4

A centered set of real numbers is a set $S$ of real numbers such that $0 \in S$ and $x \in S \rightarrow -x \in S$. Also, let $A$ and $B$ be sets of real numbers. The difference $A - B$ is the set $\{a - b | a \in A, b \in B\}$ It is easy to prove that every set of the form $A - A$ for a non-empty $A$ is a centered set. What about the converse? That is, is every centered set of the form $A - A$ for some non-empty $A$?

user107952
  • 20,508

1 Answers1

11

Let $S = \{-3, -1, 0, 1, 3\}$, so $S$ is a centered set.

If $S = A-A$, then there is $a, b\in A$ such that $a-b = 3$ (i.e., $b = a-3$). Furthermore, $a$ and $b$ can't be the only elements of $A$, so there must be $c\in A$, $c\neq a, b$. Notice that $a-c$ must be $\pm 1$ or $\pm 3$.

  • If $c-a = 1$ (i.e., $c = a+1$), then $c-b = 4\not \in S$. Contradiction.
  • If $c-a = -1$ (i.e., $c = a-1$), then $c-b = 2\not \in S$. Contradiction.
  • If $c-a = 3$ (i.e., $c = a+3$), then $c-b = 6\not \in S$. Contradiction.
  • If $c-a = -3$ (i.e., $c = a-3$), then $c = b$. Contradiction.
Alma Arjuna
  • 3,759
  • 1
    More generally, if $S=A-A$ for $|S|>4$ then for each $s\in S$ there must be non-zero $s_1,s_2\in S$ such that $s=s_1+s_2.$ This is because $|A|>2$ means given any $a,b\in A$ there is a $c\in A,$ $c\notin{a,b}$ and $a-b=(a-c)+(c-b).$ Indeed, there must be $|A|-2$ values of $c.$ If $|S|>n(n-1)+2$ then there are at least $n-1$ non-zero pairs $s_1,s_2.$ – Thomas Andrews Jul 10 '21 at 00:55
  • 1
    So you don't need all the cases, clearly there are no non-zero $s_1,s_2\in S$ so that $s_1+s_2=3.$ – Thomas Andrews Jul 10 '21 at 01:00