We consider the unit circle, and identify it with $X=\mathbb{R}/\mathbb{Z}$. Let $\alpha$ be irrational, and consider the $\mathbb{Z}$-action on the space $X$ given by $T(x)=x+\alpha \mod 1$ (respectively, $T^n(x)=x+n\alpha\mod 1$ for $n\in\mathbb{Z}$).
I want to show that there is no sequence $m_i$ of integers, such that $T^{m_i}$ converges (to some $T^k$, for example). By convergence of functions, we mean pointwise convergence - that for all $x\in X$, $\lim_{i\rightarrow\infty}T^{m_i}(x)=T^k(x)$.
Firstly, I am not sure if this is true or not (to begin with).
Secondly, I have no idea how to prove that there is no such sequence. We do have that the orbit of each point $x\in X$ is dense under irrational rotations, but I want to show that this does not happen in a `nice way', so that even if there is some $x_0\in X$ such that $\lim_{i\rightarrow\infty}T^{m_i}(x_0)=T^k(x_0)$, this will not hold for all $x\in X$. One idea was to consider what happens if I apply $T^l$ to both sides of the equality in some way, and reach a contradiction, but I don't think I can do that...
Actually, exactly the opposite happens: for any $\beta$ there exists a sequence $m_i$ such that $$\lim_{i\to\infty}T^{m_i}(x)=x+\beta\bmod1,$$ for all $x$. This follows from that fact that $T^m(x)=x+m\alpha\bmod1$ and so what you are asking is whether there exists a sequence such that $$\lim_{i\to\infty}(m_i\alpha-\beta)\bmod1=0,$$ which is an immediate consequence of $\alpha$ being irrational (and is independent of $x$). Actually, it happens in the best possible way: the sequence $m\alpha$ is uniformly distributed on the interval.