Re: Cantor Confusion

WM says...

On 21 Mai, 13:47, stevendaryl3...@xxxxxxxxx (Daryl McCullough) wrote:
WM says...

You claim the (unchanged) diagonal can have more x's than any line?
Than is wrong because the diagonal consists of line elements.

With your rule for the diagonal: replace 3 by 6, then
if we start with the following list:

0.6000
0.06000
0.006000
0.0006000
0.00006000
...

then the diagonal will be the same: 0.333...
There is one "3" for each number in the list.

What do you think why I chose the special list?

Because with a different starting list, it is more
immediately obvious that your claims are false.

In that list the diagonal number cannot have more 3's than at least
one line of the list (well, one more).

The point about mathematics is that you aren't allowed to just
say something because you feel it ought to be true. You have to
show that your claim *follows* from your assumptions.

If the diagonal number is complete in the sense that at every
position indexed by a natural number there is a 3, then there must be
a complete sequence of 3's among the list entries too.

No, that's false. It's *provably* false.

It is also provably right. See matrix M above.

A proof consists of a sequence of statements such
that each statement is either an assumption, or follows
from previous statements from the rules of inference.
There is such a proof of the claim

"For all j, the diagonal is unequal to r_j"

where r_j is the jth real in the list.

You are claiming, what, exactly? That the diagonal *is*
on the list? In other words, you are claiming

"There exists a j such that the diagonal is equal to r_j"

What's the proof of that? Give a sequence of statements,
such that each statement is either an assumption, or follows
from previous statements by the rules of inference.

For your particular
example, what is true is this (letting r_j be the jth number
in the list)

1. For each natural number n > 0, there exists an index j
such that r_j agrees with the diagonal number in the first
n decimal places.

2. For each index j, there exists a natural number n > 0
such that r_j does *not* agree with the diagonal number in
the first n decimal places.

Statement 1. says that the diagonal number can be approximated
arbitrarily well. Statement 2. says that the diagonal number does
not occur on the list. Both are true.

But they are contradictory

They are both *true*. They can't be contradictory.

If you claim they are contradictory, then it's easy enough
to prove such a claim. If A and B are contradictory, that
means that A implies the negation of B. Okay, so you need
to show that 1. above implies the negation of 2.

Statement 1 is

S1: Forall n > 0, exists j
r_j agrees with the diagonal number in the first
n decimal places.

The negation of statement 2 is

NS2: Exists j, forall n > 0, r_j agrees
with the diagonal in the first n decimal places.

So you need to prove that S1 implies NS2. Go ahead. Notice that
S1 starts with "forall", while NS2 starts with "Exists". There
is a difference between "Forall x, Exists y, Phi(x,y)" and
"Exists x, Forall y, Phi(x,y)". For example, let Phi(x,y) be
"y is the mother of x". Then

Forall x, Exists y, y is the mother of x

is true, because every person has a mother. But

Exists y, Forall x, y is the mother of x

is false, because not every person has the *same* mother.


These two statements have the logical forms:

1. forall n, exists j, Phi(r_j, d, n)
2. forall j, exists n, ~Phi(r_j, d, n)

where Phi(r_j, d, n) is the statement "r_j agrees with the diagonal
number d in the first n decimal places".

Otherwise the list is not complete, i.e., there is always a 6 at
some natural index.

The definition of the diagonal number d is the following: (letting
r[n] mean the nth decimal place of the real r)

d[n] = 9 - r_n[n]

The definition of the numbers r_j is this:

r_j[n] = 6, if j=n
= 3, otherwise

It follows that

r_n[n] = 6

Putting those together, we have

d[n] = 9 - 6 = 3

So every digit of the diagonal number is 3.

If there are infinitely many 3's in the diagonal, then there are
infinitely (- 1) many 3's in an r_n.

No. By the definition of the diagonal, d, if there are infinitely
many 3s in d, then that means that there are infinitely many
values of n such that r_n[n] = 6. And there are.

It's provably true. (I'm using d, rather than b, and I'm
using r_n instead a_n).

d[n] = 3
r_n[n] = 6

Clearly d[n] is never equal to r_n[n].

That is only one side

There is no "other side". d is not equal to r_0.
d is not equal to r_1. d is not equal to r_2. For
each natural number n, d is not equal to r_n.

of the medal which does not take into account
that hat each real number r_n is not better determined than by |r_n -
r_n(k)| < epsilon(k), or to express it opposite: An irrational real
number r_n cannot be determined better than by a sequence r_n(k) with
|r_n - r_n(k)| > anti-epsilon(k).

To the extent that I understand what you're trying to say, yes.
Every irrational number can be written as a limit of a sequence
of rational numbers. However, if r is irrational, and r_0, r_1, ...
is a sequence of rationals converging to r, then it follows that:

1. Forall n, r is not equal to r_n.
2. r = Limit as n --> infinity of r_n

Both statements are true. But only the first is relevant to
the question: Does r appear on the list r_0, r_1, ...?

I say in this case because not every sequence has a limit. For example,
the sequence

0.600...
0.330...
0.6660...
0.33300...
0.666600...
0.3333000...

has no limit. But it still has a diagonal, namely
0.363636...

But if this diagonal is to be considered a real number, then we need
the factors 10^-i. The same is true for
lim[i --> oo] (b_i - a_i) * 10^-i

That limit is zero. But so what? To say that the limit
of a sequence is zero is *different* from saying that any
element in the sequence is equal to zero.

For example, the sequence 1, 1/2, 1/3, 1/4, ... has the limit 0.
But no number in the sequence is equal to zero.

1. The number d appears on the list r_j. For this to be true,
it must be the case that Exists j, Forall n, d[n] = r_j[n].
That is *false*. Provably false.

No. It may be provably false for the diagonal sequence, but not for
the diagonal number.

What are you talking about? If you have a sequence of reals
r_0, r_1, r_2, ..., then to say that a number d appears on
the list means, by definition, that d is equal to r_n for some
number n. To say that d = r_n means that every decimal place of
d is equal to the corresponding decimal place of r_n. In other words

d is on the list
<->
exists n, d = r_n
<->
exists n, forall j, d[j] = r_n[j]

And that's false. For every n, d[n] is never equal to r_n[n].
So for every n, d is not equal to r_n. So d is not on the list.

2. The number d is the limit of the numbers r_j. For that to be
true, it must be the case that Forall n > 0, Exists j, Forall m < n,
d[m] = r_j[m]. That's true in your case.

And that is *all* we can consider in a list f n-ary expansions.

No, in the case r_0 = 0.666..., r_1 = 0.3666..., r_2 = 0.63666..., etc.,
and the case d = 0.3333..., we can prove

d is not equal to r_0
d is not equal to r_1
d is not equal to r_2
etc.

Any infinite n-ary expansion of a real number (not ending by zeros) *is
not* defined better than up to an anti-epsilon > 0 because for every
approximation r_n(k) we can find an anti-epsilon > 0 which is less
than the difference |r_n - r_n(k)|.

To know that d is not equal to r_0, we only need to look in the
first decimal place. To know that d is not equal to r_1, we only
need to look in two decimal places. To know that d is not equal to
r_2, we only need to look at three decimal places. For every n,
we only need to look at the first n decimal places to see that
d is not equal to r_n.

What do limits have to do with anything? There are two different
questions:

1. Does the diagonal number d appear on the list? The answer is no.

This answer is given at most for the diagonal sequence, not for the
diagonal number.

I repeat: Does the diagonal number d appear on the list? The
answer is no. There is no number n such that r_n = 0.3333....

Yes, that's true. 1 does not appear on the list

0.9
0.99
0.999
0.9999
0.99999
etc.

and 0.33333 does not appear on the list

0.6666...
0.36666
0.336666
0.3336666
etc.

So r does not appear on the list.

So the problem of replacing 0 by 9 is not existing.

What are you talking about? Of course the number 0.9999...
(all 9s) exists. And it is equal to 1. But it is not on the
list

0.9
0.99
0.999
0.9999
0.99999
etc.

But we have in the limit j --> oo: D_j = 0.

Yes. That's true, but irrelevant. For the diagonal to be
on the list, it must be that D_j = 0 for some finite j.

That is only required for a finite list.

The words "d appears on the list" means the SAME THING as
"there exists a j such that d = r_j", which means the SAME
THING as "there exists a j such that D_j = 0".

It's not good enough that limit of D_j = 0.

Consider Dave L. Renfro's example:

0.49999999...
0.49111111...
0.44911111...
0.44491111...
0.44449111...
. . . . . . .

Replacing 4 by 5 and 9 by 0 gives the diagonal number
0.5000...0999... =/= 0.5 for any finite j. Only in the limit j --> oo
the diagonal number is 0.5000... = 0.499..., the first entry. Here it
is good enough to consider the limit to make the proof fail?

That's right. To make sure that the diagonal number d does not appear
on the list, you need to prove that for each n, d is unequal to r_n.
If r_n ends with all 0s or all 9s, then that means that r_n has two
decimal expansions, r1_n and r2_n. You need to show that d[n] is
unequal to r1_n[n] *and* d[n] is unequal to r2_n[n]. The easiest way
is to make sure that d[n] is never 0 and is never 9. A diagonal
procedure that does this is the following:

replace 9 by 1
replace 8 by 1
replace 7 by 1
replace 6 by 1
replace 5 by 1
replace 4 by 1
replace 3 by 1
replace 2 by 1
replace 1 by 2
replace 0 by 1

--
Daryl McCullough
Ithaca, NY

.

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