# 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 $2^{\aleph_0}\leq \aleph_{\lceil\log_2{(n+1)}\rceil}$.