Re: Cantor Confusion
- From: "MoeBlee" <jazzmobe@xxxxxxxxxxx>
- Date: 1 Nov 2006 10:31:27 -0800
mueckenh@xxxxxxxxxxxxxxxxx wrote:
MoeBlee schrieb:
mueckenh@xxxxxxxxxxxxxxxxx wrote:
All entries of the list have a finite number of letters. An infinite
sequence is larger than any finite sequence. The diagonal of a list
cannot have more letters than the lines.
According to your logic the list can have infinitely many lines. But
even if that was correct it would not facilitate an infinte diagonal.
The number of diagonal elements is the minimum of columns and lines.
0
1 2
3 4 5
6 7 8 9
...............
.................
...................
infinitely downward for an infinite list of finite lists.
The diagonal is 0 2 5 9 14 ... infinitely across.
All entries in the infinite list are finite lists.
better say finite sequences or numbers or entries
'sequence' and 'list' are synonymous here.
I'm pefectly happy to use just 'sequence'. Doing so does not at all
harm my argument.
The infinite list is
longer than any finite list.
The entries surpass every finite entry. Nevertheless you call all of
them finite.
I don't know what you're trying to say. Even using just the word
'sequence', my point is correct.
We have an infinite sequence S of finite sequences. Being an infinite
sequence, the length of S is longer than the length of any finite
sequence.
The diagonal of the list is infinite.
That is your assertion. But obviously the diagonal elements are
simultaneously elements of the entries.
No, we trivially PROVE the diagonal sequnece is infinite.
Also, since you have pointed to more specific language - 'sequence'
rather than 'list' - and you are now using the word 'element', I'll
state my terms here with reasonable precision short of putting this in
the formal language:
A sequence is a function such that the domain of the function is an
ordinal. A finite sequence has a natural number as its domain. A
denumerable (countably infinite) sequence has omega as its domain. An
uncountable sequence has an uncountable ordinal as its domain.
The entries in a sequence are members of the range of the sequence.
Each entry is indexed by a member of the domain. The elements of the
sequence are ordered pairs of the form <x y> where x is a member of the
domain (which is an ordinal) and y is a member of the range of the
sequence (so the y's are the entries).
dom(S) = the domain of S.
length(S) = dom(S).
In my example S is the denumerable sequence recursively defined as
follows:
S(0) = {<0 0>}
S(n+1) = the unique finite sequence f such that length(f) =
length(S(n))+1 and such that f(0)= S(n)(max(dom(S(n)))) + 1
So S is a denumerable sequence such that each entry of S is a finite
sequence.
The diagonal of S = the unique denumerable sequence D such that, for
all n in omega, D(n) = S(n)(n).
D is a denumerable sequence.
The first four members of S are:
{<0 0>}
{<0 1> <1 2>}
{<0 3> <1 4> <2 5>}
{<0 6> <1 7> <2 8> <3 9>}
For visual simplicity, we strip the parentheses, angle brackets, and
indexes, leaving just the entries of each member (this is informal,
just for visualization, and no part of my argument depends on this):
0
1 2
3 4 5
6 7 8 9
To represent the denumerable sequence, we add ellipses (this is
informal, just for visualization, and no part of my argument depends on
this):
0
1 2
3 4 5
6 7 8 9
..............
.................
....................
The first four members of the diagonal are (this is informal, just for
visualization, and no part of my argument depends on this):
<0 0>
<1 2>
<2 5>
<3 9>
For visual simplicity, we strip the angle brackets, and indexes,
leaving just the entries of each member (this is informal, just for
visualization, and no part of my argument depends on this):
0
2
5
9
To represent the denumerable sequence, we add ellipses (this is
informal, just for visualization, and no part of my argument depends on
this):
0 2 5 9 ...
And that be formalized easily in set theory.
And I did that: diagonal of S = the unique denumerable sequence D such
that, for all n in omega, D(n) = S(n)(n).
That may be, therefore it is no wonder that set theory yields
selfcontradictions.
You just claimed that set theory "yields selfcontradictions". I suppose
by "yield" you mean that set theory contradicts your own personal
mathematizing, since the last time you claimed to have shown a
contradiction of set theory, it turned out that you had not shown a
sentence P in the language of a particular set theory such that both P
and ~P are theorems of that theory.
The diagonal elements are simultaneously elements of the entries.
Therefore the diagonal elements cannot sum up to a number which is
larger than any natural number unless also the elements of list entries
sum up to a number which is larger than any natural.
In my example, I said nothing about summing up. And I said nothing
about anything in S being larger than any natural number.
Or put it so: Every segment of the diagonal is covered by an entry.
Which 'entries'?
There is no segment which is not covered.
What is the initial segment {<0 2>}, of the diagonal, covered by? And
what does it matter?
If all entries are finite,
Yes, all entries of S are finite sequences.
then the diagonal cannot be infinite (if infinite omega is larger than
any finite n).
In this post I PROVED that the diagonal of S is infinite. The diagonal
of S is an infinite set. It is an infinite sequence. It has an infinite
domain. It is an infinite set of ordered paris. (And, by the way, it
has an infinite range.) And you've not shown that that contradicts any
theorem of any Z set theory. That it may contradict your own confused
word jumbles is not of concern to me.
MoeBlee
.
- Follow-Ups:
- Re: Cantor Confusion
- From: mueckenh
- Re: Cantor Confusion
- References:
- Re: Cantor Confusion
- From: mueckenh
- Re: Cantor Confusion
- Prev by Date: Re: Cantor Confusion
- Next by Date: Re: Continuous maps and extending into closure
- Previous by thread: Re: Cantor Confusion
- Next by thread: Re: Cantor Confusion
- Index(es):
Relevant Pages
|
