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Let $A, B \subset \mathbb{R}$. We know that $A' = B' = \varnothing$ and $(A + B)' = [0, 1]$, where $$A + B = \{a + b \; | \; a \in A, b \in B\}$$ Find $A, B $.

I am trying to solve that problem for the second day in a row and the only thing that seems reasonable to me is taking $A = B = \frac{1}{n}$. If so, we can get a fraction any close to each point from $[0, 1]$. But in this case we have $A' = B' = \{0\}$ and that ruins the solution.

Would be thankful for any help!

  • Is $A'$ the set of limit points of $A$? Must $A$ and $B$ themselves be contained in $[0,1]$? It seems extremely unlikely to me that such pairs of sets $A,B$ are unique (all expressions seem invariant under adding finitely many points, for example); are you sure that's how the problem is stated? – Greg Martin Nov 02 '20 at 17:07
  • Yes, A' is a set of limit points of A. And I also guess there are many ways of choosing A and B, but I can't find at least one – K-dizzled Nov 02 '20 at 17:10
  • $A = {-1, -2, -3, \dots, -n, \dots}$ and $B = {1 + \tfrac{1}{1}, 2 + \tfrac{1}{2}, 3 + \tfrac{1}{3}, \dots, n + \tfrac{1}{n}, \dots}$ give $0 \in (A + B)'$. Is this useful? – Nikita Skybytskyi Nov 02 '20 at 17:12
  • Not sure what did you mean by "give $0$" But I am checking your solution and it seems to work – K-dizzled Nov 02 '20 at 17:28
  • If I got you right, in your example we will only have fractions of type $\frac{1}{n}$ in $[0, 1]$ and there would be only one limit point = $0$ – K-dizzled Nov 02 '20 at 22:03
  • @AndreiKozyrev - they only gave you a hint, not a full answer. Here is another hint: $\Bbb Q\cap [0,1]$ is dense in $[0,1]$ and countable. – Paul Sinclair Nov 03 '20 at 00:39

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I managed to get the answer only a few days ago. The idea was to take the set $\mathbb{Q}$ and some function, that grows fast. For example $2^n$. Let set $A$ consist of $a_i$, where $a_i \in 2^n$ and $i$ is an index of the sequence. Let $B$ equal to $-a_i + j$, where $j \in \mathbb{Q}$ is a rational number with index j.