For a positive integer $n$, let $[n]$ denotes the set $[n]:=\{1,2,3,...,n\}$.
Let $m$ be a positive integer, then define a set $A$ as
$$A:= [2^m - 2]\big\backslash [2^{m-1}-2]$$.
Now, define the Cyclotomic Coset of $2^{m-1}-1$ as $$B:=\big\{\big((2^{m-1}-1)2^{j}\big) \bmod (2^{m}-1)\ :\ j\in\mathbb{Z}_+\big\}.$$ It is not difficult to see that $$B:= \{2^{m-1}-1\} \cup \{2^m -1 - 2^{i-1}\ :\ 1\leq i \leq m-1\},$$ and hence $B \subset A$.
Now, define a set $C$ as
$$C:= A\backslash B.$$
Now, I am trying to prove the following:
$\textbf{Preposition:}$ For all $x\in C$, there exists a positive integer $j$ such that $$(x. 2^j)\bmod(2^{m}-1)\in [2^{m-1} - 2].$$
$\textbf{Here is what I have tried}:$
Let $w = x - (2^{m-1}-1)$.Then it is not difficult to see that $$1\leq w \leq 2^{m-1}-2,\ \mbox{such that $(2^{m-1}-1 + w)\notin B$}.$$ Now, \begin{align} (x\cdot2^j)\bmod(2^m-1) &= \Big(\big((2^{m-1}-1)+ w \big)2^j\Big)\bmod(2^m-1)\\ & = \Big(\big((2^{m-1}-1)+ w\big)\bmod(2^m-1)\cdot 2^{j}\bmod(2^m-1)\Big)\bmod(2^m-1)\\ & = \Big(\big((2^{m-1}-1)+ w\big)\cdot 2^{j\bmod(m)}\Big)\bmod(2^m-1)\\ & = \Big(\big((2^{m-1}-1)+ w\big)\cdot 2^r\Big)\bmod(2^m-1),\ \mbox{where $r = j\bmod(m)$}\ \ \ (1). \end{align} When $r = 0$, equation $(1)$ becomes $$(x\cdot2^j)\bmod(2^m-1) = \Big(\big((2^{m-1}-1)+ w\big)\Big)\bmod(2^m-1) = (2^{m-1}-1)+ w = x \notin [2^{m-1}-2].$$ When $1\leq r \leq m-1$, equation $(1)$ becomes \begin{align} (x\cdot2^j)\bmod(2^m-1) & = \Big(\big((2^{m}-2)+ 2w\big)\cdot 2^{r-1}\Big)\bmod(2^m-1)\\ & = \Big(\big((2^{m}-1)+ (2w -1)\big)\cdot 2^{r-1}\Big)\bmod(2^m-1)\\ & = \Big(\big((2^{m}-1)\cdot2^{r-1}+ (2w -1)\cdot2^{r-1}\big)\Big)\bmod(2^m-1)\\ & = \big((2w -1)\cdot2^{r-1}\big)\bmod(2^m-1).\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2) \end{align} Now, I have stuck on equation (2). Somehow, I have to prove that for some $1\leq r \leq m-1$, we have $\big((2w -1)\cdot2^{r-1}\big)\bmod(2^m-1)\in [2^{m-1}-2]$.
Anybody here have a better idea of proving my preposition?
You are looking for the smallest integers in cyclotomic cosets modulo $N=2^m-1$. If $\mathcal{C}$ is such a coset and $N-x$ is the smallest element in it, then you cannot have $2x<N$ for otherwise $2(N-x)\equiv N-2x\in\mathcal{C}$ would be smaller (congruence obviously modulo $N$).
So either $x=2^{m-1}$, when you have the coset $B$ you already identified. Or, when $x>2^{m-1}$, the coset has at least one element $<2^{m-1}-1$. That is exactly what you wanted to prove.