Monthly Archives: July 2009

Functions with Very Low Symmetry and the Continuum Hypothesis

A function from \mathbb{R} to \mathbb{R} is called even if for all h\in\mathbb{R}, f(-h) = f(h).  We might call it even about the point x if, for all h\in\mathbb{R}, f(x-h) = f(x +h).

Conversely, we can call a function strongly non-even if for all x\in\mathbb{R}, h>0, f(x-h)\ne f(x+h).

Finding strongly non-even functions is easy, as any injective function provides a trivial example.  We can make things harder for ourselves by considering only functions from \mathbb{R} to \mathbb{N}.  But now, it is just as easy to show that there are no strongly non-even functions.

Therefore, let’s make the following definition: Let a function f\colon \mathbb{R}\to\mathbb{N} be non-even of order n if, for all x, |\{h>0\mid f(x-h) = f(x + h)\}|\leq n.  Thus, a strongly non-even function is non-even of order 0, and a function being non-even of order n implies that it’s non-even of order m for all m\geq n.

In this paper, the set theorists Peter Komjáth and Saharon Shelah proved:

The existence of a non-even function of order 1 is equivalent to the Continuum Hypothesis (i.e., the statement that 2^{\aleph_0} = \aleph_1).

Thus, if we assume that there is a non-even function of order 1, then we can conclude that 2^{\aleph_0} = \aleph_1.  Can we weaken the hypothesis and still conclude something interesting?  We can, as they also proved:

For any n, if there is a non-even function of order n, then 2^{\aleph_0}\leq \aleph_{\lceil\log_2{(n+1)}\rceil}.

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