Recently, I am reading the book [1]. Theorem 1.6.1 in [1] is as follows:
Theorem 1.6.1 There exists a set on the real line that is not Lebesgue-measurable.
Proof. Denote by $\xi$ any irrational number and also the number 0. Denote by $R$ a sequence $\{ r_n \}$ of all the rational numbers. Finally, denote by $E_\xi$ the set $\{ \xi + r; r \in R \}$. Clearly $E_{\xi_1} \cap E_{\xi_2} = \emptyset$ if $E_{\xi_1} \neq E_{\xi_2}$. …
I try to prove the following claim by myself:
Claim: $E_{\xi_1} \cap E_{\xi_2} = \emptyset$ if $E_{\xi_1} \neq E_{\xi_2}$.
Proof. I try to use the proof by contradiction to prove the above claim. Suppose that $E_{\xi_1} \cap E_{\xi_2} \neq \emptyset$. Then there exists an element $a$ such that $a \in E_{\xi_1} \cap E_{\xi_2}$. Therefore, $a$ can be written as
$$ a = \xi_1 + r_1 = \xi_2 + r_2, \tag{1}$$
where $\xi_1$ and $\xi_2$ are irrational numbers or 0 and $r_1$ and $r_2$ are rational numbers. From (1) we have
$$ \xi_1 – \xi_2 = r_2 – r_1. \tag{2} $$
Because $E_{\xi_1} \neq E_{\xi_2}$, $\xi_1 \neq \xi_2$. Because if $\xi_1 = \xi_2$, then $E_{\xi_1} = E_{\xi_2}$. $\xi_1 – \xi_2$ could be an irrational number or a rational number. $r_2 – r_1$ must be a rational number. I cannot derive a contradiction.
Do you know how to prove the claim? Thank you in advance.
References
[1] A. Friedman, Foundations of Modern Analysis, Dover Publications. Inc., 1982.
Best Answer
You are almost there. Because $\xi_1 - \xi_2 = r_2 - r_1 $ and $r_1, r_2$ are rational, it follows that $\xi_1 - \xi_2$ must be rational. This means that $\xi_1 = q+ \xi_2$ for some rational $q$. Thus, $\xi_1 \in E_{\xi_2}$ and analogous $\xi_2 \in E_{\xi_1}$. This already shows that both sets are equal.