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.**