On nowhere weakly symmetric functions and functions with two-element range
Volume 168 / 2001
Abstract
A function $f : {\mathbb R}\to\{0,1\}$ is weakly symmetric (resp. weakly symmetrically continuous) at $x\in{\mathbb R}$ provided there is a sequence $h_n\to 0$ such that $f(x+h_n)=f(x-h_n)=f(x)$ (resp. $f(x+h_n)=f(x-h_n)$) for every $n$. We characterize the sets $S(f)$ of all points at which $f$ fails to be weakly symmetrically continuous and show that $f$ must be weakly symmetric at some $x\in{\mathbb R}\setminus S(f)$. In particular, there is no $f : {\mathbb R}\to\{0,1\}$ which is nowhere weakly symmetric.
It is also shown that if at each point $x$ we ignore some countable set from which we can choose the sequence $h_n$, then there exists a function $f : {\mathbb R}\to\{0,1\}$ which is nowhere weakly symmetric in this weaker sense if and only if the continuum hypothesis holds.
