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?

  • $\begingroup$ 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

$A+B=\mathbb{Q}$ holds with $A=\mathbb{Z}$ and $B=\{q_n-n\}$, where $q_1,q_2,\ldots$ is an enumeration of $\mathbb{Q}\cap[0,1)$.

  • 2
    $\begingroup$ 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

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 $$ K+K=\mathbb R. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.