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.
@article{bwmeta1.element.bwnjournal-article-doi-10_4064-fm168-2-3,
author = {Krzysztof Ciesielski and Kandasamy Muthuvel and Andrzej Nowik},
title = {On nowhere weakly symmetric functions and functions with two-element range},
journal = {Fundamenta Mathematicae},
volume = {167},
year = {2001},
pages = {119-130},
zbl = {0986.26005},
language = {en},
url = {http://dml.mathdoc.fr/item/bwmeta1.element.bwnjournal-article-doi-10_4064-fm168-2-3}
}
Krzysztof Ciesielski; Kandasamy Muthuvel; Andrzej Nowik. On nowhere weakly symmetric functions and functions with two-element range. Fundamenta Mathematicae, Tome 167 (2001) pp. 119-130. http://gdmltest.u-ga.fr/item/bwmeta1.element.bwnjournal-article-doi-10_4064-fm168-2-3/