I'm trying to prove that following :
Let $(X,T)$ by a dynamical system such that $T$ is an isometry. Then $(X,T)$ est uniquely ergodic iff $(X,T)$ is minimal.
So far, I've been unsuccessful on both sufficient and necessary condition.
On the "then" statement, I did the following: let $\nu$ be the unique invariant measure on $(X,T)$. Suppose $x \in X$ has not a dense orbit in $X$. Then, there exists some open set $V$ such that $T^n(x)$ never encounters $V$ as $n$ runs through $\mathbb{N}$. By considering $\mu_n = (1/n) \sum_{k = 0}^{n-1} \delta_{T^kx}$, then there exists a subsequence $(\mu_{n_k})_k$ converging to $\mu$ a $T-$invariant measure. By unique ergodicity, this means that $\mu = \nu$. From this, it follows easily that $\nu(V) = 0$. At first, I would have liked to use that because $T$ is isometric, any of its invariant measure must have total support, but this seems to be false anyway...
Thanks a lot!