On the Self-Consistency of the Steady-State Cosmology: Reply to David Hawkins

Abstract

In my judgment Professor Hawkins makes persistent error in his relating of the paths to Thus, he writes, referring to counting scheme as illustrated by his Figure 1, that every path which represents particle consists of a finite sequence of digits terminating with 1 for the nth digit, where the nth is the generation in which the given particle is born; and this 1 is followed by an infinite sequence of 0's. Other infinite sequences of the set (0, 1) represent lines of descent but do not correspond to particles. There are two respects in which the quoted statement is wrong. First, it is not true that path which represents particle necessarily consists of finite sequence of digits preceding the sequence of 0's which indicate the existence of the particle through the represented by those 0's (we recall that Prof. Hawkins uses 0 to indicate continuation of particle from the preceding generation, 1 to indicate birth of particle in the generation to which the 1 belongs). For, if new particles are being produced throughout infinite time, we may of course have particle appearing an infinite number of generations after an initial particle, and hence there may be an infinite sequence of digits preceding the 1 that is followed (as Hawkins correctly asserts) by 0's only. This sequence of 0's may be finite or infinite, depending on whether the particle appeared at finite or infinite time in the past. Second, every infinite sequence of the set (0, 1) represents particle. This must be so, since in any given generation every path with terminal 0 represents particle also existent in the immediately preceding generation, and every path with terminal 1 represents particle created in that generation. If the penultimate as well as the terminal digit is 0 the particle was also existent in the penultimate generation; if the penultimate digit is 1 and the ultimate digit is 0 the particle of interest was created in the penultimate generation. Finally, if both the terminal and penultimate digits are 1, the particle created in the penultimate generation (as indicated by the 1) does exist in the terminal generation by virtue of the path with terminal digit 0 that must be constructed for every particle that exists in the penultimate generation. And so on; there is no way in which an n-digit lineage path can fail to represent particle existing in the nth generation. Hence, the isomorphism between lineage path labels and infinite binary sequences immediately implies (Cantor's diagonal proof) the existence of nondenumerable infinity of The set F + which Hawkins constructs is not fair representation of the set of