A question on a problem in set theory.

Question

PROBLEM: Prove/disprove that for every countable subset $A \subset \mathbb R$, $\exists a \in \mathbb R$, such that $(a+A) \cap A = \emptyset$, where we define the set $(a + A) = \{a + x \,|\, \forall x \in A\}$.

MY ATTEMPT:

My idea was to prove the statement. Considering the set $S$ to contain the pairwise absolute difference of any two elements $a_i , a_j \in A$, I defined an $\epsilon = \inf(S)$, and defined $a = \epsilon/2$. Now it is quite trivial to see that this definition of $a$ gives the required result when added to $A$, provided $A$ is not very dense. That is, it fails for example in the case $A = \mathbb Q$, as $a = 0$ in this case.

Can anyone please suggest an universal proof for the statement? Or at least provide some hints.

Do you think of $\emptyset$ instead of $\Phi$? And you should use ${~}$ for sets and not $[~]$. — Mundron Schmidt, Nov 26 '17 at 11:40
Possible duplicate of https://math.stackexchange.com/questions/355049/let-a-be-any-uncountable-set-and-let-b-be-a-countable-subset-of-a-prove. Since $\mathbb{R}$ is uncountable, but $A$ here is countable. — mavavilj, Nov 26 '17 at 11:53

José Carlos Santos · Accepted Answer · 2017-11-26T15:21:00.773

2

Suppose that, for each $a\in\mathbb R$, $(a+A)\cap A\neq\emptyset$. For each $a\in\mathbb R$, let $b_a,c_a\in A$ be such that $a+b_a=c_a$. The function$$\begin{array}{rccc}f\colon&A^2&\longrightarrow&\mathbb R\\&(a_1,a_2)&\mapsto&a_1-a_2\end{array}$$is then surjective, since $f(c_a,b_a)=a$, for each $a\in\mathbb R$. But $A^2$ is countable and there are no surjective maps form countable sets onto $\mathbb R$.

edited Nov 26 '17 at 15:21

answered Nov 26 '17 at 11:40

José Carlos Santos

427,504

I think you mistakenly wrote $a \in A$ instead of $a \in R$ – Lelouch Nov 26 '17 at 11:45
@Lelouch I've edited my answer. Thank you. – José Carlos Santos Nov 26 '17 at 11:47

A question on a problem in set theory.

1 Answers1