Does there exist a uniformly antisymmetric function f:R-->R with range being (a) finite? (b) bounded?

by
Krzysztof Chris Ciesielski

A function f:R-->R is weakly symmetrically continuous at point x if there is a sequence $h$_n-->0 such that

$f(x+h$_n)-f(x-h_n))-->0 as n tends to infinity.

A function f:R-->R is uniformly antisymmetric if it is nowhere weakly symmetrically continuous. In 1993 Ciesielski and Larson constructed a uniformly antisymmetric f:R-->N. (Uniformly antisymmetric functions, Real Anal. Exchange 19 (1993-94), 226-235.) Also I have proved, (On range of uniformly antisymmetric functions, Real Anal. Exchange 19 (1993-94), 616-619) that the range of such a function must have at least 4 elements. See also section 2 of the survey Set Theoretic Real Analysis for more on this subject.

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A partial solution to the problem

by

K. Ciesielski and S. Shelah

Comment added May 6, 1998.

There exists a uniformly antisymmetric function f:R-->R with bounded countable range.