Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I finished my test and there is a question I completely failed but that my teacher did not go over, so I was hoping someone could post a correction of it, so that I understand what I was supposed to do for next time.

Suppose that A is a countable set of real numbers. Show that there exists a real number a such that $(a+A)\cap A=\varnothing$. (Note: By definition, $a+A= \{a+r\mid r\in A\}$.)

share|improve this question

2 Answers 2

up vote 6 down vote accepted

Let $B = \{ x \mid (x + A)\cap A \neq \varnothing \} \subseteq \{ x-y \mid x,y \in A \} = C$. Of course $C$ is countable, hence is $B$, but $\mathbb{R}$ in uncountable and therefore $\mathbb{R} \backslash B$ is not empty.

share|improve this answer

Let $a$ be such that $(a+A) \cap A$ is nonempty, i.e. there are $b,c \in A$ such that $a + b = c$.

Then $a = b-c$ lies in the set $\{b - c : b,c \in A\}$.

Since this set is countable with $A$ being countable, it is not all of $\mathbb{R}$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.