Prove that if a sequence satisfies $\lim\limits_{n\to\infty}|a_{n+1} - a_n|=0$ then the set of its limit points is connected.
My professor once mentioned a proposition likewise but I cannot find the exact version of it. Can anyone help?
Prove that if a sequence satisfies $\lim\limits_{n\to\infty}|a_{n+1} - a_n|=0$ then the set of its limit points is connected.
My professor once mentioned a proposition likewise but I cannot find the exact version of it. Can anyone help?
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
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
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:
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.