Date: Nov 12, 2012 4:05 AM Author: Zaljohar@gmail.com Subject: Cantor's first proof in DETAILS 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 on-line 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

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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_i-1 < b_i < b_i-1

But also i>0 -> a_i-1 < a_i < b_i-1

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_i-1 and < b_i-1, 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_i-1

i>0 -> a_i < a_i+1 < b_i-1

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_i-1, 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