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Suppose that $\{\alpha_k\}$ is a bounded sequence of real numbers satisfying the condition $\displaystyle\lim_{k\rightarrow\infty}|\alpha_k-\alpha_{k+1}|=0$. Let $\displaystyle m = \varliminf_{k\rightarrow\infty}\alpha_k$ and $\displaystyle M = \varlimsup_{k\rightarrow\infty}\alpha_k$. Prove that the cluster point set of the sequence $\{\alpha_k\}$ is the whole segment $[m;M]$.

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You can see a proof in the following blog post: mathproblems123.wordpress.com/2009/09/09/property-of-a-sequence –  Beni Bogosel Mar 17 '12 at 18:11
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Please accept some answers to you questions before asking for more. –  Beni Bogosel Mar 17 '12 at 18:51
    
Thank you for your solution. How about the extension of this result in $\mathbb{R}^n$. If $\{x^k\}\subset \mathbb{R}^n$ is bounded and such that $\displaystyle\|x^k-x^{k+1}\|\rightarrow 0$. Could we conclude that the set of accummulation points of $\{x^k\}$ is nonempty, closed and connected? –  impartialmale Mar 17 '12 at 21:22
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No, we can't. There could be only two accumulation points and the sequence could go back and forth between them along infinitely many different paths. The result in one dimension only comes about because there's only a single path connecting two accumulation points. –  joriki Mar 17 '12 at 23:18
    
Joriki, your explanasion does not satisfy me. Please give a counterexample to express your ideas. Thank you for your comments. –  impartialmale Mar 18 '12 at 20:35
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up vote 3 down vote accepted

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.

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The fact that the set of cluster points of such sequence is connected for sequences in compact metric spaces (and, consequently, for bounded sequences in $\mathbb R^n$) is shown, for example in a paper by Asic and Adamovic; see here for the exact reference. –  Martin Sleziak Jun 7 '13 at 16:14
    
@Martin Oops. Unsurprisingly, the sequence constructed here is unbounded... I modified my post. Thanks. –  Did Jun 7 '13 at 17:13
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