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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.

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I think that this problem is somehow related with the problem of construct a set $A$ such that $A+A=A$, but there is some $\alpha>0$ such that $\alpha A\neq A$. –  Tomás Feb 16 '13 at 14:33
    
If you have a such set $A$ and $v \in A$, then must $A$ contain $\lambda v$ for $\lambda > 0$? Because I can't imagine a such set $A$ without this property. –  Diego Silvera Feb 16 '13 at 15:25
    
That's why I asked in the linked post if the set $A$ is a convex cone, because if $\alpha A= A$ for every $\alpha >0$ and $A+A=A$, then $A$ is a convex cone. –  Tomás Feb 16 '13 at 16:23
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@Tomás That would not be hard to do: take positive rational numbers, for example. –  user53153 Feb 17 '13 at 2:34
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2 Answers 2

up vote 7 down vote accepted

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.

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Nice! Needs a lot less detail than the fibonacci construction. +1 –  coffeemath Feb 17 '13 at 10:33
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Let $F_n$ denote the $n^{th}$ fibonacci number, with a few terms having nonpositive subscripts, so that $$F_{-2}=-1,\ F_{-1}=1,\ F_0=0,\ F_1=F_2=1,\ F_3=2,\ F_4=3,...$$ Define the points $v_n$ in the $xy$ plane by $$v_n=[(-1)^n F_{n-3},\ (-1)^n F_{n-1}],$$ so that $$v_1=[1,0],\ v_2=[1,1],\ v_3=[0,-1],\ v_4=[1,2],\ v_5=[-1,-3],\ v_6=[2,5],...$$ Note that these points have the property that $v_n=v_{n+1}+v_{n+2}$ for each $n \ge 1$. In fact they may be generated from this relation, after $n=2$, taking $[1,0],\ [1,1]$ as the first two points and thereafter defining $v_n=v_{n-2}-v_{n-1}.$

Define the set $A$ in the plane as consisting of all (finite) liner combinations of the $v_n$ with nonnegative integer coefficients. Then $A+A=A$. This is because each $v_n$ is the sum of the next two, and $A$ is the closure of the set of $v_n$ under addition.

Now the closure of $A$ can only contain integer pairs, so to show this closure does not contain $[0,0]$ is the same as showing that $[0,0] \notin A.$

Suppose we have a nontrivial relation

[1] $\sum_{k=1}^{m+1} c_k v_k=[0,0].$

Then using the relation $v_n=v_{n+1}+v_{n+2}$ we can transform the sum until it only involves the highest two indices of the sum, namely $k=m,m+1.$ In so transforming the relation, the coefficients remain nonnegative (one only adds such coefficients). So the relation [1] would imply there are nonegative $x,y$ for which $$x v_m + y v_{m+1}=[0,0]$$ Factoring out $\pm 1$ from this relation we have the left side as $$x[F_{n-3},F_{n-1}]+y[-F_{n-2},-F_n]=[u,v],$$ where $u=xF_{n-3}-yF_{n-2}$ while using the recurrence for the fibonacci numbers, $$v=xF_{n-3}+xF_{n-2}-yF_{n-2}-yF_{n-1}.$$ Then $v-u=xF_{n-2}-yF_{n-1}.$ So from $[u,v]=[0,0]$ would follow the two equations $$xF_{n-3}-yF_{n-2}=0,\\ xF_{n-2}-yF_{n-1}=0.$$ But the determinant of this system is $-F_{n-3}F_{n-1}+F_{n-2}^2,$ which a fibonacci identity says is always $\pm 1$. This implies that $x,y$ are both zero, which means the supposed relation [1] was trivial after all.

We conclude that $[0,0]$ is not in the set $A$, so that this is an example in two dimensions of a set $A$ for which $A+A=A$, and yet $[0,0]$ is not in the closure of $A$.

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