Can someone please tell me how this proof works/how to fill the gaps ?
I need to show that the set $C$ of all cluster points of the sequence $(i_n)_ {n\in \mathbb{N} }$, where $i_n=T^n(x)$ and $T:X\rightarrow X$ is a continuous mapping from a compact $T_2$ space $X$ to itself, is nonempty and closed.
The proof with which I have problems goes like this: "Define $C_n$ as the closure of $\{ i_k| k \geq n\}$: $C_n=cl(\{ i_k| k \geq n\})$. Then by the definition of a cluster point $C=\cap_{n \in \mathbb{N}} C_n$ and because all $C_n$'s are closed, $C$ is closed too and because the closed nonempty sets $C_n$ make up a desceding sequence and $X$ is compact $C$ is nonempty."
I managed to prove the closedness of $C$ somewhat differently (Obviously $C\subseteq cl(C)$. Choose an arbitrary element $c$ from $ cl(C)$ and WLOG an open neighborhood $U$ of $c$. This has to satisfy $U\cap C \neq \emptyset$, by the definition of $cl(C)$, so there is $c_U \in U\cap C$. Since $c_U$ is a cluster point of $(i_n)_ {n\in \mathbb{N} }$, for every neighborhood of $c_U$ and for all $N\in \mathbb{N}$ there is an $m\geq N$ such that $i_m \in U$. Because $U$ is a neighborhood for $c$, since $U$ was open, it follows that $c$ is a cluster point of $C$, so $cl(C)\subseteq C$.), but I still don't understand why the equality $C=\cap_{n \in \mathbb{N}} C_n$ holds or the argument why $C$ should be nonempty.