
Re: Uncountability of the Real Numbers Without Decimals
Posted:
Dec 4, 2013 8:08 PM


On Monday, December 2, 2013 7:30:41 PM UTC8, Ross A. Finlayson wrote: > On Monday, December 2, 2013 12:09:39 AM UTC8, Zeit Geist wrote: > > > The following is a Proof of the Uncountability of the Set of real Numbers. Please note that it avoids the use the Representation of the Real Numbers as Infinite Decimals. It uses only the Property of Completeness (that every Set of Real Numbers that is bounded above has a Least Upper Bound which is a Real Number); and some Properties due to the Ordering of the Real Numbers, such as if x is Real Number then there exist Real Numbers a and b such that a < b and x ~e [ a, b ]. > > > > > > > > > > > > The proof proceeds by Contradiction. We assume the Set of Real Numbers is Countable, and thus can be exhausted in a Sequence. > > > > > > > > > > > > Take any such Sequence of Real Numbers, X = { x_n  n e N }. We begin by choosing Real Numbers, a_1 and b_1, such that a_1 < b_1 and x_1 ~e [ a_1, b_1 ]. Next, we choose Real Numbers, a_2 and b_2, such that a_1 < a_2 < b_2 < b_1 and x_2 ~e [ a_2, b_2 ]. We continue by choosing Real Numbers a_k and b_k for every k e N, such that for every k e N, we have > > > > > > a_k1 < a_k < b_k < b_k1 and, of course, x_k ~e [ a_k, b_k ]. > > > > > > > > > > > > Doing so for every Natural Number, we define A = { a_n  n e N }. Now, A is a Set of Real Numbers that is bounded above, since any b_n is an upper bound of A. Hence, a = sup(A) is a Real Number. Since our Sequence, X, exhausts all Real Numbers, a e X and there is a Natural Number, m, such that x_m = a. > > > > > > > > > > > > Now, we have previously defined Real Numbers, a_m and b_m, such that a = x_m ~e [ a_m, b_m ]. However, we know a_m <= a, since a = sup(A) and a e A; and a <= b_m, since any b_n is an upper bound of A. These together give us that we must have x_m e [ a_m, b_m ]. This results in a Contradiction. Hence, we must have that the Set of Real Numbers is Uncountable. > > > > > > > > > > > > qed > > > > > > > > > > > > I find the proof rather straight forward. Question, comments, suggestions and corrections are welcome. > > > > > > > > > > > > ZG > > > > > > > > Hello ZG, > > > > > > Thank you for the effort, I'm always interested in proofs of uncountability to > > show how the equivalency function is with it. Here it was interesting to start > > with the completeness properties, then it was as to nested intervals never > > containing only two sequential points. rather that is nested intervals. > > > > > > 1: X = { x_n = EF(n) : n e N } > > > > x_0 = 0 x_oo = 1 ranges uniformly [0,1] > > > > 2: x_1 ~= 0, so, a_1, b_1 > 0, or, a_1, b_1 < 0 > > > > Then, choosing all the other values, to have x_N which goes to one not in a_N, > > b_N, then a_N, b_N, is outside of R_[0,1] that is the range. > > > > > > The values x_m in the range are [0,1], and A is [0,1]. Then, the numbers > > defined, a_m and b_m, are outside the unit interval, and disjoint with A. > > Then, that a_m < a, the supremum of the values of zero and one, that's either > > zero or one, or, zero through one. No element of A is in [a_0, b_0]. > > > > > > Then the contradiction as noted doesn't apply. > > > > > > Note that where you say "Sequence of Real Numbers (of any of the set of real > > numbers)", I admit, "Sequence of real numbers from [0,1]", or rather, "here > > sequence of real numbers from [0,1], that then any segment of R is equivalent > > to R". Then though that's the simple definition of the function as to R[0,1]. > > A is bounded above. Here it is that A is from [0,1] and [a_m, b_m] is disjoint > > A, and, courtesy the least _upper_ bound property, in [1, oo) instead of (oo, > > 0]. > > > > > > Then, A and X are where, a_n could be x_n+1. Then X = A (and b_oo > 1). Then, > > to be greater than x_oo = 1, a_oo would have that a_n might be less than one > > but a_oo or Sup(A) >= 1. > > > > > > Then otherwise X int A = {}, where a_m <=/= a, rather, a <= a_k < b_k. Then, > > the contradiction does not follow. > >
Then, you might consider constructing [a_k, b_k] in [0,1] also. Here, x_0 = 0, so a_k > 0. For a_k going to 1, but b_k also going to zero from one, then a_oo = b_oo < 1, while the range of the function goes through one. How does f go through that b_oo never ranges more than a zero distance under 1? Then it should be a matter of construction that while satisfying a < a+ < b+ < b , that b++ infinity or b_oo (b sub infinity) < 1 but a_oo  1b_oo ~= 1, where a_oo goes to b_oo. Then A is still [0,1], and the conditions hold as x_k < a_k, that b_0 is in the range, but basically only at the end of the range, that it is defined in asymptotics on a, which is a convergent sequence from zero to one.
Then what would be the natural reflective properties that (as the skies above move a second a century) that would concisely define functions of sequences, where this is so? As to existence, there's trivial existence of convergent sequences as for each, then the convergent sequences that have b increasing but not having a nonzero limit. Let's see, b can not be increasing, have a limit, start at zero, and not have a nonzero limit. Here then there's recourse to Bourbaki or the increasing and strictly increasing, yet, perhaps that is part of it's feature, for a simpler formalist notion that, even infinitely summed, the quantity under is less than a finite sum. As infinitesimals it is for example the fluxion in the totally ordered, or as the hyperreals of the real field theory, as to value in sequence. Here then these are as to rates of slow convergence of a, though with that they rapidly converge to one. a is slow to b, but a is fast to x, here that x is always technically behind any standard convergence (that x converges as the sweep).
That would then construct for example with convergent sequences between zero and one, then as to convergent real sequences or real numbers > 1.
Here it is that a converges so much faster than b, that ab would appear to converge here that that is the scale of the usual reductions of the error term. This is though just a general plan, basically that a converges to one (and no less) while b, converging to a value different than one, converges to a value not equal to one, that in the related framework when ran f = a_oo, that a_oo = 1 (and b is decreasing).
Then, it is similar to the antidiagonal case for base b > 3, where it would seem ran f is [0,1/b] or to less than one. Then it is to that, then so on and recursively. Here the point is that a and b, each with the general tendency to be increasing if not strictly increasing, has then that the more a is a sequence with a_0 >= 0 and a_k >> 1b_k (at least, as between zero and one it can be), then transitively it is to that 1f or here REF, that it starts at one, would have that it's gone to zero. Otherwise it is like the antidiagonal case and binary. It is that it would be the same to say a_oo is zero, and any other, then that here, b_oo < 1. The function is basically scaled under those bounds, capped at one.
This is the bit less formal than usual, for the properties of the real numbers, of (Lebesgue or nilpotent or nilsquare) analysis and of the field.
Regards, Ross Finlayson

