5

Let $A$ and $B$ be two closed subsets of the set of real numbers.

Define $A+B=\{a+b\in\mathbb{R}:a\in A ,b\in B\}$.

Is it true that $A+B$ is closed in $\mathbb{R}$?

If not, could you give a counter-example?

User
  • 607
  • 5
  • 12

2 Answers2

14

Let $A$ be the set of negative integers.

Let $B$ be the set of all $n+\frac{1}{2^n}$ where $n$ ranges over the positive integers.

Then $A$ and $B$ are closed.

But $A+B$ is not closed, since it contains numbers arbitrarily close to $0$ but does not contain $0$.

André Nicolas
  • 507,029
3

The result is not true. Take $A=\{\pi+n+1\colon n\in\mathbb N\}$, and $ B=\left \{-n-1+\frac1 {n+2}\colon n\in \mathbb N\right\} $. Both sets are closed, $\pi $ is a limit point of their sum, but is not on their sum.

If both $ A, B $ are compact, so is their sum (Since $ A+B $ is the image of $ A\times B $ under the continuous function $(x, y)\mapsto x+y $). Can you see what happens when one set is compact and the other isn't?

Andrés E. Caicedo
  • 79,201
  • May I ask why you used "$n+1$" in the argument of the set $A$ and "$-n-1$" in the argument of the set $B$? – User Aug 22 '14 at 22:43
  • 1
    (To see how one could find the counterexample, think about the obvious approach: If $ x $ is a limit point of the sum, there are $ a_n\in A $ and $ b_n\in B $ with $ a_n+ b_n\to x $. If the sets are bounded, both sequences have subsequences that converge to points whose sum converges to $ x $. The issue is then if the sequences are unbounded and yet $ a_n+b_n $ somehow manages to converge.) – Andrés E. Caicedo Aug 22 '14 at 22:47
  • @Yk26 It is not strictly needed, simply to avoid having to decide what happens if we allow $ n=0$. – Andrés E. Caicedo Aug 22 '14 at 22:54
  • (Typo in the comment: whose sum is.) – Andrés E. Caicedo Aug 22 '14 at 22:56