Disjoint nondenumerable sets of irrationals

[This page went online Feb. 28, 2002.]

I wish to thank Amherst logician Dan Velleman, author of 'How to Prove It,' for not permitting any slack or insecure arguments. He has not vouched for the final version of this theorem and proof. The theorem itself is nothing special. But the proof is amusing in that it gives an inkling of how vast is the sea of irrationals with respect to the rationals.

Theorem: There exists a denumerable infinite family of sets such that each family member is a nondenumerable set of irrationals that is disjoint from every other member.

Proof:

We establish a monotonically climbing differentiable real function h(x), where x is continuous, and where the first derivative exists and is neither a constant nor a periodic (as in cos(ax)). We use only the integer values h(n) in our proof.

We denote a rational q's infinite decimal digit string by S_0 and its recurring finite substring by s. We see that the first digit of s, which has k digit places, recurs every (nk)th digit place. We denote the initial digit place of s by p. We pair p with d_2, which symbolizes a specific numeral digit. Initially, we are assuming a base 10 system.

We then establish the erase-and-replace function f by locating the h(n)th (p, d_2) and replacing d_2 with d_3.

The altered string, S_x, is forever aperiodic.

Example:

s = (01), k = 2, h = n^2

S_0 = 0.010101010101010101...

S_x = 0.210101210101010121...

We know the set of h functions is infinite and we presume that that set, and hence the set of f functions, is denumerable. So we form the indexed set U f_i.

However, this permits us to form a 'diagonal' function g, such that g is formed by the sequence f_1(1), f_2(2), f_3(3) ... ad infinitum. We then alter the g(n)th (p,d_1) by use of a third digit symbol, writing (p,d_3).

Since g is aperiodic and cannot be indexed, we have established a nondenumerable set of irrationals, which we denote R.

In general, we will write R_a to mean the set R derived from the rational q_a.

Indexing the denumerable set of rationals by j, we question whether R_1 intersection R_2 is nonempty. If empty, we proceed consecutively until we find -- if ever -- R_1 intersection R_j is nonempty.

In that event, we form a new function t_i and require use of (p, d_4) on the subset of R_j strings, denoted R'_j, that are identical to R_1 strings. R_j U R'_j we call T_j (which still holds if R_j U R'_j = R_j). We then proceed to the next instance where R_j intersection R_(j+n) is nonempty. Again, we insert digit d_5, obtaining T_(j+n).

Obviously, with a base 10 system, this procedure can be repeated 9 times.

As the number of digit places in s is irrelevant, we can increase our set of available digits with a higher base number system. In that case, we apply induction: If a base m system yields m-1 replacement digits, then a base m+1 system, yields m replacement digits.

With m e M = N, and N denumerable, we find that the family U( T_j) exists, where J = N, requiring that U( T) is denumerable and that for all j e J, n e N(T_j is disjoint from T_(j+n).

Proof concluded.

Remark: It is curious that the set of writable functions is considered denumerable because each expression is composed of a finite string of symbols. For example, 2 + log x, contains 9 symbols, including spaces. Hence, each such expression can be counted, and the entire set of such expressions must also be countable. Since our diagonal function is writable as g(n) = f_n(n), we have shown a nondenumerable subset of writable functions.