
Re: Cantor's first proof in DETAILS
Posted:
Nov 13, 2012 7:52 PM


On Nov 12, 12:00 pm, Zuhair <zaljo...@gmail.com> wrote: > On Nov 12, 5:09 pm, CharlieBoo <shymath...@gmail.com> wrote: > > > > > > > On Nov 12, 4:05 am, Zuhair <zaljo...@gmail.com> wrote: > > > > Apologies beforehand for this long proof, and for any possible errors, > > > typos, mistakes that most possibly would be there with such a long > > > draft. I'v written this with the intention to give what I think it to > > > be the complete story of Cantor's first proof. So the following is my > > > view of this proof, it came from reading online proofs other than the > > > original one, since I don't have the original article of Cantor. > > > References given below. > > > > If a mistake in this proof is noticed, then please feel free to > > > outline it. > > > > CANTORS FIRST PROOF OF UNCOUNTABILITY OF REALS > > >   > > > > Statement: There is no bijection between the set N of all naturals > > > and the set R of all reals. > > > > Proof: > > > We prove that for every injection (x_n) from N to R, there > > > exist a real J such that J not in the range of (x_n). > > > > Notation: for every x_i, i shall be called the place of x_i in (x_n), > > > while x is the value of x_i. Whenever mentioned in this article > > > symbols < , > , = and =/= are comparisons of the values of entries of > > > sequences mentioned, while the places of those entries shall be > > > compared by "lies before" , "lies after" , is the first entry, is the > > > last entry, in the same place, etc.. > > > > (x_n) is said to have the Intermediate Value property (IVP) iff > > > for every two entries x_i,x_j of (x_n) there exist an entry x_k > > > of (x_n) such that: x_i < x_k < x_j or x_i > x_k > x_j > > > > If (x_n) don't possess IVP, then it is easy to find J. > > > > If (x_n) possess IVP, then we construct sequences (a_n), (b_n) > > > in the following manner: > > > > Let a_0 = x_0 > > > Let b_0 be the first entry in (x_n) such that b_0 > a_0. > > > Let a_i+1 be the first entry in (x_n) such that a_i < a_i+1 < b_i. > > > Let b_i+1 be the first entry in (x_n) such that a_i+1 < b_i+1 < b_i. > > > > We notice that for every i,j: j>i > (a_j > a_i) & (b_j < b_i) > > > i.e. (a_i) is an increasing sequence, and (b_i) is a decreasing > > > sequence. > > > > We also notice that for every i: a_i < b_i > > > since a_0 < b_0, and since by definition > > > for all i. a_i+1 < b_i+1, then by > > > induction: > > > > for all i: a_i < b_i. > > > > From that we have the following result: > > > > Result 1: for all i. for all a_i. Exist a_i+1 (a_i < a_i+1). > > > Result 2: for all i. for all b_i. Exist b_i+1 (b_i+1 < b_i) > > > > Result 3: for all i,j: a_i < b_j > > > Proof: either i=j, or i<j or i>j > > > if i=j then a_i < b_i & b_i = b_j so by identity a_i < b_j > > > if i<j then a_i < a_j & a_j < b_j then by transitivity a_i < b_j > > > if i>j then a_i < b_i & b_i < b_j then by transitivity a_i < b_j > > > > Define (lies before): for all i,k. a_i lies before x_k in (x_n) iff > > > Exist m. a_i = x_m & m < k > > > > Define (lies after): for all i,k. a_i lies after x_k in (x_n) iff > > > Exist m. a_i = x_m & k < m. > > > > Similar definitions applies to b_i. > > > > Result 4: for all i: b_i lies after a_i in (x_n) > > > > Proof: b_0 lies after a_0 by definition. > > > for all i. i>0 > a_i1 < b_i < b_i1 > > > But also i>0 > a_i1 < a_i < b_i1 > > > Now if we suppose that b_i lies before a_i in (x_n) > > > then a_i will no longer be the first item in (x_n) > > > that is > a_i1 and < b_i1, unless b_i lies at > > > the same place of a_i in (x_n). which is impossible > > > since a_i < b_i and (x_n) is an injection. > > > > Result 5: for all i: a_i+1 lies after b_i in (x_n) > > > > Proof: a_1 lies after b_0 in (x_n). > > > i >0 > a_i < b_i < b_i1 > > > i>0 > a_i < a_i+1 < b_i1 > > > Now if a_i+1 lies before b_i in (x_n), then > > > b_i would no longer be the first item in (x_n) > > > that is > a_i and < b_i1, unless a_i+1 and b_i > > > lie at the same place in (x_n), which is impossible > > > since a_i+1 < b_i and (x_n) is an injection. > > > > Result 6: for all i: a_i+1 lies after a_i & b_i+1 lies after b_i > > > Since at each i. a_i+1 lies after b_i which lies after a_i > > > then by transitivity a_i+1 lies after a_i > > > Similarly b_i+1 lies after a_i+1 which lies after b_i. > > > > Informally as i increase each a_i,b_i is coming from a deeper > > > and deeper place in (x_n). > > > > Define (external): for all k. x_k external in (x_n) iff > > > x_k an item of (x_n) & > > > x_k not an item of (a_n) & > > > x_k not an item of (b_n). > > > > Result 7: For every x_k. x_k external in (x_n) > > > > Exist i. (a_i lies before x_k in (x_n) & a_i+1 lies after x_k in > > > (x_n)) > > > > Proof: x_0 (which is a_0) lies before x_k, and if the above > > > doesn't hold then for every a_i lying before x_k in (x_n) > > > a_i+1 would lie also before x_k in (x_n), since the place > > > of each a_i is a natural number and so is k, then this would > > > entail the existence of infinitely many naturals before k > > > which is absurd. > > > > Define : a_i is last of (a_n) lying before x_k in (x_n) iff > > > (a_i lies before x_k in (x_n) & ~ (a_i+1 lies before x_k in (x_n))) > > > > Result 8: For all k. x_k external in (x_n) > > > > for all i. (a_i is last of (a_n) lying before x_k in (x_n) > > > > b_i lies before x_k in (x_n) or b_i lies after x_k in (x_n)) > > > > Proof: properties of natural numbers, and definition of external. > > > > Define (intervene): for all k,i. x_k intervene a_i,b_i in (x_n) iff > > > x_k is external in (x_n) & > > > a_i is last of (a_n) lying before x_k in (x_n) & > > > b_i lies after x_k in (x_n). > > > > Define (passed): for all k,i. x_k passed a_i,b_i in (x_n) iff > > > x_k is external in (x_n) & > > > a_i is last of (a_n) lying before x_k in (x_n) & > > > b_i lies before x_k in (x_n). > > > > Result 9: for all k. x_k is external in (x_n) > > > > Exist i. x_k intervene a_i,b_i in (x_n) or x_k passed a_i,b_i in > > > (x_n). > > > > Proof: Results 7.8 and definitions above. > > > > Lemma 1: for all k,i. x_k intervene a_i+1,b_i+1 in (x_n) > > > > x_k < a_i+1 or x_k > b_i > > > > Proof: if not then x_k would be an item of (x_n) that is> a_i+1 & < b_i and since it lies before b_i+1 in > > > > (x_n) then this violates the definition of b_i+1. > > > > Lemma 2: for all k,i. x_k passed a_i,b_i in (x_n) > > > > x_k < a_i or x_k > b_i > > > > Proof: if not then x_k would be an item of (x_n) > > > that is > a_i & < b_i and since it lies before a_i+1 > > > in (x_n) then this violates the definition of a_i+1. > > > > let L be the least upper bound on (a_n), that is > > > > (for all i. a_i =< L) & for all X. (for all i. a_i =< X) > L =< X. > > > > Theorem 1. for all i. a_i =/= L > > > > Proof: assume there exist t such that a_t = L > > > then a_t+1 > a_t, but from definition of L we > > > must have L >= a_t+1, and since L=a_t > > > thus we'll arrive at a_t >= a_t+1 > a_t > > > which is absurd. > > > > Theorem 2. for all i. L < b_i > > > > Proof: assume there exist r such that b_r =< L > > > then b_r+1 < L and b_r+1 >= a_i for all i > > > thus L is not the "least" upper bound of (a_n). > > > A contradiction. > > > > Theorem 3: for all i,j: a_i < L < b_j > > > > Proof: Definition of L and Theorem 1,2. > > > > Theorem 4. for all i. x_i =/= L > > > > Proof: > > > Suppose that x_k = L, then x_k is external in (x_n) (Th.1,2) > > > > for all i If x_k intervene a_i+1,b_i+1 in (x_n), then > > > x_k < a_i+1 or x_k > b_i , But a_i+1 < L < b_i. > > > A contradiction. > > > > If x_k intervene a_0,b_0, then x_k < a_0 > > > But a_0 < L and L=x_k. > > > A contradiction. > > > > If x_k passed a_i, b_i in (x_n), then > > > x_k < a_i or x_k > b_i, But a_i < L < b_i. > > > A contradiction. > > > > Let J=L > > > > QED > > > > Corollary: > > > For every injection (x_n*) from N* to R, where > > > N* is bijective to N. Then (x_n*) misses a real > > > from its range. > > > > Proof: Let (g(n*)) be a bijection from N* to N. > > > Define (x_n) as {y_n Exist n*: y_n* in (x_n*) & g(n*) = n} > > > so range of (x_n) = range of (x_n*) > > > But domain of (x_n) is N. > > > So (x_n) is an injection from N to R. > > > Thus it misses a real. (above proof), > > > so (x_n*) misses a real too, since it has > > > the same range of (x_n). > > > > QED > > > > References: > > > > [1]http://www.math.jhu.edu/~wright/Cantor_Pick_Phi.pdf > > > > [2]http://www.proofwiki.org/wiki/Real_Numbers_are_Uncountable/ > > > Cantor's_First_Proof > > > > [3]http://en.wikipedia.org/wiki/Cantor's_first_uncountability_proof > > > > Zuhair > > > This is very sloppy. You need to give a high level explanation > > first. Otherwise, people have to waste time going through formalisms > > when the problem is with the logic and can be exposed by examining the > > logic alone. You immediately go into your formalisms. This is like > > trying to figure out someone's computer programs without the specs > > (higher level) or documentation (redundant statements that serve only > > to clarify by expressing something in a different way.) > > > CB > > What the references are for then? you'll get the informal level of > this argument there. > To go explicate every bit and piece informally as well as formally > would be really taxing. > > Zuhair
The first link has no high level description, the second link is empty and the 3rd one does not show a parallel proof at a high level.
I can hardly imagine a person writing a proof without first developing it at a high level then working out the details. Like any good programmer, one can include the high level in parallel with the details. If you want to be a bad programmer, then you have raised the cost of one contributing (analogous to mainintain someone else's software) which is your decision. Good luck. :)
CB

