1
$\begingroup$

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 $\mathbb N=\{0,1,2,...\}$

  • 0
    @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

1 Answers 1

1

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 $t=2\left(2^{r-1}(2s-1)\right)-1=2^r(2s-1)-1\in X_r$