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Cantor's first proof in DETAILS
Posted:
Nov 12, 2012 4:05 AM
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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
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