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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?

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  • $\begingroup$ You can make $A$ to be the integers and $B= \sqrt{2} A$ $\endgroup$ – PenasRaul Sep 1 '14 at 21:58
  • $\begingroup$ @PenasRaul This does not seem to work because $A+B\neq\mathbb{Q}$ $\endgroup$ – user2097 Sep 1 '14 at 22:01
  • $\begingroup$ @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. $\endgroup$ – egreg Sep 1 '14 at 23:25
  • $\begingroup$ @egreg The OP wants $A+B=\mathbb{Q}$ if I understand the question correctly. $\endgroup$ – user2097 Sep 1 '14 at 23:29
  • $\begingroup$ @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?” $\endgroup$ – egreg Sep 1 '14 at 23:33
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$A+B=\mathbb{Q}$ holds with $A=\mathbb{Z}$ and $B=\{q_n-n\}$, where $q_1,q_2,\ldots$ is an enumeration of $\mathbb{Q}\cap[0,1)$.

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    $\begingroup$ Instead of 'denotes', I would say 'enumerates' (or even better, 'is an enumeration of'). But +1. $\endgroup$ – TonyK Sep 1 '14 at 22:18
  • $\begingroup$ Sorry if I'm missing something obvious, but why is $B$ nowhere dense? $\endgroup$ – David Mitra Sep 1 '14 at 22:55
  • $\begingroup$ @DavidMitra Because $[-n,-n+1)$ contains exactly one point from $B$. $\endgroup$ – user2097 Sep 1 '14 at 22:59
  • $\begingroup$ Ah, of course. Thanks. $\endgroup$ – David Mitra Sep 1 '14 at 23:03
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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 $$ K+K=\mathbb R. $$

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