TITLE: Construct an example of set $A$ for which $A+A=A $ but $0∉cl(A)$
QUESTION [14 upvotes]: How to prove that convexity is necessary condition in this question? Need to construct an example of set $A$ for which $A+A=A$ but $0 \notin cl(A)$. From the linked question follows that $A$ must be non-convex.

REPLY [7 votes]: This is merely a distilled version of the answer by coffeemath (hence CW). Pick an irrational number $\alpha$ and let $A=\{(m,n)\in\mathbb Z^2:m\alpha+n>0\}$. This is a discrete set, it does not contain $(0,0)$, and obviously satisfies $A+A\subseteq A$. To prove the reverse inclusion $A\subseteq A+A$, note that for any $(m,n)\in A$ there exists $(m',n')$ such that $0<m'\alpha+n'<m\alpha+n$, and therefore $(m,n)=(m',n')+(m-m',n-n')\in A+A$. 
This construction comes up in complex analysis, since the linear combinations of the functions $\{z^mw^n: (m,n)\in A\}$ form an algebra on the torus $|z|=|w|=1$ (here $z,w$ are complex). However, in this case one throws in $(0,0)$ (hence, the constant function) in order to have a unital algebra. See, e.g., Banach spaces of analytic functions by K. Hoffman, pp 54-57.