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 $(0,\infty)$.
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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 $k \geq 2l$.
Diego Silvera
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1Does $A$ must be a convex cone? – Tomás Feb 16 '13 at 11:47
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1Do you mean "$k<r<2l$" instead of "$r<k<2l$"? – Matemáticos Chibchas Feb 16 '13 at 19:51
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No, it's right in that way. – Diego Silvera Feb 16 '13 at 22:30
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1@DiegoSilvera If $r<k$, then by definition of $k=\sup K$, where $K={\epsilon>0: B_\epsilon(0)\cap(A+A)=\emptyset}$, there is some $s\in K$ such that $r<s$, and so $B_r(0)\cap(A+A)\subseteq B_s(0)\cap(A+A)=\emptyset$. On the other hand, if you assume $k<r\leq 2l$, then $r\notin K$, and so the existence of the elements $v,w$ follows. – Matemáticos Chibchas Feb 17 '13 at 11:28
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Yes, you are OK. Thanks a lot. – Diego Silvera Feb 17 '13 at 12:10
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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 $k\to\infty$, $0\in\text{cl}(A)$.
Sean Eberhard
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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.
Michael Greinecker
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1Good, you gave me a Pavlovic conditioning that whenever I see your name I think about Rebel Yell and Billy Idol. :-) – Asaf Karagila Feb 16 '13 at 11:46
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@Tomás Not sure. $\operatorname{cl}(A)$ is certainly a convex cone. – Michael Greinecker Feb 16 '13 at 17:36