The question:
If $\ A,B\ $ are countable, nowhere dense subsets of $\ \mathbb{R},\ $ and furthermore $\ A\ $ is bounded, then is $\ A+B = \{a+b:\ a\in A,\ b\in B \}\ $ nowhere dense in $\ \mathbb{R}\ ?$
Firstly, I know from another question I asked that if $\ C\ $ is the ternary Cantor set, then $\ C + C = [0,2].\ $ $\ C\ $ is nowhere dense. However, it is uncountable, and this question is about countable nowhere dense subsets of $\ \mathbb{R}. $
Secondly, without the boundedness restriction, we could let $\ \{\alpha_n\}_{n\in\mathbb{N}}\ $ be an enumeration of the rationals between $\ 0\ $ and $\ \frac{1}{10}.\ $ Then if we let $\ A = \{ 1+\frac{1}{2}+ \frac{\alpha_1}{2},\ 2+ \frac{1}{3}+ \frac{\alpha_2}{2},\ 3+ \frac{1}{4}+ \frac{\alpha_3}{2},\ \ldots\}\ $ and $\ B = \{ -1-\frac{1}{2}+ \frac{\alpha_1}{2},\ -2- \frac{1}{3}+ \frac{\alpha_2}{2},\ -3- \frac{1}{4}+ \frac{\alpha_3}{2},\ \ldots\},\ $ then $\ \{\alpha_n\}_{n\in\mathbb{N}}\ \subset A+B,\ $ and so $\ A+B\ $ is dense in $\ \left[ 0, \frac{1}{10}\right].$
But with the restriction of $\ A\ $ being bounded, I don't think this process works without making $\ A\ $ and $\ B\ $ both dense somewhere in $\ \mathbb{R}\ $ which is not allowed, but I am not sure.
Obviously both $\ A\ $ and $\ B\ $ must have limit points due to $\ B-W,\ $ but I don't see how this helps answer the question.
Both $A$ and $B$ being the set of the rationals in the Cantor set is a counterexample. $\overline{\Bbb Q\cap C}=C$, therefore $$\overline{(\Bbb Q\cap C)+(\Bbb Q\cap C)}\supseteq C+C= [0,2].$$