EN
A function f: ℝ → {0,1} is weakly symmetric (resp. weakly symmetrically continuous) at x ∈ ℝ provided there is a sequence hₙ → 0 such that f(x+hₙ) = f(x-hₙ) = f(x) (resp. f(x+hₙ) = f(x-hₙ)) 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 ∈ ℝ∖S(f). In particular, there is no f: ℝ → {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 hₙ, then there exists a function f: ℝ → {0,1} which is nowhere weakly symmetric in this weaker sense if and only if the continuum hypothesis holds.