Can any real number be expressed as the sum of a rational number with a number of the Cantor set?

Question

I only ask for either a proof that this is true or a counterexample of a real number that can NOT be expressed as the sum of a rational + a number in the Cantor Set

Thanks

Answer 1

No, and it follows from a dimension argument. Let $C$ be the cantor set. Then your claim is that $C + \mathbb{Q} = \mathbb{R}$, i.e. that every real is a sum of an element of $C$ and a rational. We know that $\dim_{H}(C) = \frac{\log 2}{\log 3}$. But \begin{align*} C + \mathbb{Q} & = \{c + q : c \in C, q \in \mathbb{Q} \} \\ & = \bigcup_{q \in \mathbb{Q}} (C + q) , \end{align*} But we know since $\dim_{H}$ is translation-invariant that $\dim_{H}(C + q) = \dim_{H}(C)$, and we know that since the dimension of a countable union of sets is the supremum of the dimensions of the sets that $\dim_{H}(C + \mathbb{Q}) = \dim_{H}(\bigcup_{q \in Q} C + q) = \dim_{H}(C) = \frac{\log 2}{\log 3}$. But $\dim_{H}(\mathbb{R}) = 1$, so we know that $C + \mathbb{Q} \neq \mathbb{R}$.

EDIT: This also follows from a simpler measure argument that since Lebesgue measure is countably sub-additive, and $C$ has null Lebesgue measure, we have $\lambda(C + \mathbb{Q}) = 0$, while $\lambda(\mathbb{R}) = + \infty$. More generally, the former argument show that if you have some set $F$ where $\dim_{H}(F) < 1$, and some countable set $S$, then $F + S \neq \mathbb{R}$, where the latter is slightly more general in that we need only assume $\lambda(F) = 0$, a weaker assumption that $\dim_{H}(F) < 1$.

Answer 2

Take any closed subset $E$ of $\mathbb R$ with empty interior, positive measure or not. Then $E + \mathbb Q \ne \mathbb R.$ Proof: Baire.