# The set of clusterpoint of the sequence consisting of iterations of a map is nonempty and closed

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.

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As far as I can tell, none of the proofs (yours, the one you're trying to understand, or Brian's) ever use the fact that the sequence is generated by iterating some mapping, let alone a continuous mapping. Isn't this just a statement about sequences in general? – joriki Nov 1 '11 at 13:57

Suppose that $x\in C$. Fix $n\in \mathbb{N}$. Let $U$ be any open nbhd of $x$. Then there is a $k\ge n$ such that $i_k\in U$; this is just the definition of cluster point. It follows that $U\cap \{i_k:k\ge n\}\ne \varnothing$. Thus, every open nbhd of $x$ meets the set $\{i_k:k\ge n\}$, and by definition $x\in\operatorname{cl}\{i_k:k\ge n\} = C_n$. Since $n$ was an arbitrary element of $\mathbb{N}$, $x\in\bigcap_{n\in\mathbb{N}}C_n$, and since $x$ was an arbitrary element of $C$, $C\subseteq \bigcap_{n\in\mathbb{N}}C_n$.
Conversely, suppose that $x\in\bigcap_{n\in\mathbb{N}}C_n$, and let $U$ be any open nbhd of $x$. For each $n\in\mathbb{N}$, $x\in C_n = \operatorname{cl}\{i_k:k\ge n\}$, so $U \cap \{i_k:k\ge n\}\ne \varnothing$, and there must be some $k\ge n$ such that $x_k \in U$. But this is exactly what it means for $x$ to be a cluster point of the sequence, so $x\in C$. Since $x$ was an arbitrary element of $\bigcap_{n\in\mathbb{N}}C_n$, $\bigcap_{n\in\mathbb{N}}C_n \subseteq C$. Combining results, we have $C = \bigcap_{n\in\mathbb{N}}C_n$.
The fact that $C\ne\varnothing$ follows from the compactness of $X$. Recall that the sets $C_n$ are closed. It should also be clear that they’re nested: $C_0\supseteq C_1\supseteq C_2\supseteq\dots$. For each $n\in\mathbb{N}$ let $V_n=X\setminus C_n$, and let $\mathscr{V}=\{V_n:n\in\mathbb{N}\}$. Then $$X\setminus \bigcup\mathscr{V}=X\setminus\bigcup_{n\in\mathbb{N}}V_n=\bigcap_{n\in\mathbb{N}}(X\setminus V_n)=\bigcap_{n\in\mathbb{N}}C_n=C\;,$$ so if $C = \varnothing$, $\mathscr{V}$ is an open cover of $X$. $X$ is compact, so $\mathscr{V}$ has a finite subcover, say $\{V_{n_1},\dots,V_{n_k}\}$. The sets $V_n$ are increasing, so $V_{n_1}\subseteq V_{n_2}\subseteq\dots\subseteq V_{n_k}$, and therefore $X = V_{n_1}\cup V_{n_2}\cup\dots\cup V_{n_k} = V_{n_k}$. But that means that $C_{n_k} = X\setminus V_{n_k} = \varnothing$, which is impossible: an infinite sequence in a compact space always has a cluster point. (Exercise: If you had an infinite sequence with no cluster point, $X$ could not be compact.) Thus, $C$ cannot be empty.