Originally Posted by

**Defunkt** The suggestion that the thread be locked was because you were not reading the posts explaining the proofs.

Here is a detailed explanation of Fernando's proof. If there is something you do not understand, **say exactly what is not clear**.

Let $\displaystyle \lambda \in \mathbb{R}$ be an arbitrary irrational number, and let $\displaystyle T_{\lambda} : S^1 \to S^1$ be a function which maps an element of the unit circle to an element of the unit circle.

Define the mapping by $\displaystyle T_{\lambda} (e^{i \theta}) = e^{i ( \theta + 2 \pi \lambda ) }$ and note that, since each element of $\displaystyle S^1$ is uniquely determined by its angle $\displaystyle \theta$ with the positive x axis, this mapping is the same as the mapping that Fernando has defined.

Note that we will use the notation $\displaystyle T_{\lambda}^k(e^{i \theta})$ to abbreviate $\displaystyle \left( T_{\lambda}(e^{i \theta}) \right)^k =$ $\displaystyle = T_{\lambda}(e^{i \theta}) \circ T_{\lambda}(e^{i \theta}) \circ \overbrace{\ldots}^{\text{k times}} \circ T_{\lambda}(e^{i \theta})$

First, we will show that if $\displaystyle m,n$ are distinct integers ($\displaystyle m \ne n$), then $\displaystyle T_{\lambda}^m(\theta) - T_{\lambda}^n(\theta) \ne 0 \text{ or } 2 \pi k, \ k \in \mathbb{N} $:

$\displaystyle T_{\lambda}^m(\theta) - T_{\lambda}^n(\theta) = (\theta + 2 \pi m \lambda) - (\theta + 2 \pi n \lambda) = \theta - \theta + 2 \pi (m - n) \lambda = 2 \pi (m-n) \lambda$

Now, that equals 0 or $\displaystyle 2 \pi k$ iff $\displaystyle m=n$ or $\displaystyle m - n$ is irrational, but we chose them so that $\displaystyle m \ne n$ and $\displaystyle m - n \in \mathbb{N}$, and so $\displaystyle T_{\lambda}^m(\theta) - T_{\lambda}^n(\theta) \ne 0 \text{ or } 2 \pi k, \ k \in \mathbb{N}, \ \forall m \ne n \in \mathbb{N}$.

Now, this means that all the elements of the sequence $\displaystyle \{T_{\lambda}^k(\theta)\}_{k=1}^{\infty}$ are distinct, and therefore this sequence has a limit point (by sequential compactness), ie. a convergent subsequence.

Since that subsequence is cauchy, for any $\displaystyle \epsilon>0$ there exist integers $\displaystyle m<n$ such that $\displaystyle |T_{\lambda}^n(\theta) - T_{\lambda}^m(\theta)| < \epsilon$, but then note that $\displaystyle T_{\lambda}^n(\theta) - T_{\lambda}^m(\theta) = \theta + 2 \pi n \lambda - \theta - 2 \pi m \lambda = 2 \pi (n-m) \lambda = T_{\lambda}^{n-m}(\theta) - \theta$, so let $\displaystyle k = n-m$ to have $\displaystyle |T_{\lambda}^k(\theta) - \theta| < \epsilon$.

__Now, note that__ $\displaystyle T_{\lambda}$ is a length-preserving map (ie. it maps an interval of length $\displaystyle t$ to an interval of length $\displaystyle t$).

Also, note that $\displaystyle T_{\lambda}^k$ maps the arc connecting the point $\displaystyle \theta$ with $\displaystyle T_{\lambda}^k(\theta)$ to the arc connecting $\displaystyle T_{\lambda}^{k}(\theta)$ with $\displaystyle T_{\lambda}^{2k}(\theta)$.

Consequently, since $\displaystyle |T_{\lambda}^k(\theta) - \theta| < \epsilon$ and $\displaystyle T_{\lambda}$ preserves lengths, we have that $\displaystyle |T_{\lambda}^{2k}(\theta) - T_{\lambda}^k(\theta)| < \epsilon$, and by the same fashion $\displaystyle |T_{\lambda}^{mk}(\theta) - T_{\lambda}^{(m-1)k}(\theta)| < \epsilon$ for any $\displaystyle m \in \mathbb{N}$.

__Now, since all elements of __$\displaystyle \{T_{\lambda}^k(\theta)\}_{k=1}^{\infty}$ are distinct, we have that for every $\displaystyle \theta \in S^1$ there exists an integer $\displaystyle n \in \mathbb{N}$, such that $\displaystyle |\theta - T_{\lambda}^{nk}(\theta)| < \epsilon$, and so we have that $\displaystyle \{T_{\lambda}^{nk}(\theta)\}_{n=1}^{\infty}$ is dense in $\displaystyle S^1$, and in particular any set containing it is dense in $\displaystyle S^1$ - and such is $\displaystyle \{T_{\lambda}^m(\theta)\}_{m=1}^{\infty}$.

Now, if we take $\displaystyle \lambda = \frac{1}{2 \pi}$ (which is clearly irrational) and take $\displaystyle \theta = 0$, we get that $\displaystyle \{T_{\lambda}^m(\theta)\}_{m=1}^{\infty} = \{e^{im}\}_{m=1}^{\infty}$ is dense in $\displaystyle S^1$, as conjectured.