
Re: Uncountability of the Real Numbers Without Decimals
Posted:
Dec 2, 2013 10:30 PM


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.
Regards, Ross Finlayson

