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

Volume 168 / 2001

Krzysztof Ciesielski, Kandasamy Muthuvel, Andrzej Nowik Fundamenta Mathematicae 168 (2001), 119-130 MSC: Primary 26A15; Secondary 03E50, 03E35. DOI: 10.4064/fm168-2-3


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.


  • Krzysztof CiesielskiDepartment of Mathematics
    West Virginia University
    Morgantown, WV 26506-6310, U.S.A.
  • Kandasamy MuthuvelDepartment of Mathematics
    University of Wisconsin-Oshkosh
    Oshkosh, WI 54901-8601, U.S.A.
  • Andrzej NowikInstitute of Mathematics
    Polish Academy of Sciences
    Abrahama 18, Sopot, Poland
    Department of Mathematics
    Gda/nsk University
    Wita Stwosza 57
    80-952 Gda/nsk, Poland

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