### 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.

by

K. Ciesielski and S. Shelah

Comment added May 6, 1998.

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