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

Question

$$\mathbb{Z}_+ \cup \{ -1 + \frac{1}{2}, -2+\frac{1}{3}, -3+\frac{1}{4}, -5+\frac{1}{6}, \cdots \}$$ 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. — Alex Provost, Apr 06 '13 at 18:55
I think $S+S$ means the set consisting of sums of $2$ elements of $S$. — Dylan Yott, Apr 06 '13 at 18:58
@Silencer It is the Minkowski sum. $A+B:={a+b:a\in A,b\in B}$ — Pedro, Apr 06 '13 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 06 '13 at 18:58
Hi, your account @user71284 is not registered, which is why you seemed to have lost control to the question. Please create a new account, register it, and then visit this page to reclaim ownership of this question. — Willie Wong, Apr 12 '13 at 14:53

score 2 · Accepted Answer · answered Apr 06 '13 at 19:00

2

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

answered Apr 06 '13 at 19:00

Did

279,727

Excellent, so it seems very important that these sets are closed but NOT bounded. – user71284 Apr 06 '13 at 19:12
@user71284, yes, because in a topological group the sum of a closed and a compact subset is closed. (See http://math.stackexchange.com/questions/80974/closed-sum-of-sets) – Andreas Caranti Apr 06 '13 at 19:15
@AndreasCaranti Thanks. – Did Apr 06 '13 at 19:35

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

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