There is no strongly locally antisymmetric set (Q1189107)

A set \(A\subset R\) is said to be strongly locally antisymmetric if for every \(x\in R\) there exists a \(\delta_ x>0\) such that for each \(h\), \(0<h<\delta_ x\), \(x+h\in A\) if and only if \(x-h\notin A\). Answering a question asked by the reviewer [Comput. Math. Appl. 17, 103-115 (1989; Zbl 0703.26003)], the author proves that no strongly locally antisymmetric set exists; in the proof, it is used a construction introduced by \textit{B. S. Thomson} [Real Anal. Exch. 6, 77-93 (1981; Zbl 0459.26004)]. The author introduces the following polar opposite of locally symmetric functions: A function \(f\) is said to be locally antisymmetric if for each \(x\in R\) there is a \(\delta_ x>0\) such that \(| f(x+h)-f(x-h)|\geq\delta_ x\) holds for each \(h\), \(0<h<\delta_ x\). It is an open problem whether such a function really exists.