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Let $$X_{r}=\{2^{r}(2s-1)-1:s=1,2,3,...\}.$$

Show that $X_{n} \cap X_{m}=\emptyset$ for all $n\ne m$ and also the union of $X_{i}$ $(i\in \mathbb N)$ is $2\mathbb N+1$.

NB: $n$ is fixed. For a fixed natural number $n=1$

$$X_{1}=\{2(2s-1)-1:s=1,2,3...\}$$ Also I assume here $\mathbb N=\{0,1,2,...\}$

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    Oh look, a question with sets! It must be [set-theory]!2012-03-03
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    Hint: $2^r(2s-1)-1=2^t(2u-1)-1$ if and only if $2^r(2s-1)=2^t(2u-1)$.2012-03-03
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    Actually, $$\bigcup_{i \in \mathbb{N}} X_i = \begin{cases} \mathbb{N} \cup (-2^{\mathbb{N}} - 1) & 0 \in \mathbb{N}, \\ 2\mathbb{N} - 1 & 0 \notin \mathbb{N}. \end{cases}$$ With your choice of $\mathbb{N}$, the union should also include $1 \in X_1$, while $1 \notin 2\mathbb{N} + 1$.2012-03-03
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    @AsafKaragila: May be it is.2012-03-04
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    @Hassan: No, it's not a set theory question at all.2012-03-04
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    @ArturoMagidin: I understand my mistake. Actually the set was given within the proof of Sierpinskis theorem of existence of 2 transformation from $\mathbb N_{+}$ to itself, such that their composition is another transformation from $\mathbb N_{+}$ to itself.2012-03-04

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Suppose $\,r\geq a\,\,,\,\,a,b,r,s\in\mathbb N\,$:

$$2^r(2s-1)-1=2^a(2b-1)-1\Longrightarrow 2^{r-a}(2s-1)=2b-1\Longrightarrow r=a$$as the RHS is odd and so must be the LHS $\,\Longrightarrow 2s-1=2b-1\Longrightarrow b=s\,$ , and uniqueness has been proved , so that $\,X_r\cap X_a=\emptyset\,\,\,,\,\,\forall\,r\neq a\,$

Also, if $\,t\,$ is an odd natural, then $\,t=2k-1\,$ , for some natural $\,k\,$ . Let $\,r-1\in\mathbb N\,$ be the maximal power of $\,2\,$ that divides $\,k\,$ , then $$t=2(2^{r-1}m)-1$$and since $\,m\,$ is odd, $\,m=2s-1\,\,,\,\,s\in\mathbb N\,$ , so $$t=2\left(2^{r-1}(2s-1)\right)-1=2^r(2s-1)-1\in X_r$$