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The question :

Let $A \subseteq \mathbb{R}$ be a countable set of Real numbers. For all $\alpha \in \mathbb{R}$ we define a set $A_\alpha \subseteq \mathbb{R}$ as follows : $$A_\alpha:=\{x+\alpha : x\in A\}$$ (i) Prove that there exists $\alpha \in \mathbb{R}$ such that $A \cap A_\alpha=\emptyset$.
(ii)Prove that there exists $\alpha \in \mathbb{R}$ such for all two pairs of $m,n \in \mathbb{N}$ , $$A_{m\alpha}\cap A_{n \alpha}=\emptyset$$

What i have tried for part one:

Let $\alpha \in \mathbb{R}$ if $A \cap A_\alpha \neq\emptyset$ then there exist an element that is in $A$ and in $A_\alpha$ therefore we can write it in one way as $x+\alpha : x\in A$ and in another as $y \in A$ therefore we get that : $$y=\alpha +x$$ $$\alpha=y-x$$ therefore we define $B:=\{\alpha : A \cap A_\alpha\neq\emptyset\}$ and we notice that $B$ is contained in $C:= \{y-x \in \mathbb{R} | x,y \in A\}$ but i got no where from here, i think we need to show that $C$ is countable (i dont know how) and then we get that $B$ is countable and beacuse $\mathbb{R}$ is not countable then there exist $\alpha \in \mathbb{R}$ that is not in $B$ and therefore $$A \cap A_\alpha=\emptyset$$

and for part two i think its from part one but i am not sure.

wiliam
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2 Answers2

1

Let us prove (ii). [$m=0,n=1$ gives (i)].

If this is not true, then, for any $\alpha$ there is a point $x \in A_{n\alpha}\cap A_{m\alpha}$. Assume $n \neq m$. [(ii) is false for $m=n$]. We can write $x=a+n\alpha, x=b+m\alpha$ with $a, b \in A$. Thus, $\alpha=\frac {a-b}{m-n}$. The map $\alpha \mapsto (a,b,n,m)$ is an injective map into a countable set since a finite product of countable sets is countable. See: The cartesian product of a finite amount of countable sets is countable.

This makes $\mathbb R$ countable, a contradiction.

Lucenaposition
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Kavi Rama Murthy
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1

(i) Assume $A$ is sorted with $\delta$ being the smallest absolute difference between the consecutive numbers in $A$. Then any positive $\alpha < \delta$ would translate $A$ into a set that has no common elements with $A$.

(ii) Let $A_k$ be $A$ translated by a natural number $k$. Let $\delta_m$ and $\delta_n$ be the smallest differences between consecutive elements in $A_m$ and $A_n$ respectively. Then for any $\alpha < \min(\delta_m/m, \delta_n/n)$, $n\alpha$ and $m\alpha$ will translate $A$ into two sets with no common elements.

NOTE: This reasoning is not general enough since it does not work for all cases. As noted in the comment below, minimal $\delta$ need not exist, as in the sequence ${1/n}$, for example.

dnqxt
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