I am going through a book on topological dynamics, and in establishing their definitions they mention the following 'fact,' leaving it as an exercise:
If $X$ is a compact Hausdorff space, $x_0 \in X$ and $f : X \rightarrow X$ is a continuous function, then the following two conditions are equivalent:
1) $x_0 \in \cap f^n(X)$, and
2) there is an infinite sequence $x_{-n}$ with $f(x_{-n }) = x_{-n+1}$
In other words, $x_0$ has a complete 'reverse orbit' iff it is contained in the image of every iterate of $f$. I am not convinced that (1) implies (2) even if $X$ is compact metric. To see this, attach a straight line segment $A_n$ to the origin in the plane, such that the length of $A_n$ is $\frac{1}{n}$ with radian angle $\frac{\pi}{n}$. Then their union is compact metric.
Now, let $B_n = \lbrace b_1^n, b_2^n, \dots, b_n^n \rbrace \subset A_n$ be a subset consisting of $n$ points, each containing the origin. Let $X = \cup B_n$. It is compact metric, and we can define a continuous function on each $B_n$ separately via $f(b_k^n) = b_{k+1}^n$, where we assume $b_n^n = b_{n+1}^n = \cdots = 0$, i.e. the origin is a sink for the action on each $B_n$. Then isn't it true that $0$ satisfies (1) but not (2)?
To avoid the constant sequence being a solution, we can assume that there is a finite tail after $0$ which then becomes a periodic orbit (or anything else with compact phase space). So these points will pass through $0$ before dumping into a finite periodic orbit.
Thanks for checking!
The claim of equivalence is true and employs the compactness of $X$ as a critical assumption.
Per assumption of 1), $x_0\in \bigcap_{n=1}^\infty f^n(X)$, which means that pre-images $\tilde x_{-n}$ of any depth $n$ exist, $x_0=f^n(\tilde x_{-n})$. By compactness, the sequence $f^{n-1}(\tilde x_{-n})$ has an accumulation point $x_{-1}$, which means that there is a sub-sequence $\tilde x_{-n_k}$ with $\lim_{k\to\infty} f^{n_k-1}(\tilde x_{-n_k})=x_{-1}$. By continuity $$f(x_{-1})=f\left(\lim_{k\to\infty} f^{n_k-1}(\tilde x_{-n_k})\right)=\lim_{k\to\infty} f^{n_k}(\tilde x_{-n_k})=x_0.$$
It remains to show that also $x_{-1}\in \bigcap_{n=1}^\infty f^n(X)$. Fix some $N\in \Bbb N$, pick any accumulation point $\tilde x_{-1,N}$ of the sequence $$\left(f^{n_k-1-N}(\tilde x_{-n_k})\right)_{k\in \Bbb N,\, n_k>N+1},$$ which exist because of the compactness of $X$. We get that $x_{-1}$ is the image of $\tilde x_{-1,N}$ under $f^N$, again by continuity. Thus, $x_{-1}\in f^N(X)$, for any $N\in \Bbb N$.
Now apply the same process on $x_{-1}$ to obtain one $x_{-2}$ etc. to obtain a sequence of points with $f(x_{-k-1})=x_{-k}$ as claimed in 2).