Center of distances and Bernstein sets
Mateusz Kula
TL;DR
This work resolves Filipczak's question by showing that for every $A\subset [0,\infty)$ with $0\in A$ there exists $X\subset R$ with center of distances $S(X)=A$, and moreover that $X$ can be chosen as a Bernstein set. The authors develop an average-operator framework $T_C$ and a transfinite, Hamel-basis–driven construction to achieve surjectivity of the center-of-distances operator, embedding the prescribed centers into the complement of the dyadic-averaging closure $T_C^\infty(Y)$. They then refine the construction to produce Bernstein sets realizing $S(X)=A$, including the full case $S(X)=[0,\infty)$. Overall, the results give a complete answer to the posed problem and expand techniques for fabricating sets with prescribed distance-centered structures on the real line.
Abstract
We show that for any subset $A\subset [0,\infty)$, where $0\in A$, there exists a Bernstein set $X\subset \mathbb R$ such that $A$ is the center of distances of $X$.
