Cantor's First Proof
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From: RussellE on 26 Jun 2010 18:56 Why are there still any questions about infinity? We have been able to prove infinity is inconsistent for millennia: http://en.wikipedia.org/wiki/Zeno's_paradoxes Even Aristotle couldn't refute Zeno's arguments. A more modern proof of this inconsistency: http://arxiv.org/PS_cache/arxiv/pdf/1004/1004.4461v1.pdf The author describes several proofs of inconsistency I have been posting to this newsgroup for years. NESTED-SET INCONSISTENCY Let A be a denumerable set. Let A_0 = A. Define a sequence of sets: A_i+1 = A_i - {a_i} Is the empty set a member of this sequence? Assume it is. Then A_z = {} and z-1 is the largest natural number. Assume the empty set is not a member of this sequence. Then we can prove by induction there exists an element, a_z, that is a member of every A_i. This proves A is not denumerable. We can even use Cantor's first proof of the uncountability of the reals to prove the rational numbers are uncountable. http://en.wikipedia.org/wiki/Cantor's_first_uncountability_proof Since Cantor proves the algebraic reals are countable in the same article, his paper proves infinity is inconsistent. He proves the algebraic reals are both countable and uncountable. Assume we have an interval of rational numbers, (a,b), and an enumeration of the rationals in this interval. I like to use the factorial base for my enumeration. Every rational number has a unique finite representation in base factorial. http://en.wikipedia.org/wiki/Factorial_number_system Let a=0 and b=1. My enumeration of the rationals for interval (0,1) is to "count" in the factorial base. Positions in factorial base: 1/2!, 1/3!, 1/4!, 1/5!, ... Factorial = Decimal ..1 = 1/2! = 1/2 ..01 = 1/3! = 1/6 ..11 = 1/2! + 1/3! = 2/3 ..02 = 2/3! = 1/3 ..12 = 1/2! + 2/3! = 5/6 ..001 = 1/4! = 1/24 .... The gives us the enumeration: 1/2, 1/6, 2/3, 1/3, 5/6, 1/24, 13/24, 5/24, 17/24, 3/8, 7/8, 1/12, 7/12, 1/4, 3/4, 5/12, 11/12, 1/8, 5/8, 7/24, ... Take the first two terms of the enumeration. Let the smallest be a0 and the largest be b0. a0 = 1/6, b0 = 1/2 Create the interval (a0,b0) = (1/6, 1/2). Create a sequence of nested intervals by finding the next two nonequal terms in the sequence inside the previous interval. (a0,b0) = (1/6, 1/2) (a1,b1) = (5/24, 1/3) (a2,b2) = (1/4, 7/24) .... Assume this sequence is finite. Let (az,bz) be the last interval. By construction, az =/= bz. Since the rationals are dense, there are an infinite number of rationals in the interval (az,bz) not in the sequence. Assume the sequence is infinite. Since the rationals are dense, ai =/= bi for all i, and the intervals are nested, there exists rational, z, such that z is inside every interval. This proves z is not in the sequence of rationals. Russell - 2 many 2 count
From: Virgil on 26 Jun 2010 23:24 In article <4c7b682a-f17b-4f97-9f7a-cb0845bf407c(a)b33g2000prd.googlegroups.com>, RussellE <reasterly(a)gmail.com> wrote: > Why are there still any questions about infinity? > We have been able to prove infinity is inconsistent for millennia: > http://en.wikipedia.org/wiki/Zeno's_paradoxes > > Even Aristotle couldn't refute Zeno's arguments. > > A more modern proof of this inconsistency: > http://arxiv.org/PS_cache/arxiv/pdf/1004/1004.4461v1.pdf > > The author describes several proofs of inconsistency > I have been posting to this newsgroup for years. > > NESTED-SET INCONSISTENCY > > Let A be a denumerable set. > Let A_0 = A. > Define a sequence of sets: A_i+1 = A_i - {a_i} > Is the empty set a member of this sequence? > > Assume it is. Then A_z = {} and z-1 is the > largest natural number. > > Assume the empty set is not a member > of this sequence. Then we can prove by > induction there exists an element, a_z, > that is a member of every A_i. No you can't. What you can prove is that for each A_i there is some a_z in it, but yor cannot prove it is the same a_z in all of them, as your statement claims. > This proves A is not denumerable. No it doesn't. > > We can even use Cantor's first proof > of the uncountability of the reals to prove > the rational numbers are uncountable. That would require that every sequence of rationals that is bounded above has a rational exact upper bound, which is not the case. > http://en.wikipedia.org/wiki/Cantor's_first_uncountability_proof > > Since Cantor proves the algebraic reals > are countable in the same article, his > paper proves infinity is inconsistent. Nonense again! > He proves the algebraic reals are both > countable and uncountable. Nonense again! The rest snipped as irrelevant.
From: RussellE on 27 Jun 2010 05:48
On Jun 26, 8:24 pm, Virgil <Vir...(a)home.esc> wrote: > In article > <4c7b682a-f17b-4f97-9f7a-cb0845bf4...(a)b33g2000prd.googlegroups.com>, > > Assume the empty set is not a member > > of this sequence. Then we can prove by > > induction there exists an element, a_z, > > that is a member of every A_i. > > No you can't. > > What you can prove is that for each A_i there is some a_z in it, but yor > cannot prove it is the same a_z in all of them, as your statement claims. I may not be able to name the element, but I can prove such an element exists. > > This proves A is not denumerable. > > No it doesn't. Yes it does.This element can't be associated with a natural number, proving A is not denumerable. This is a contradiction since we assumed A is denumerable. > > We can even use Cantor's first proof > > of the uncountability of the reals to prove > > the rational numbers are uncountable. > > That would require that every sequence of rationals that is bounded > above has a rational exact upper bound, which is not the case. Cantor proves there is a non-empty interval not in the sequence of rational numbers. > >http://en.wikipedia.org/wiki/Cantor's_first_uncountability_proof > > > Since Cantor proves the algebraic reals > > are countable in the same article, his > > paper proves infinity is inconsistent. > > Nonense again! > > > He proves the algebraic reals are both > > countable and uncountable. > > Nonense again! > > The rest snipped as irrelevant Ignoring a proof doesn't make it go away. Russell - 2 many 2 count |