On nowhere weakly symmetric functions and functions with two-element range

Krzysztof Ciesielski¹ ; Kandasamy Muthuvel² ; Andrzej Nowik³

¹ Department of Mathematics West Virginia University Morgantown, WV 26506-6310, U.S.A.
² Department of Mathematics University of Wisconsin-Oshkosh Oshkosh, WI 54901-8601, U.S.A.
³ Institute of Mathematics Polish Academy of Sciences Abrahama 18, Sopot, Poland and Department of Mathematics Gda/nsk University Wita Stwosza 57 80-952 Gda/nsk, Poland

Fundamenta Mathematicae, Tome 168 (2001) no. 2, pp. 119-130

Résumé

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.

Détail

DOI : 10.4064/fm168-2-3

Keywords: function mathbb weakly symmetric resp weakly symmetrically continuous mathbb provided there sequence n x h resp n x h every characterize sets points which fails weakly symmetrically continuous weakly symmetric mathbb setminus particular there mathbb which nowhere weakly symmetric shown each point ignore countable set which choose sequence there exists function mathbb which nowhere weakly symmetric weaker sense only continuum hypothesis holds