The question Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function such that $\lim_{n \to \infty}f^{n}(x)$ exists $\forall x$. Define $S = \{\lim_{n \to \infty}f^{n}(x): x \in \mathbb{R}\}$ and $T = \{x \in \mathbb{R}: f(x)=x\}$. Then conclude what can be said about the inclusion relation between $S$ and $T$, either $S$ is properly contained in $T$ or vice versa, or $S=T$.


I somehow feel that the answer should be $S$ is contained in $T$, as the sequence will converge to a stable point, and not every stable point needs to be an attracting point. Though I am not sure how to prove this. When the function is contractive, there is Banach fixed point theorem to help, but when it's just continuous I don't know how to conclude anything concretely.

  • $\begingroup$ You can't really say anything at all unless $f$ is continuous. $\endgroup$ Commented Apr 29 at 15:28
  • $\begingroup$ For example, your intuition/thoughts are true only if $f$ is continuous. $\endgroup$ Commented Apr 29 at 15:32
  • $\begingroup$ In the title I mentioned that f is continuous, I added it to the body as well now. Any thoughts now? @ThomasAndrews $\endgroup$
    – Debu
    Commented Apr 29 at 15:33
  • 2
    $\begingroup$ It seems to me that $S=T$: If $x \in T$ then trivially $\lim_{n \to \infty}f^n(x)=x$, hence $x \in S$. If $y \in S$, then for some $x$ we have $y= \lim_{n \to \infty}f^n(x)$. Since $f$ is continuous $y$ is a fixed point of $f$. Hence $y \in T$. $\endgroup$
    – Gerd
    Commented Apr 29 at 15:36
  • $\begingroup$ The only way to have $S\subsetneq T$ is if you define $S$ differently, like: $$S=\{\lim f^n(x)\mid x\notin T\}.$$ $\endgroup$ Commented Apr 29 at 15:42

1 Answer 1


I claim that $S = T$. Let us prove the inclusion $\supseteq$ first. If $x \in T$, then $f(x) = x$ and inductively $f^n(x) = x$. It follows that

$$ \lim_{n \to \infty} f^n(x) = x, $$ therefore $x$ appears as a limit of iterating $f$ and consequently, $x \in S$. For the converse direction, let $L \in S$, so there exists $x \in \mathbb{R}$ such that $\lim_{n\to\infty}f^n(x) = L$. We want to show that $f(L) = L$, so that we may conclude that $L \in T$. We have

$$ f(L) = f(\lim_{n \to \infty} f^{n}(x)) = \lim_{n\to \infty} f(f^n(x)) = \lim_{n \to \infty} f^{n+1}(x) = L, $$ which finishes the proof. Note that for the second equality, continuity of $f$ was needed.

  • $\begingroup$ Thanks a lot. That was very elegant. $\endgroup$
    – Debu
    Commented Apr 29 at 15:39

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