# Construct an example in $\mathbf R$ where $A$ and $B$ are two nowhere dense sets but closure of $A + B$ is $\mathbf R$ itself.

Any answers will be appreciated. I know $\mathbf Q$ has closure $\mathbf R$, but can $\mathbf Q$ be expressed as the sum of two nowhere dense sets? Or will some other example work?

• You can make $A$ to be the integers and $B= \sqrt{2} A$ – PenasRaul Sep 1 '14 at 21:58
• @PenasRaul This does not seem to work because $A+B\neq\mathbb{Q}$ – user2097 Sep 1 '14 at 22:01
• @user2097 An additive subgroup of $\mathbb{R}$ is either cyclic or dense. Since both $\mathbb{Z}$ and $\sqrt{2}\mathbb{Z}$ are additive subgroups and their sum is not cyclic, their sum is dense. On the other hand both are closed with empty interior. – egreg Sep 1 '14 at 23:25
• @egreg The OP wants $A+B=\mathbb{Q}$ if I understand the question correctly. – user2097 Sep 1 '14 at 23:29
• @user2097 My impression is that the title and the question don't match and that asking about $\mathbb{Q}$ is just an attempt. And there is “Or will some other example work?” – egreg Sep 1 '14 at 23:33

$A+B=\mathbb{Q}$ holds with $A=\mathbb{Z}$ and $B=\{q_n-n\}$, where $q_1,q_2,\ldots$ is an enumeration of $\mathbb{Q}\cap[0,1)$.
• Sorry if I'm missing something obvious, but why is $B$ nowhere dense? – David Mitra Sep 1 '14 at 22:55
• @DavidMitra Because $[-n,-n+1)$ contains exactly one point from $B$. – user2097 Sep 1 '14 at 22:59
A well-known property of the Cantor set (which is nowhere dense) is that $C+C=[0,2]$. (See here.)
Now, let $K$ be the 1-periodic continuation of Cantor set, i.e., $$K=\bigcup_{k\in\mathbb Z}k+C.$$ Then $K$ is nowhere dense, and $$K+K=\mathbb R.$$