Re: Cantor Confusion


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

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"

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.

Maybe, if you say so. But omega is not the maximum of all finite
sequences. Therefore the width of the list is less than 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.

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.

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".

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. Without a sequence of 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.

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. 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.

Regards, WM

.

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