Does there exist a subset $S\subset\mathbb R$ such that inf $\{a>0:S+a=\mathbb R-S\}=0$? I founded the following question a good challenge in real analysis and topological properties of real line...

Does there exist a subset $S\subset\mathbb R$ such that inf $\{a>0:S+a=\mathbb R-S\}=0$ ?

Observe that,
$$S+a=\{s+a:s\in S\}$$
$$\mathbb R-S=\{x\in\mathbb R:x\notin S\}$$
Can anybody prove or disprove it ?

The answer of question is

 Yes, there is such a set.

Reference: Problems in Mathematical Analysis-Piotr Biler,Alfred Witkowski
 A: For a positive sequence $a_n\to0$, we
 want to achieve that for $x\in S\iff x+a_n\notin S$.
For a single $a$ this is not difficult: Let $R$ be  system of representatives of $\mathbb R/a\mathbb Z$ and let $S=\{\,x+2ka\mid x\in R,k\in\mathbb Z\,\}$.
Here's how to handle the infinite case:
Let $A_n=\bigoplus_{i=1}^n\mathbb Z$, $A=\bigoplus_{i=1}^\infty\mathbb Z=\bigcup_{n=1}^\infty A_n$.
Let $C$ be the kernel of $A\stackrel \sigma\longrightarrow\mathbb Z\stackrel{\text{can.}}\longrightarrow\mathbb Z/2\mathbb Z$, i.e. the set of elements with even sum.
Define injective linear maps $f_n\colon A_n\to\mathbb R$ recursively (starting from the zero map $f_0\colon 0\to \mathbb R$):
Given $f_{n-1}$, pick $a_n\in(0,\frac1n)\setminus \bigcup_{n\in\mathbb N}\frac1nf_{n-1}(A_{n-1})$ and define $f_n(x_1,\ldots,x_n)=f_{n-1}(x_1,\ldots,x_{n-1})+x_na_n$.
The choice is possible because we remove only countably many points from an (uncountable) open interval, and the removal guarantees that $f_n$ is injective.
Since $f_n|_{A_{n-1}}=f_{n-1}$ at each step, we obtain an injective linear map $A\to\mathbb R$.
Let $R$ be a system of representatives of $\mathbb R/f(B)$ and let 
$$ S=R+f(C)=\{\,x+f(c)\mid x\in R, c\in C\,\}.$$
One verifies that for each $a_n$ we have $x\in S\iff x\pm a_n\notin S$, i.e. 
$$a_n\in\{\,a>0\mid S+a=\mathbb R-s\,\}$$
As $a_n<\frac 1n$, we conclude 
$$\inf\{\,a>0\mid S+a=\mathbb R-s\,\}=0.$$

Remark: We could make the choice of $a_n$ constructive, e.g. let $a_n$ be the reciprocal of the $n$th prime. But for $R$ we really need the Axiom of Choice.
A: Let
$$
A=\{a_n=e^{-n}\,|\,n\in \mathbb N\}
$$
The number $e^{-1}$ is transcendental, hence the set $A$ is independent on rational numbers. Now by Zorn's lemma, there is a Hamel basis $H$ including $A$.
For every real number $x=\sum_{\alpha_i\in H} q_i \alpha_i$, with $q_i$ rational and only finitely many of them nonzero, define
$$
\pi:\mathbb R\to \{0,1\},
\pi (x)\equiv\sum_{\alpha_i\in A} \lfloor q_i\rfloor\pmod 2
$$
Now define $S=\pi^{-1}(\{0\})$ and observe that, for every natural number $n$
$$
S+a_n=\mathbb R-S
$$
