An analysis problem about convergence Suppose that $f$ is a continuous function from $[a,b]$ to $[a,b]$. Let $x_0\in [a,b]$, and define by induction that $x_{n+1}=f(x_n)$. Show that $$\lim_{n \rightarrow \infty} (x_{n+1}-x_n)=0$$ implies $$\lim_{n \rightarrow \infty}x_n$$ exists.
（This problem is from a analysis book, and the author tells us the answer can be found in American Mathematical Monthly, volume 83(1976), page 273, which I have no access to, so you can either offer the reference or give the sketch of the proof. Thanks!）
 A: Some preliminary warnings. Hopefully the following proof is correct. The risk is however big that I have made some silly error. In that case I apologize in advance. Quite a lot of details are left out. 
Below I will say that a point $x$ is $\epsilon$-close to a set $A$, if there is a point $a\in A$ such that $|x-a|<\epsilon$
Let $x:\mathbb{N} \rightarrow  [a,b]$ a sequence. If $g$ is a strictly increasing sequence $g: \mathbb{N} \rightarrow \mathbb{N}$, then $x\circ g$ is called a subsequence. I will sometimes use the notation $x_{g(k)}$  for a subsequence to to stress the dependence on $g$.
We first prove the following lemma: 
Lemma If $x_{n_K}$ is a convergent subsequence of $x_n$ then the limit point of that sequence is a fixed point of $f$.
$Proof$ Denote by $x$ the limit point of $x_{n_k}$, then $x=\lim_{k\rightarrow \infty}x_{n_K}=\lim_{k\rightarrow \infty}f(x_{n_K-1})=f(\lim_{k\rightarrow \infty}x_{n_K-1})=f(x)$. The second equality follows by the fact that if $x_{n_k}$ converges to $x$  then  $x_{n_k-1}$ converges to $x$. To see this, use that $x_{n+1}-x_n \rightarrow 0$ and the triangle inequality. The third equality follows by continuity of $f$ and the last equality follows since $x_{n_k}$ and $x_{n_k-1}$ converges to the same point $x$. 
Proof of main statement:
Now assume by contradiction that $x_n$ is not convergent. Then there are two limit points $x,y$, $x<y$ of the sequence, both belonging to $[a,b]$. (This is the only place where the compactness of [a,b] is used). Choose subsequences such that $x_{g(k)}$ converges to $x$ and $x_{h(k)}$ converges to $y$. Now let $n\in \mathbb{N}$ and $\frac{1}{n}<\frac{y-x}{2}$ and choose a $K\in \mathbb{N}$ ($K$ depends on $n$) such that $|x_{g(K)}-x|<\frac{1}{n}$ and $|x_{h(K)}-y|<\frac{1}{n}$ and such that  $|x_{k+1}-x_k|<\frac{1}{n}$ all $k\geq K$. 
Now  w.l.o.g. assume that $g(K)<h(K)$  and note that $g(K)\geq K$ and $h(K)\geq K$. Let $A_n=\{x_s |g(K) \leq s\leq h(K)\}$. 
We claim that $A_n$ is $\frac{1}{n}$-close to every point $z\in [x,y]$.  Let $z\in (x,y)$ (if $z=x$ or $z=y$ then the claim is trivial)and let  $k$ be the smallest integer such that $x_k \in A_n$ and $x_k\geq z$, then it follows that $x_{k-1}<z\leq x_k$. Implying that $z$ is $\frac{1}{n}$-close to $x_k$. (Note that if $x_k< z<y$ for all $x_k \in A_n$, then in particular $x_{h(K)}< z$, implying that $z$ is $\frac{1}{n}$- close to $x_{h(K)}$ ) 
In the passing we note that $\cup_{n=k} ^{\infty}A_n$ is countable infinite for all $k$. This follows since $A_n \neq  \emptyset$ for  all $n\in \mathbb{N}$.
We claim that for every $z\in [a,b]$  there is a subsequence $x_{n_k}$ converging to $z$.  Suppose $x_{n_k}$ $1\leq k\leq K$ has been choosen such that $|z-x_{n_k}|<\frac{1}{k}$ . Now since $A_{n}$ for $n\geq K+1$ is $\frac{1}{K+1}$ close to $z$, we may pick a $z\in \cup_{n=K+1}^{\infty} A_n$.  In particular, we choose $z$ such that $z=x_m$ where $m>n_K$. This is possible since $\cup_{n=K+1}^{\infty} A_n$ is countably infinite. 
But as every point of $[x,y]$ is a limit point of $\{x_n\}$, we have a contradiction by lemma 1.  (Since then all points of $[x,y]$ are fixed points of $f$, implying in particular that $x_n=f(x_n)$ some $n\in \mathbb{N}$ implying that $x_n$ converges. )
