# A Set That is Closed but For Which $S + S$ is Not

$$\mathbb{Z}_+ \cup \{ -1 + \frac{1}{2}, -2+\frac{1}{3}, -3+\frac{1}{4}, -5+\frac{1}{6}, \cdots \}$$ 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?

• What is S + S? Union? The union of two closed sets is always closed. Apr 6, 2013 at 18:55
• I think $S+S$ means the set consisting of sums of $2$ elements of $S$. Apr 6, 2013 at 18:58
• @Silencer It is the Minkowski sum. $A+B:=\{a+b:a\in A,b\in B\}$
– Pedro
Apr 6, 2013 at 18:58
• 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.
– Orat
Apr 6, 2013 at 18:58
• @Taro your revision was correct - it looks much better now. Apr 6, 2013 at 19:07

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 $\mathrm{cl}(S+S)$.