Given $A\subset \mathbf{R}\setminus \{0\}$ with $|A| = n$, show that there is a $B\subset A$ with $|B| > n/3$ so that $(B+B)\cap B =\emptyset$ I am asked to show that for any $A\subset \mathbf{R}\setminus \{0\}$ with $|A| = n$, there is a subset $B\subset A$ of more than $n/3$ numbers such that no $a_1,a_2,a_3\in B$ satisfy $a_1+a_2=a_3$. I am aware of this question, indeed I managed to prove this part using the probabilistic method. However I have no idea how to extend this property to irrational numbers.
I tried to slightly perturb every $a\in A$ by a tiny number to produce a set of rational for which the result then holds. But then a subset of those rationals that is sum free does not necessarily map back to a set of the irrationals that is sum free.
Maybe we can partition $A$ into sets $A_1,\ldots, A_k$ such that elements in $A_i$ have rational ratios with each other and then.... ?
Any ideas? Thanks!
Update I think we might be able to do it as follows. If $a_i+a_j - a_k\neq 0$ then it exceeds in magnitude some rational number $q_{ijk}$. For $ijk$ such that $a_i+a_j -a_k = 0$, set $q_{ijk} = 0$. Then define the set of conditions on a set $B = \{b_1,\ldots, b_n\}$:
\begin{align}
b_i + b_j + b_k = 0 &\iff q_{ijk} = 0,\\
b_i + b_j + b_k \geq  q_{ijk} &\iff q_{ijk} \neq 0.\\ 
\end{align}
This is a set of $3^n$ conditions. It seems like a linear programming question without any optimization part. I know little about these kinds of problems but maybe there is some theorem that we may use to show that the above has a solution with rational $b_1,\ldots, b_n$. Then we would be done as $b_i+b_j = b_k$ then implies $a_i + a_j = a_k$.
 A: The real line $\mathbb{R}$ is a infinite-dimensional $\mathbb{Q}$-vector space. You can let $A=\{\alpha_1, \dotsc, \alpha_n\}$, and $\ell$ be any "rational line" (i.e. $\ell = \{q \cdot r \mid q \in \mathbb{Q} \}$ for some $r \in \mathbb{R}$), and consider $A'$ the set obtained by projecting every $\alpha_i$ in this line. To construct the projection, start with the set $\{ r\}$ and add vectors until it becomes a basis $\mathcal{B}$, similarly to the finite-dimensional case (well, it is not obvious you can do it, since it requires the Axiom of Choice). Define $Pr=r$ and $Pt=0$ for any $t \in \mathcal{B}$ different from $r$. By a famous result from linear algebra (which generalizes to infinite-dimensional spaces) every linear transformation is uniquely determined by its values on the basis, therefore $P$ defined above extends to a unique linear transformation in $\mathbb{R}$ as a $\mathbb{Q}$-vector space. Since this set is isomorphic to the rational number, we can use the result previously prove for rational numbers to conclude there exists a $B'$ such that $\lvert B' \rvert > n/3$ and $B'$ is sum-free. Let $B$ be the pre-image of $B'$ by the projection. If $b \in (B + B) \cap B$, then there exists $b_1$, $b_2 \in B$ such that $b_1 + b_2 = b$. Applying the projection $P$, this means $Pb_1 + Pb_2 = Pb_3$. However $Pb_1$, $Pb_2$, $Pb \in B'$. Contradiction, since this set is sum-free.
