Problem
A set that shifts into its complement
For a set \(S\subseteq\mathbb R\) and a real number \(a\), write
\[S+a=\{s+a:s\in S\}.\]Does there exist a set \(S\subseteq\mathbb R\) such that
\[\inf\{a>0:S+a=\mathbb R\setminus S\}=0?\]In other words, can a set become its complement under arbitrarily small positive translations?
The solution uses vector spaces over \(\mathbb Q\) and the axiom of choice.
Hint
Choose positive numbers \(a_n\to0\) that are linearly independent over \(\mathbb Q\), and extend them to a basis. Adding \(a_n\) then increases just one coordinate by \(1\). Use \(\lfloor q+1\rfloor=\lfloor q\rfloor+1\) to reverse a parity.
Solution
Yes. The idea is to give every real number a parity that flips when any one of a sequence of small translations is applied.
Let \(a_n=e^{-n}\) for \(n\geq1\). These numbers are linearly independent over \(\mathbb Q\): a nontrivial relation
\[\sum_{n=1}^{N}c_ne^{-n}=0, \qquad c_n\in\mathbb Q,\]would, after multiplication by \(e^N\), give a nonzero polynomial with rational coefficients vanishing at \(e\). This contradicts the transcendence of \(e\).
Extend \(A=\{a_1,a_2,\ldots\}\) to a Hamel basis \(H\) of \(\mathbb R\) over \(\mathbb Q\). This is the step that uses the axiom of choice, through Zorn’s lemma. Every real number has a unique expansion
\[x=\sum_{h\in H}q_h(x)h, \qquad q_h(x)\in\mathbb Q,\]with only finitely many nonzero coefficients. Define the integer
\[p(x)=\sum_{n=1}^{\infty}\lfloor q_{a_n}(x)\rfloor\]and put
\[S=\{x\in\mathbb R:p(x)\text{ is even}\}.\]The sum defining \(p(x)\) is finite for each \(x\), since all but finitely many coordinates are zero.
Adding \(a_n\) increases \(q_{a_n}(x)\) by \(1\) and leaves every other coordinate unchanged. Consequently,
\[p(x+a_n)=p(x)+1.\]Thus \(x\in S\) if and only if \(x+a_n\notin S\). This gives both inclusions in \(S+a_n=\mathbb R\setminus S\): a point of \(S\) translates outside \(S\), and every point \(y\notin S\) has \(y-a_n\in S\).
Since \(a_n>0\) and \(a_n\to0\), the required infimum is zero.
The same construction works for any positive sequence tending to zero that is linearly independent over \(\mathbb Q\).
I encountered this problem as Problem 1.32 in Piotr Biler and Alfred Witkowski’s Problems in Mathematical Analysis (1990). I posted the question on Math StackExchange in 2014 and added this construction in 2016.
Extension
Could such a set be Lebesgue measurable?
No. Suppose \(S\) were measurable and \(S+a_n=\mathbb R\setminus S\) for some positive sequence \(a_n\to0\). Write \(m\) for Lebesgue measure. The set \(S\) cannot have measure zero: translation invariance would make its complement null as well.
By the Lebesgue density theorem, there is a bounded interval \(I\) such that, for \(E=S\cap I\),
\[m(E)>\frac34m(I).\]Choose \(n\) with \(a_n<\frac12m(I)\). The sets \(E\) and \(E+a_n\) are disjoint, since one lies in \(S\) and the other in its complement. Their union lies in \(I\cup(I+a_n)\), so
\[\begin{aligned} \frac32m(I)&<2m(E)\\ &=m\bigl(E\cup(E+a_n)\bigr)\\ &\leq m(I)+a_n\\ &<\frac32m(I), \end{aligned}\]a contradiction. Nonmeasurability is forced by the problem itself.