Re: Cantor Confusion

mueckenh@xxxxxxxxxxxxxxxxx wrote:
MoeBlee schrieb:


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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.

The digits of the numbers written down in your list (above) are not
bounded by a finite number.

S is a denumerable sequence of finite sequences. But, yes, of course,
every natural number is a member of the range of exactly one of the
finite sequences that are in the range of S. So, yes, of course, the
union of the ranges of the finite sequences is unbounded.

So lets call that union, 'E' (which is, as I understand, what you would
call the set of entries).

Maybe, if you say so. But omega is not the maximum of all finite
sequences.
Yes, since omega is not a sequence at all, let alone being a finite
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 width of the list is the number of digits of the number with most
digits. As such a number does not exist, he width is the supremum,
namely omega.

First, my proof says nothing about digits or numbers of digits. My
proof doesn't even require that we have proven the basis representation
theorem.

Then, to make your definition precise, this is how I take it:

The width of S is the number of digits in the element of E that has
more digits than any other element of E, if such an element of E
exists, and otherwise the width of S is the supremum (by ordinal
ordering) of the set of numbers of digits of elements of E.

And, so, yes, under that definition, the width of S is omega.

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.

I did not introduce a number omega which is larger than every natural
number.
But IF such a number is introduced, THEN we should be allowed to use
the inequality omega > n for every natural number n, i.e. for the n
digits of the n-th list entry.

If '>' stands for the standard ordinal ordering, then, yes, of course,
An(n in omega -> n < omega).

But that doesn't entail that your argument by analogy has any merit
whatsoever.

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".

So you do not man that omega > n holds fo every n e N? Then we have no
dissent.

No, of course, if '>' stands for the standard ordinal ordering, then
omega > n holds for every n in omega. But all I said was that my proof
does not mention that. My proof does not require mentioning that. Just
because my proof does not mention it does not mean that my proof
contradicts it.

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.

But the list has a diagonal which, if mapped on a line, has infinite
length.

First, my proof does not require mentioning mapping to lines (or "on a
line"). Second, the diagonal is a sequence, so I don't know what you
mean by mapping a sequence "on a line" (and I take it that you mean the
real line or some segment of the real line), unless you mean an
injection from the RANGE of the sequence into a line. But that is
trivial to do, so I suspect that is not what you mean either. So maybe
you have in mind some particular order preserving mapping. Or, perhaps
what you're driving at is the fact that the diagonal is infinite. Yes,
indeed, the diagonal is denumerable, which is what I proved, and that
does not contradict anything about the real line or any mappings from
any set to the set of real numbers or any order preserving mappings.

Look, I claimed that the diagonal is infinite. As I understand, you
disputed that. So I proved that the diagonal is countably infinite. Now
what is it that you are disputing? That the diagonal is infinite? That
the diagonal is countable? I mean, you seem to be AGREEING with me that
the diagonal is infinite? So what are you trying to argue now? That the
infinitude or the countability of the diagonal shows a contradiction in
set theory? If so, then please state the sentence P in the language of
set theory such that you believe you have proofs in set theory of P and
of ~P.

If my proof is incorrect as a proof in Z, then either I've not used
first order reasoning correctly in some step of my proof or I've used
some premise that is not derivable with first order reasoning from the
axioms of Z. But you don't point out anything like that. Instead, you
raise considerations that are not in my proof. So I can only surmise
that you think these considerations somehow contradict my proof, thus,
since you have not refuted my proof, but have only shown what you think
are contradictions between my proof and other considerations, then, if
these considerations are part of Z, then Z is inconsistent (if the
considerations are not part of Z, then, as I've said, of course I make
no warranty that my proof is consistent with your personal, informal
notions). So, if that is the case, again, what proof do you think you
have, in set theory, of a sentence P & ~P?. Moreover, IF set theory is
inconsistent, then every sentence in the language of set theory is a
theorem of set theory, so it would make no sense for you to claim of
any sentence in set theory, including that the diagonal of S is
denumerable, that it is not a theorem of set theory.

So, please say which step in my proof you believe is not justified by
first order logic applied to the Z axioms, or please state what proof
you think you have, in set theory, of a sentence P & ~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.

Omega is used in se theory. It is the supremum of the squence of

I mentioned nothing about supremum. And the fact that omega is the
supremum (by the standard ordinal ordering) of any particular set does
not contradict anything in 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.

Now map it on a line. It is longer than any line entry.

The real line? What is your definition of 'length' (since you use
'longer than') of entries in the real line? And 'entry' on the real
line is something yet again we need to define. But I suppose here that
you just mean that the entries on the real line are the real numbers?
So what is the 'length' you have in mind for real numbers? The length
of certain sequences that represent those real numbers? But that
wouldn't seem to go with the idea of mapping the diagonal "on a line"
and comparing lengths of entries. So, you'll have to define your terms.

But all line
entries which can exist are already there. Hence the mapping of the
diagonal cannot exist.

Here, I'll repeat what I wrote earlier in this post, since, though
somewhat lengthy, these are remarks that do bear repeating:

First, my proof does not require mentioning mapping to lines (or "on a
line"). Second, the diagonal is a sequence, so I don't know what you
mean by mapping a sequence "on a line" (and I take it that you mean the
real line or some segment of the real line), unless you mean an
injection from the RANGE of the sequence into a line. But that is
trivial to do, so I suspect that is not what you mean either. So maybe
you have in mind some particular order preserving mapping. Or, perhaps
what you're driving at is the fact that the diagonal is infinite. Yes,
indeed, the diagonal is denumerable, which is what I proved, and that
does not contradict anything about the real line or any mappings from
any set to the set of real numbers or any order preserving mappings.

Look, I claimed that the diagonal is infinite. As I understand, you
disputed that. So I proved that the diagonal is countably infinite. Now
what is it that you are disputing? That the diagonal is infinite? That
the diagonal is countable? I mean, you seem to be AGREEING with me that
the diagonal is infinite? So what are you trying to argue now? That the
infinitude or the countability of the diagonal shows a contradiction in
set theory? If so, then please state the sentence P in the language of
set theory such that you believe you have proofs in set theory of P and
of ~P.

If my proof is incorrect as a proof in Z, then either I've not used
first order reasoning correctly in some step of my proof or I've used
some premise that is not derivable with first order reasoning from the
axioms of Z. But you don't point out anything like that. Instead, you
raise considerations that are not in my proof. So I can only surmise
that you think these considerations somehow contradict my proof, thus,
since you have not refuted my proof, but have only shown what you think
are contradictions between my proof and other considerations, then, if
these considerations are part of Z, then Z is inconsistent (if the
considerations are not part of Z, then, as I've said, of course I make
no warranty that my proof is consistent with your personal, informal
notions). So, if that is the case, again, what proof do you think you
have, in set theory, of a sentence P & ~P?. Moreover, IF set theory is
inconsistent, then every sentence in the language of set theory is a
theorem of set theory, so it would make no sense for you to claim of
any sentence in set theory, including that the diagonal of S is
denumerable, that it is not a theorem of set theory.

So, please say which step in my proof you believe is not justified by
first order logic applied to the Z axioms, or please state what proof
you think you have, in set theory, of a sentence P & ~P.

MoeBlee

.

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