$$\mathbb{Z}_+ \cup \{ -1 + \frac{1}{2}, -2+\frac{1}{3}, -3+\frac{1}{4}, -5+\frac{1}{6}, \cdots \}$$ Is apparently an example because $0$ is not in $S + S$. I am unclear as to why it is not though.

$Edit$: I am the user that originally posted this question (I should have made an account) but I realized I am not sure why S itself is closed either. Isn't zero a boundary point of S not contained in S?

  • $\begingroup$ What is S + S? Union? The union of two closed sets is always closed. $\endgroup$ Apr 6, 2013 at 18:55
  • $\begingroup$ I think $S+S$ means the set consisting of sums of $2$ elements of $S$. $\endgroup$
    – Dylan Yott
    Apr 6, 2013 at 18:58
  • $\begingroup$ @Silencer It is the Minkowski sum. $A+B:=\{a+b:a\in A,b\in B\}$ $\endgroup$
    – Pedro
    Apr 6, 2013 at 18:58
  • $\begingroup$ I rewrite your question with LaTeX. Please check whether I don't change the meaning or not. To write LaTeX by yourself, please see here for example. $\endgroup$
    – Orat
    Apr 6, 2013 at 18:58
  • $\begingroup$ @Taro your revision was correct - it looks much better now. $\endgroup$
    – user71284
    Apr 6, 2013 at 19:07

1 Answer 1


Elements of $S+S$ are either sums of two positive integers, hence positive, or sums of two negative numbers, hence negative, or sums $-n+\frac1{n+1}+k$ with $n\geqslant1$ and $k\geqslant1$. These last numbers are sums of an integer and $\frac1{n+1}$, hence not zero. Finally, no element of $S+S$ is $0$.

But $-n+\frac1{n+1}+n=\frac1{n+1}$ is in $S+S$ for every $n\geqslant1$ and $\frac1{n+1}\to0$ when $n\to\infty$, hence indeed $0$ is in $\mathrm{cl}(S+S)$.

  • $\begingroup$ Excellent, so it seems very important that these sets are closed but NOT bounded. $\endgroup$
    – user71284
    Apr 6, 2013 at 19:12
  • $\begingroup$ @user71284, yes, because in a topological group the sum of a closed and a compact subset is closed. (See math.stackexchange.com/questions/80974/closed-sum-of-sets) $\endgroup$ Apr 6, 2013 at 19:15
  • $\begingroup$ @AndreasCaranti Thanks. $\endgroup$
    – Did
    Apr 6, 2013 at 19:35

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