An additional question for the one I asked here I thought it will be better to ask it separately.
While looking at all the subsets of $\mathbb{R}$ with the following property: $a,b\in A \Rightarrow a+b\not \in A$, I thought about trying to find the maximum set of all those subsets. Is it possible to prove that there must be a maximum set that follows the property?
Yes, there must be a $\subseteq$-maximal such set -- by Zorn's lemma:
Let $\mathcal A$ be the set of all of these sets. For any chain $X \subseteq \mathcal A$, we have $\bigcup X \in \mathcal A$ as a $\subseteq$-upper bound for $X$. Hence Zorn's lemma applies.
(Note, however, that this maximal element of $(\mathcal A; \subseteq)$ is not unique -- which is why I wouldn't call it a maximum.)
Stefan already reasoned non-constructively that such a set exists. My answer is some comment on how such a subset might not look like, and some generalizations. I think it might be interesting to investigate whether there exist explicitely constructible "maximal sum-free subsets".
Definition. A set $R$ is called sum-free if it holds $a,b\in R\implies a+b\not\in R$.
Note that I formulated this definition pretty general. We do not have to restrict to subsets $R\subset\Bbb R$, but this definition makes sense for all groups, e.g. $R\subset\Bbb Z$.
So given a group $G$ and a sum-free subset $R\subset G$, then we call $R$ maximal if there is no proper sum-free superset of $R$ in $G$.
My first approach for finding explicit maximal sum-free subsets was using cosets. If $R$ is a subgroup of $(\Bbb R,+)$ and $r\not\in R$, then $R+r$ is a (proper) coset of $R$ and it is not hard to see that it is sum-free. And this works for all groups. The cosets of subgroups are always sum-free.
However, it turns out that such a cotset is maximal sumfree if and only if $R$ is a maximal subgroup. Unfortunately, $\Bbb R$ and $\Bbb Q$ and similar groups do not have a maximal subgroup. This means, no maximal sum-free set in $\Bbb R$ (or $\Bbb Q$) can be a coset.
However, this works for other groups, e.g. $\Bbb Z$ as the maximal subgroup $2\Bbb Z$. Hence $2\Bbb Z+1$ (the set of odd integers) is maxmal sum-free.
