is now intruding on most moments of my life.

Answer 1

This problem actually is a consequence of the following theorem of Furstenberg.

Theorem:(Furstenberg) Let $T:X\to X$ be an uniquely ergodic homeomorphism of a compact metric space $X$ and with invariant measure $\mu$. Let $G$ be a compact group with Haar measure $m_G$, and let $c:X \to G$ be a continuous map. Define $$ \begin{array}{rcl} F: X\times G& \longrightarrow & X\times G\\ (x,g) & \mapsto & (T(x), c(x).g). \end{array} $$ If $F$ is ergodic with respect to $\mu \times m_G$ then $F$ is uniquely ergodic.

Applying this theorem, one only needs to check that the Lebesgue measure on $\mathbb{T}^2$ is ergodic for $F$. The proof of this is more standard and is done using Fourier series to prove that any $F$-invariant function of $L^2(Leb)$ is constant almost everywhere. Unique ergodicity implies minimality of the map.

You can find the proof of Furstenberg's theorem, for instance, on the book : "Ergodic Theory with a view towards Number Theory" of Manfred Einsiedler and Thomas Ward, Theorem 4.21 (page 114) and Corollary 4.22, where in this corollary they prove the ergodicity for the Lebesgue measure for the type of transformation you want, which sometimes are called skew-shifts.

Answer 2

I have finally come up with a simple solution to this question. Pick any $x\in \mathbb T^2$. As the torus is metrizable, we can define $B_n$ to be the open ball centred at $x$ with radius $1/n$. For each $n$ define $A_n=\cup_{i=0}^\infty f^i(B_n)$. As $f$ is a homeomorphism we know each $A_n$ is open, and it is obviously forward $f$ invariant. As mentioned in the question this means that each $A_n$ is dense (the question doesn't specify forward invariance, but this follows from a fairly standard mixing argument, which I might add as an answer to the linked post at a later stage.) Now the torus is a compact Hausdorff space, so the Baire Category Theorem gives us that $A=\cap_{n=1}^\infty A_n$ is dense in the torus, but $A$ is precisely the forward orbit of $x$ under $f$, which gives the result.