Michel Mendès-France: The diagonal of Cantor

This is a joint work with Srecko Brlek, Mike Robson and Martin Rubey.

Consider the infinite tableau where rows consist of the expansion in basis $b>1$ of all real algebraic numbers of the the open unit interval. Cantor observed that the diagonal, if disturbed, represents the $b$-expansion of a transcental number. We show that the ``undisturbed'' diagonal is the expansion of a transcendental number. By permuting the rows of the tableau in all possible ways, we therefore obtain a set of transcendental numbers.

Theorem 1: These numbers all contain infinitely many times each one of the digits $0,1,2,\ldots,b-1$.

In particular, if $b>2$, then the existance of Liouville numbers proves that we do not get all transcendental numbers of the unit interval. However if $b=2$ the situation changes.

Theorem 2: In basis 2, the set of diagonal numbers consists exactly of all transcendental numbers of the unit interval.

We also concern ourselves with finite square $n\times n$ tableaux with elements in a finite alphabet $A=\{a_1,a_2,\ldots,a_s\}$. There are obviously $s^{n^2}$ tableaux of size $n\times n$. Consider one such tableau together with all those obtained by permuting the rows. If none of the diagonals are equal to any one of the rows, the tableau will be called ``Cantorian". Let $C(n)$ be the number of $n\times n$ Cantorian tableaux.

Theorem 3: Let $a<1$ and $b>1$. If $s<an/\log n$, then $C(n)/s^{n^2} \to 0$ when $n$ increases to infinity. If $s>bn/\log n$ then $C(n)/s^{n^2} \to 1$.

In other words, $n/\log n$ is a critical size for the alphabet $A$: ``almost no" tableaux are Cantorian if the alphabet A is ``small" and ``almost all" are Cantorian if the alphabet is large.

Back to the Index