Set of cluster points of a bounded sequence

Solution 1:

Here is an example showing that in higher dimension the cluster point set of an unbounded sequence $(x_n)$ such that $\|x_{n+1}-x_n\|\to0$ may be disconnected, for the reason explained by @joriki (naturally, this cluster point set is still closed).

In words, consider the curves $\gamma_n$, each making a vee joining the points $a=(0,0)$ and $b=(2,0)$ through the bottom point $(1,-2^n)$, and the curve $\gamma$ which concatenates them, joining $a$ to $b$ through $\gamma_{2n}$, then $b$ to $a$ through $\gamma_{2n+1}$, and so on, at speed roughly $1$.

In maths, for each $n\geqslant0$, $\gamma_n:[0,2^n]\to\mathbb R^2$ is defined by $$ \gamma_n(t)=(2^{1-n}t,|2^n-2t|-2^n), $$ and the curve $\gamma:[0,+\infty)\to\mathbb R$ is defined by $\gamma(2^{2n}+t)=\gamma_{2n}(t)$ for every $t$ in $[0,2^{2n}]$ and by $\gamma(2^{2n+1}+t)=\gamma_{2n+1}(2^{2n+1}-t)$ for every $t$ in $[0,2^{2n+1}]$.

Now, consider $x_n=\gamma(\sqrt{n})$ for every $n\geqslant0$. Since $\sqrt{n+1}-\sqrt{n}\to0$ and $\|\gamma'\|$ is bounded, $\|x_{n+1}-x_n\|\to0$. Since $\sqrt{n}\to\infty$, the sequence $(x_n)$ passes near $a$ and near $b$ infinitely often (in fact $x_{2^{4n}}$ is exactly $a$ and $x_{2^{4n+2}}$ is exactly $b$, for every $n\geqslant0$).

The set of limit points of $(x_n)$ is $\{a,b\}$, which is not connected. Since the union of the paths $\gamma_n$ is unbounded, $(x_n)$ is unbounded.