If a fixed point is a limit of a subsequence of iterates, must the whole sequence converge to it?

Question

Say $X$ is a compact metric space, with $f:X \to X$ continuous. Now, for any $x_0$, $f^n(x_0)$ must have a convergent subsequence, say $f^{n_i}(x_0) \to x_\infty$. If we know that any such limit is a fixed point of $f$, that is, $f(x_\infty) = x_\infty$ for any such convergent subsequence, does it force the whole sequence $f^n(x_0)$ to converge to $x_\infty$?

If needed, we may assume $f$ has only finitely many, and hence discrete fixed points.

I see that each translate of the subsequence ($f^{n_i+r}(x_0)$) converges to $x_\infty$, and this would have solved the problem if the differences between consecutive $n_i$ were bounded. Also, I'm pretty sure that this, combined with the fixed points being isolated, forces any convergent subsequence to have the same limit, but that doesn't seem to prove the result either.

Edited: I realized my wording was wrong on a crucial point, and since the sole existing answer was not perfectly correct, I decided to put in the change. Previously we only knew the limit to be a fixed point for one particular subsequence, now we now this for any convergent subsequence of iterates.

Answer 1

Let $X$ be the subset of the Euclidean plane $\mathbb R^2$ consisting of the following points. For each $n=1,2,\dots$, let $X$ contain the points $(\frac kn,\frac 1n)$ for $k=0,1,\dots,n$; in addition. let $X$ contain all the points $(x,0)$ for $0\leq x\leq 1$. Define $f:X\to X$ as follows. Each of the points $(x,0)$ is fixed by $f$. For the remaining points, $f(\frac kn,\frac 1n)=(\frac{k+1}n,\frac1n)$ as long as $k<n$ while $f(1,\frac1n)=(0,\frac1{n+1})$. Then for any $p\in X$ not of the form $(x,0)$, the sequence $f^n(p)$ has subsequences converging to all the fixed-points $(x,0)$.

EDIT: Thanks to dfeuer for pointing out the error in my argument and for suggesting a correction. My construction was defined with $X$ included in the unit square $[0,1]^2$, and the error is that points of the form $(1,\frac1n)$ at the right edge are mapped by $f$ all the way over to the left edge, while their limit, $(1,0)$ is fixed by $f$. The problem goes away if we (1) identify the left and right edges of the square to form a cylinder and (2) delete the now redundant points on the right edge. To be specific, we'll have $f$ mapping each of the $n$ points "at level $n$", namely $(\frac kn,\frac 1n)$ for $k=0,\dots,n-1$, to the next one except that the last one $(\frac{n-1}n,\frac1n)$ gets mapped to $(0,\frac1{n+1})$, which is now, thanks to the wrap-around in the cylinder, a very small distance away.

Answer 2

The following has been edited to complete and improve it and to incorporate ideas from bryanj's comment. If I haven't made any terrible errors, it now proves that the set of accumulation points of $(f^n(x_0))$ will always be connected. It is likely that this argument can be simplified/generalized/strengthened in some fashion, and any suggestions for doing so will be appreciated and acted upon.

Fix $x_0$ and let $x_n = f^n(x_0)$ for each $n > 0$.

Let $A$ be the set of accumulation points of $(x_n)$.

Suppose for the sake of contradiction that $A$ is not connected.

By (ZF) Prove 'the set of all subsequential limits of a sequence in a metric space is closed. or similar, $A$ is closed in $X$.

Since $A$ is disconnected and closed, it is separated by closed, and hence compact, sets $C$ and $D$.

As a consequence of If $A$ is compact and $B$ is closed, show $d(A,B)$ is achieved or similar bog-standard results, there exists an $\epsilon > 0$ such that for all $x \in C$ and $y \in D$, $d(x,y) > 2\epsilon$.

Let $W_C$ be the union of all the open $\epsilon$-balls about elements of $C$ and define $W_D$ similarly. By the definition of accumulation point, $(x_n)$ is frequently in $W_C$ and frequently in $W_D$.

Suppose for the sake of contradiction that $\exists x \in W_C \cap W_D$. Then there must be $p\in C$ and $q\in D$ such that $d(x,p)<\epsilon$ and $d(x,q)<\epsilon$. By the triangle inequality, $d(p,q)<2\epsilon$, a contradiction. Thus we conclude that $W_C \cap W_D = \varnothing$.

Since $X$ is compact and $f$ is continuous, $f$ is uniformly continuous by the Heine-Cantor theorem. Thus there is a $\delta$ such that $0 < \delta < \epsilon$ such that for all $x,y \in X$, $d(x,y) < \delta \implies d(f(x),f(y)) < \epsilon$.

Let $U$ be the union of all the open $\delta$-balls about elements of $A$.

Then $(x_n)$ is eventually in $U$: if not, then there is a subsequence in $X\setminus U$, which is closed, hence compact, so there is an accumulation point in $X\setminus U$, a contradiction.

Let $N$ be such that $n \ge N \implies x_n\in U$.

Let $n > N$.

Since $\delta < \epsilon$, $x_n \in U \subseteq W_C \cup W_D$.

Assume without loss of generality that $x_n \in U \cap W_C$.

By the definition of $U$, there is an element $a \in A$ such that $d(x_n,a)<\delta<\epsilon$. Since $W_C$ and $W_D$ are disjoint, $x_n \notin W_D$, so $a \notin D$, so $a \in C$.

By the choice of $\delta$, $d(f(x_n),f(a))<\epsilon$. Thus $d(x_{n+1},a)<\epsilon$, so $x_{n+1} \in W_C$.

Thus we have shown that $(x_n)$ is eventually in $W_C$ and thus is not frequently in $W_D$, contradicting the fact that $W_D$ is an open set containing elements of $A$.

As this contradiction arose from the assumption that $A$ is disconnected, we conclude that $A$ is connected.

Answer 3

Here's another answer in the spirit of the previous answer, in which continuity near the fixed points (except maybe $(1,0)$) relied on the fact that the horizontal change in $f(x,y)$ becomes small as $y$ becomes small.

Let $\gamma(x) = \text{sin}(\frac{1}{x})$ be the Topologist's sine curve, on $(0,1]$. For each $x \in (0,1]$ we can define $g(x)$ as the point in $(0, 1]$ with $g(x) < x$ such that the distance from $(x, \gamma(x))$ to $(g(x), \gamma(g(x)))$ along the Topologist's sine curve equals $x$. In other words $g(x)$ satisfies $$\int_{g(x)}^{x}\sqrt{1+\gamma'(x)^2} dx = x$$ Then define the metric space $X$ to be the graph of the curve $\gamma(x)$, along with the portion on the $y$-axis $\{(0,y)\}$ where $-1 \le y \le 1$.

Define the function $f:X \to X$ by $f(x, \gamma(x)) = (g(x), \gamma(g(x)))$, when $x > 0$, and $f(0, y) = (0, y)$.

I think this is continuous away from the $y$-axis. On the $y$-axis, it is continuous. If $(x, \gamma(x))$ is within $\frac{\epsilon}{2}$ of $(0, y)$, then the distance along the the Topologist's curve from $(x, \gamma(x))$ to $f(x, \gamma(x))$ is $x \le \frac{\epsilon}{2}$. Then $|(x, \gamma(x))$ - $f(x, \gamma(x))| \le \frac{\epsilon}{2}$.