How to prove that if convex $A \subset \mathbb{R}^n$ and $A+A=A$ then $0 \in cl(A)$? How to prove that if convex $A \subset \mathbb{R}^n$ and $A + A = A$ then $0 \in cl(A)$? 
For a example of $A$ which holds a condition but $0 \notin A$ consider $(0,\infty)$.
 A: If $0 \not\in cl(A)$ then there exist $\varepsilon > 0$ such that $B_{\varepsilon}(0) \cap A =\emptyset$. Now take $$l = \sup\{\varepsilon > 0\text{ such that }B_{\varepsilon}(0) \cap A =\emptyset\}.$$
Clearly $B_l(0) \cap cl(A) \neq \emptyset$ and then you can get an absurd using that $A + A = A$. If $k = \sup\{\varepsilon > 0\text{ such that }B_{\varepsilon}(0) \cap (A + A) =\emptyset\}$ then $k \geq 2l$ since if not, let $k < r < 2l$, there exist $v,w \in A$ such that $$v + w = 2\left(\frac{1}{2}v + \frac{1}{2}w\right) \in B_{r}(0).$$
Since $A$ is convex $\frac{1}{2}v + \frac{1}{2}w \in B_{\frac{r}{2}}(0) \cap A$ and this is absurd since $\frac{r}{2} < l$. Now, since $A + A = A$ then $l$ should be $k \geq 2l$.
A: Assume $A\neq\emptyset$. Let $x\in A$. Then since $A+A\supset A$ there exists $y,z\in A$ such that $x=y+z$. But then by convexity $\tfrac{1}{2}x = \tfrac{1}{2}(y+z)\in A$. Repeating this argument with $\tfrac{1}{2}x$, $\tfrac{1}{4}x$, and so on, you get $\tfrac{1}{2^k} x\in A$. Letting $k\to\infty$, $0\in\text{cl}(A)$.
A: We have to assume $A\neq\emptyset$. Suppose $0\notin\operatorname{cl}(A)$. Then there exists $p\neq 0$ and $r>0$ such that $px\geq r$ for all $x\in\operatorname{cl}(A)$ and $ px^*= r$ for some  $x^*\in\operatorname{cl}(A)$ by the separating hyperplane theorem. We can write each $x\in\operatorname{cl}(A)$ as $x=a+b$ and hence $px=pa+pb\geq r+r=2r$, in particular $px^*\geq 2r$, which is impossible.
