Consider $T_a(x)=x+a$ where $T:[0,1)/\{0=1\}$ if $a \not \in \Bbb Q$ then prove that $O_T(0)=\{na\pmod 1: n \in \Bbb Z\}$ is dense in $[0,1)$.

Now $ma=na\pmod 1$ iff $m=n$. Next, how to preceed?

  • $\begingroup$ $O_T(0)$ (I guess it should be $O_T(a)$) is an additive subgroup of $[0,1)\pmod{1}$ with elements arbitrarily close to zero, since if $\frac{p}{q}$ is a convergent of the continued fraction of $a$ we have $|qa-p|\leq\frac{1}{q}$. Density follows. $\endgroup$ Commented Feb 24, 2018 at 19:00

1 Answer 1


Hint: See the the elements of $O_T(0)$ as a sequence $s_n=e^{inθ}\in S^1\subset\mathbb{C}$ where $nθ\neq 2π$, $\forall n\in \mathbb{Z}$ (this is the usual trick of seeing $\mathbb{T}$ both as the circle and the interval with the ends identified).

Now prove that if we have $|s_n-s_m|=ε$ then we can "fill" the circle with points that are have distant from each other approximately $ε$.

Since $S^1$ is compact you can make $ε$ as small as you want.

I don't give a detailed answer since the concept of this site is to try to learn from others not just blindly copy proofs.

If you want a more detailed account, this is also known as Kronecker's Theorem.


You must log in to answer this question.