We say an action of $\mathbb{Z}$ on a compact Housdorff space $X$ minimal if every orbit of the action is dense in $X$. We assume the action is free and $X$ has no isolated points.
Then in this case, Why the half-orbit, say $\mathbb{N}x$ and $\mathbb{-N}x$, are also dense?
I can prove one of them is dense, but I do not how to prove both of them are dense.
Thank you for all helps!
Consider the half-orbit $\mathbb{N} x$. Let it's accumulation set be $C$. The set $C$ is closed. The set $C$ is also invariant by the whole group action. So if $\mathbb{N} x$ were not dense, then any point contained in $C$ would have a nondense orbit, a contradiction.
So why is $C$ invariant by the whole group action? I'll show you why it is invariant under negative elements of the group action. Suppose $y \in C$, we need to show $(-1) \cdot y \in C$. Then there exists a sequence of natural numbers $n_i \to +\infty$ such that $n_i \cdot x$ has limit $y$. By continuity of the map $p \mapsto (-1) \cdot p$, it follows that $(n_i - 1) \cdot x$ has limit $(-1) \cdot y$. And if $i$ is sufficiently large then $n_i > 1$, so $n_i-1 \in \mathbb{N}$, hence $(-1) \cdot y \in C$.