If $x_{n+1}=f(x_n)$ and $x_{n+1}-x_n\to 0$, then $\{x_n\}$ converges Let $f:[0,1] \rightarrow [0,1]$be a continuous function. Choose any point $x_0 \in [0,1]$ and define a sequence recursively by $x_{n+1}=f(x_n)$. Suppose $\lim_{n \rightarrow \infty}x_{n+1}-x_n =0$, does this sequence converge?

Yes. Let $I = \lim_{k \rightarrow \infty} I_k$ where $I_k$ is the closure of $\{x_k,x_{k+1},...\}$. Note that because $x_{k+1}-x_k \rightarrow 0$, $I$ is connected, thus either a singleton or an interval. If $I$ is a singleton, we are done. If $I$ is an interval $(x_-,x_+)$, then $f(x)=x$ on this interval. It then follows, unless $x_k$ is constant for sufficiently large $k$, that $x_k$ is strictly monotone in $k$; and thus that the sequence must converge to some point within $I$.
