A function f:R-->{0,1} is weakly symmetric (weakly symmetrically continuous) at x in R provided there is no sequence h_n-->0 such that f(x+h_n)=f(x-h_n)=f(x) (f(x+h_n)=f(x-h_n), respectively) for every n. We will 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 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 h_n, 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.

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Last modified June 22, 2001.