Re: Cantor Confusion
- From: "MoeBlee" <jazzmobe@xxxxxxxxxxx>
- Date: 2 Nov 2006 12:12:35 -0800
mueckenh@xxxxxxxxxxxxxxxxx wrote:
MoeBlee schrieb:
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.
a list is an injective sequence
Then that is your definition. Usually, injection is not required. But
for this particular discussion it doesn't matter, since all the
sequences I've used are injections.
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.
Because you did not read what I wrote. I defined it above: "better say
finite sequences or numbers or entries"
No, I read it over a few times. When I say I don't understand
something, you can take me at my word that I mean just that - I read it
a few times, thought about it, and don't understand it. Thus, you can
save yourself the wasted typing of saying false things such as that I
didn't read what you wrote.
I don't know what you mean by entries SURPASSING every finite entries.
What entries surpass which other entries? What does 'surpass' mean? If
you give me ordinary discourse, then I'll have a better chance of
understanding you, just as I defined each of my terms, 'sequence',
'entry', etc. in my own remarks.
Yes, since omega is not a sequence at all, let alone being a finiteEven 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.
Maybe, if you say so. But omega is not the maximum of all finite
sequences.
sequence, let alone being the maximum of all finite sequences.
Therefore the width of the list is less than omega.
My argument does not mention 'width of the list'. If YOU want to refer
to 'width of the list', then YOU need to define it. And that means
first proving that there exists a unique object that meets the
description.
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 sequence is infinite.
You may also prove that the maximum of numbers less than 5 is 5.
Nevertheless it is false.
No, I can't prove that.
The diagonal of a list of sequences with less than 5 terms is less than
5.
The diagonal of a list of sequences with less than omega terms is less
than omega.
This simple truth should convince you that ZFC is not acceptable.
You claim it is a simple truth without proving it. And your claim is
not even compatible with the simple intuitive picture that uses
ellipses. So not only do you not have a mathematical proof of your
claim, you don't have an intuitive explanation, except an argument by
ANALOGY in which you analogize between the finite and infinite, only
assuming, as a form of question begging, that what holds for a finite
sequence must hold for an infinite sequence.
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.
You said the domain is omega. You said "we trivially PROVE the diagonal
sequence is infinite". omega is larger than any natural number.
"Infinite" means "larger than any natural number".
The common definition of 'is infinite' I use is:
x is infinite <-> ~En(n is a natural number & x is equinumerous with n)
which in turn reduces to:
x is infinite <-> ~Enf(n is a natural number & f is a bijection between
x and n)
No mention there of "larger".
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.
Without a maximum.
Yes, if you mean that there is no entry has a greater length (notice,
by the way, that 'greater' here is just the usual 'greater than'
relation among natural numbers; i.e., finite) than all other entires.
Without a sequence of infinite length.
Correct. No entry of S has infinite length.
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.
You have derived a nice contradiction.
No, I haven't. Please state the sentence P in the language of set
theory such that you think I derived P and the negation of P.
The diagonal is an infinite sequence. So the diagonal is longer than
any of the finite sequences. But the diagonal consists of elements of
the finite sequences. So it cannot be longer than the maximum of the
finite sequences.
There is NO "maximum of the finite sequences". If you want to use "the
maximum of the finite sequences" in your argument, then you need to
prove that there exists an object that meets that description.
If this maximum does not exist, you cannot take the
supremum omega for it, because the supremum is not a member of the
sequences and does not supply elements of the diagonal.
I said nothing about taking a supremum.
Please address the proof I gave; not a strawman of my proof.
If your next response is just more strawman and use of descriptions not
proven to properly refer, then I may very well just note that rather
than waste my time yet again explaining your own errors to you.
/
For reference, here is my proof:
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).
The diagonal of S is a denumerable sequence.
Therefore, there exists a denumerable sequence S of finite sequences
such that the diagonal of S is denumerable.
/
MoeBlee
.
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