If a sequence satisfies $\lim\limits_{n\to\infty}|a_{n+1} - a_n|=0$ then the set of its limit points is connected

Solution 1:

If this is about sequences in $\mathbb R$, then the statement boils down to this:

Let $(a_n)$ be a sequence in $\mathbb R$ such that $\lim_{n\to \infty}|a_{n+1}-a_n|=0$. If $x<y<z$ and $x,z$ are limit points, then $y$ is a limit point.

The proof is simple: For $\epsilon>0$, find $M$ such that $a_{n+1}-a_n<\epsilon$ for all $n>M$. Let $N$ be a natural number $>M$. Because $x$ is a limit point, there is an $r>N$ such that $a_r<y$. Because $z$ is a limit point, there is an $s>r$ such that $a_s>y$. Therefore, among the numbers $r,\ldots, s-1$ there exists one $n$ such that $a_n<y$ and $a_{n+1}\ge y$. Because $n\ge r>M>N$, we have $y>a_n>a_{n+1}-\epsilon\ge y-\epsilon$. In summary, given $\epsilon>0$ and $N\in\mathbb N$ we have found $n>N$ with $|y-a_n|<\epsilon$. In oher words: $y$ is a limit point.

As a corollary using closedness: If $(a_n)$ is a sequence in $\mathbb R$ such that $\lim_{n\to \infty}|a_{n+1}-a_n|=0$, then the set of limit points of $(a_n)$ has one of the following forms:

• $\emptyset$: Take for example $a_n=\sqrt n$
• $\{a\}$: Take for example $a_n=a+\frac1n$.
• $\mathbb R$: Let $a_n = f(n)$ where $f(x)=\sqrt[3]x\sin\sqrt x$; observe that $a_{n+1}-a_n= f'(\xi)=\frac13\xi^{-\frac23}\sin\sqrt\xi+\xi^{-\frac13}\cos\sqrt\xi$ for some $\xi\in[n,n+1]$, hence $|a_{n+1}-a_n|\to 0$. For arbitrary $x\in\mathbb R$ one readily sees that $a_n>x$ infinitely often and $a_n<x$ infinitely often; as in the prof above one conlcudes from this that $x$ is a limit point.
• $(-\infty,a]$: With $f$ as just defined in the preceeding item, let $a_n=\min\{f(n),a\}$
• $[a,\infty)$: This time let $a_n=\max\{f(n),a\}$
• $[a,b]$ with $a<b$: This time finally let $a_n=\min\{\max\{f(n),b\},a\}$

To merge this with my previous comment: The above statement is in general not true in other spaces. As a simple example consider $\mathbb R^2$ and the curve $\gamma\colon[0,\infty)\to \mathbb R^2$, $t\mapsto (\cos t,(1+ t)\sin t)$. Note that $\gamma$ passes through both $(1,0)$ and $(-1,0)$ infinitely often. Let $\tilde\gamma$ be the reparametrization of $\gamma$ by arc length. That is: $\tilde\gamma(t)=\gamma(h(t))$ for some continuous and monotonuously increasing function $h:[0,\infty)\to[0,\infty)$ and $|\tilde\gamma'(t)|=1$ for all $t$. Set $a_n = \tilde\gamma(\sqrt n)$. Then $|a_{n+1}-a_n|\le \sqrt{n+1}-\sqrt n=\frac1{\sqrt{n+1}+\sqrt n}\to 0$, but $a_n$ is unbounded.

Then $(1,0)$ is a limit points of $(a_n)$: Let $\epsilon>0$ and $N\in \mathbb N$ be given. We have to exhibit $n>N$ such that $|a_n-(1,0)|<\epsilon$. Wlog. $n>N$ implies $\sqrt{n}-\sqrt{n-1}<\epsilon$. There is $k\in N$ such that $2 k\pi>h(\sqrt N)$. Let $t_k=h^{-1}(2k\pi)$ and $n$ minimal with $n>t$. Then $n>N$ and $|a_n-(1,0)|=|\hat\gamma(\sqrt n)-\hat\gamma(t_k)|\le \sqrt n -t_k\le \sqrt n-\sqrt{n-1}<\epsilon$.

Similarly, one proves that $(-1,0)$ is a limit point.

If the set of limit points is connected, there must be a limit point of the form $(0,y)$. But $|a_n-(0,y)|<\frac12$ implies (with $t:=h(\sqrt n)$) that $|\cos t|<\frac12$, hence $|\sin t|>\frac 12$ and $|(1+t)\sin t|<|y|+1$, i.e. $t<2|y|+1$ and finally $n<\left(h^{-1}(2|y|+1)\right)^2$. Thus $(0,y)$ is not a limit point and finally the set of limit points is not connected.