behavior of a sequence Imaging a sequence $ a_{k} \in \Omega $ with $ \Omega \subset \Bbb{R} $ closed, $ \lim\limits_{k \to \infty} \| a_{k+1} - a_{k} \| = 0 $.
My Professor said that because of this the sequence would converge to a Point in $ \Omega $, but i think this can't be true if you take something like $ a_{k} = \log(k) $.
But could it be that this proposition is true if i allow $\infty$ to be a limit too?
 A: Since $\Omega\subset \mathbb R$ and $\infty\notin \mathbb R$, even allowing $\infty$ as a limit won't make the statement true.
Note that we might have $\Omega=\mathbb R$, or minimalistically $\Omega=\{\ln n\mid n\in\mathbb N\}$ as closed subset to allow your sequence $a_k=\ln k$. 
A sequence with $\lVert a_{k+1}-a_k\rVert\to 0$ need not be Cauchy, as you correctly noticed. The situation would change if $\Omega$ were not only closed, but compact.
A: You can do that modification. That is working on $\bar{\mathbb{R}}=\mathbb{R}\cup{\pm\infty}\cong [-1,1 ]$, but in this case your metric need to be modified accordingly.
A: Let $\alpha_n=n^{-1}$. We know that the series $\sum\limits_{k=1}^\infty \alpha_k$ diverges to $+\infty$, hence we alwayas can find arbitrary large segment of this series.
Now set $\beta_k=(-1)^{f(n)}\alpha_n$ for some functions $f$ in such a way that partial sums of $a_n=\sum\limits_{k=1}^n \beta_k$ will run from $1$ to $0$ and backwards infinitely many times. The sequence $\{a_n:n\in\mathbb{N}\}$ is bounded, and satisfy rrestriction $\lim\limits_{n\to\infty}|a_{n+1}-a_n|=0$, but its limit doesn't exist.
