Consider the interval $[1,2]$ suppose I consider the sequence like this,first term $1$,then move to $2$,then come back to $1+1/2$,then return to $1$,then move to $1+1/4$,then to $1+1/2$ then $1+3/4$ then $2$,then again $2-1/8,2-2/8,...,2-7/8,1$ then again start from $1$ with step length $1/16$ and so on.I think for this sequence the set of all subsequential limits is the set $[1,2]$ and the sequence satisfies $\operatorname{lim}(u_{u+1}-u_n)=0$ but the sequence is bounded oscillating.are all my claims correct?
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You are correct that the set of subsequential limits is $[1,2]$.
On the $N$'th run-through (counting from $0$), your sequence gives all numbers of the form $1+\frac{k}{2^N}$ , where $0\le k \le 2^N$. Let $x\in[1,2]$ and $\varepsilon>0$ be given. Then we can choose $N$ and $k$ so that $1+\frac{k}{2^N} \in (x-\varepsilon, x+\varepsilon)$. This is because the dyadic rationals are dense in $\mathbb R$. Since we can find $N$ and $k$ for any $\varepsilon>0$, we can construct a subsequence that converges to $x$ by letting $\varepsilon =\frac1n$ and $n\to \infty$.
Milten
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