Re: Cantor and the binary tree

Tony Orlow (aeo6) wrote: 
Matt Gutting said:

Tony Orlow (aeo6) wrote:

David Kastrup said:


Tony Orlow (aeo6) <aeo6@xxxxxxxxxxx> writes:



David Kastrup said:


Tony Orlow (aeo6) <aeo6@xxxxxxxxxxx> writes:



David Kastrup said:


It assumes no such thing.  It assumes that the list is indexed by
natural numbers, and that the digits can be indexed by natural
numbers.


By the full entire set of natural numbers, both vrtically and
horizontally? the it is square.

Last time I looked, a square has four corners, not one.

Yeah, like any grid. 

Any grid has four corners?  Do you even read the lunacies you are
babbling?

Look, i suppose you can imagine an infinite grid with NO corners, but when you are talking about a list of digital numbers, you are talking about some kind of rectangle. That shouldn't be debatable. 

Part of the definition of a square, or of a rectangle, is the existence of
four sides, and therefore four corners. A rectangle has a top and a bottom,
a right and a left. Otherwise, it is not a rectangle. That shouldn't be
debatable.

And when you decalre that the number of digits is the same as the number of naturals, then you have a list aleph_0 wide, by aleph_1 long. 

You see one corner, and continuations to infinity horizontally and
vertically, and can't imagine an infinite square?


A square has four sides and four corners.  A quarterplane has two
sides and one corner.

So, you don't know whether there are aleph_0 or aleph_1 digits? Think of it this way: If there are aleph_0 bits, and you are supposedly listing all the reals, then there are aleph_1 strings, which is much larger. The grid is greatly elongated, and diagonal traversal does not cover all strings.

"Elongated" only applies to entities which have finite length.


Wrong. What is the ration of length vs width for a rectangle that is x wide by 2x long, as x approaches infinity? It's still twice as long as wide. Do the math, and stop playing bad logic games, and declaring nonexistent differences between the finite and infinite. Most of them are nonsense. You can't just declare them different. Use some real math occasionally.


I am using real math. "As x approaches infinity" is a shorthand for "as x
increases without bound". As x increases without bound, x remains finite. You
are thus noting a property of finite numbers, and generalizing the property
to something that is not finite.

Must you look on the finite local level and pretend there is no
structure to this list, after having asserted that there is? Do you
still think the Earth is flat? This list is the limit as n->oo of
any list of all digital numbers of length n. Any such list is
exponentially longer than it is wide in digits. This is simple
stuff.

You are babbling nonsense. Again. 

Nice declaration, but don't talk to yourself. People will think you're nuts.


Otherwise, a diagonal traversal on an elongated rectagular grid will
obviously miss many lines of the grid.


It is nonsensical to talk about "elongated rectangular" for  a quarter
plane.  A rectangle has four corners, not one.

It is a Nx2^N, or Nx10^N, rectangle.

No, it isn't. If N is something you can calculate with, it is finite. 

Not true, sorry.


If you are saying that there are infinite natural numbers with the the same
properties (including calculation) as finite numbers, your statement entails rejection of the Axiom of Choice. Is this what you intend?

Absolutely. Axiom of choice is only necessary because of the paucity of real truth in the system. It's a kludge, a band-aid. 

I am glad you cleared that up. If you are denying the Axiom of Choice, then you
are not doing standard math, and we're working at cross purposes.


That the digits can be indexed by natural numbers is obvious since
they obey the laws:

  For every digit with significance 10^-n, there is a following digit
  with significance 10^-{n+1}.

  Digits with different significances have different significances of
  the following digit.

  There is a digit with significane 10^-1.

  The digit with significance 10^-1 has no preceding digit.

  Any set including the digit with significance 10^-1, and for each of
  its digits also containing the following digit contains all digits of
  the number in question.

And that is in obvious correspondance with the Peano laws for the list
indices.  So it is easy to establish a bijection between list indices
and digits.


yes, but if it is a digital number, then there are S^L of them. If L
is the same as |N|, and S is 2 (binary, you have a grid N wide, but
2^N long. In decimal it's worse, being 10^N long. In any case, the
list is much longer than it is wide, and connot be fully traversed
diagonally.


Nonsense.  It is equally wide as long since the sequence of digit
positions and the sequence of list positions obeys exactly the same
laws.


That is absolutely incorrect. Here is a list of all 3-digit binary numbers
000
001
010
011
100
101
110
111

Are there three of them? No, there are eight. Why? Because 2^3=8. Wake up.
If you allow N bits, do you get N binary numbers? No. You get 2^N.


You are again confusing 3 and the size of the set of naturals.
Again.  Stop your wishful thinking.


No, I am giving a trivial example that shows you're wrong.


I am not aware that it shows him to be wrong, since 3 and infinity cannot
act the same way.

They can in many ways, but that's not what we're discussing. Digital number systems share the property of N=S^L, for any N, S and L, finite or infinite. N depends on S and L.

The list cannot be "fully" traversed either across, diagonally, or
downward anyway, since it has no end in either direction.


The the proof doesn't prove anything at all, does it? What do you
think the logical argument IS in that proof? What exactly does it
demonstrate, in your mind?


It shows that there can be a number (as a limit of a sequence) that
for every finite place in the list will differ by at least a finite
difference from the number at the place.


So what? How does that conflate to an uncountable set of reals?


Um, you can make a complete list of any countable set. (That is,
you can index all the members of the set by using the natural numbers.)


So you ARE assuming an NxN grid, as I said. (N is equivalent to aleph_0)


I'm not assuming a grid; I'm assuming an N-length list of numbers, each of
which is the limit of an infinite series, whose elements are countable.

If you can show that there is no way of indexing *all* the members
of the set using the natural numbers, then the set is not countable
by definition, and therefore it is uncountable.

That is a leap. Just because they are counted differently, doesn't mean they are uncountable. If they are a larger set than the naturals, then that is a valid conclusion, perhaps, but to say they can't be enumerated like the naturals (with the very same representation, no less), is just wrong. In fact, a list of N naturals also doesn't require N digits, but log(N), since the very same symbolic string arithmetic applies. All the proof proves is that the list is longer than it is wide, and so there are different infinities. That's significant, but why stop there? There are an infinite number of infinities between N and R, an infinite number less than N, and an infinite number grreater than R, which can all be placed in realtive order using actual math, once you drop your hocus-pocus approach. 

Again, you can't treat N as a natural number (by, for example, taking log(N))
unless you are denying the Axiom of Choice. This is non-standard math, which is
just fine, but it can't be extended to work in the system most mathematicians
work in most of the time. This doesn't mean that either is wrong, simply that
they are incompatible - conclusions drawn from either system can't be applied
to the other.


"Squares" do not come into play here at all, neither do
"diagonals".  Only some convenient partial visualization shows
something akin to a "diagonal", but it is not relevant to the
proof.


It is absolutely crucial. The conclusion of the proof is that the
numbers can't be listed, because a diagonal traversal can be used
to form a number, the antidiagonal, which not on the line of
traversal, and therefore not on this list.


The proof does not depend on the name "diagonal", nor is there
anything leading from one corner to another corner.  The proof
relies on a 1:1 correspondence/bijection of digits and list places.
Both obey they laws of natural numbers, and so can be put into
correspondence.


That is so far off base I can't even imagine what you're
thinking. it assumes something that is simply not true,

What? 

That the diagonal traversal covers all strings in the list.


It doesn't *assume* this, it *proves* it.

(sigh) Incorrect, again. It assumes it, then proves by contradiction that it is NOT the case that the traversal covers all items in the list, through the use of the antidiagonal, which is not in the list traversed. That original assumption, though, rather than being considered disproved, is clung to, and a different conclusion s drawn, that a list cannot be generated. It's very faulty logic. 

No; the initial assumption is not that the diagonal traversal "covers" all
strings in the list. The initial assumption is that a countable and complete
list of real numbers (not "strings") can exist. This assumption, as you note,
is disproved, and the appropriate conclusion is drawn - that such a list cannot
exist (not that it cannot be "generated"; the proof says nothing about
generating the list). That's perfectly reasonable and standard logic: assume
something is true, show that the assumption leads to a contradiction, deny
the assumption.


and jumps to conclusions about the significance of a diagonal
traversal and inversion of bits, which you think has nothing to do
with the diagonal. Maybe you NEED to smoke something.

Maybe you should stop smoking your stuff. 

Ouch!


This rests on the assumption that the diagonal covers every
digital number on the list, which is clearly not the case, since
the grid is longer than it is wide.


It does not end horizontally, and it does not end vertically.  So
you can walk on in diagonal direction (or any other direction) as
long as you want to without exhausting either dimension.


But, what does the diagonal traversal and the generated antidiagonal
signify to you, if you don't believe the diagonal has exhausted all
the numbers in the set, and yet still has some left? That is the
whole point of the proof. As far as I can see, you don't know WHAT
it's supposed to mean.


Uh, it appears you don't know what it means.  The proof shows that for
every _finite_ place in the list, the constructed number will differ
by at least non-zero finite value from the number at the place in the
list.

Yes, so? Is that the conclusion of the proof, the big significant discovery of Cantor? Maybe you should ask your classmates for their notes, because you obviously weren't paying attention that day. 

Yes, it is "the big significant discovery of Cantor". That's the whole
point of the proof.

The significance of the discovery si simply that there are different infinities, which IS significant. Don't conflate it with uncountability, though. That's an unwarranted leap. 

???


No, it is correct thinking. How many different values can you
represent with three decimal digits? Three? A thousand? Who's
blowing smoke now?


You are.  Because "three" is a fixed finite number, and n can take
on any value.


For any n-length decimal number, there are 10^n possible
values.

oo is not n-length. No payoff.

So, while the rectangles get more and more elongated as n approaches oo, they suddenly become perfectly sqaure at infinity. That's just ridiculous. 

What is this "at" infinity? You can only speak of something being in a given
state "at infinity" if it can actually reach infinity. Which, by definition,
is not possible. Similarly, n does not "approach oo", it merely increases
without bound (maintaining finite values all the time).

What is the limit of N^x/x, for positive N, as x approaches oo? Is it 1? No? Then why do you think the aspect ratio of the infinite grid is 1? It's infinite, "at infinity". 

There's no such thing as "at infinity", and there's no such thing as
"approaching infinity". There's only "arbitrarily large (but finite)", and
"increasing without bound (but remaining finite)" - at least if you're talking
about natural numbers.

I didn't say anything about "aspect ratio", that's your term. As I understand
aspect ratio, it applies only to those rectangular objects with finite length
and height. This is not the case for a complete list of reals.

In any event, as I posted elsewhere, it's possible to prove Cantor's result
without reference to diagonals, aspect ratios, grids, rectangles, or any other
geometric analogies.


In any case, as David has said, neither "rectangle" nor "square" describes
the grid/list we have, as the list doesn't fulfill the requirements of
either geometric figure.

Every complete list of digital numbers forms a rectangle as I have described. You are purposely closing your eyes to this fact.


Rectangles have left and right edges; they have top and bottom edges. Where
is the right edge to your rectangle? Where is the bottom?

Three is an example of n. Take any n>0, and try it out. See whether
the realtionship changes as you have bigger sets. Ask yourself why
this constant equality would suddenly disappear at infinity,


There is no "suddenly".  Infinity is a different ballpark.  You are
still applying your wishful thinking.

Yeah, it's like me saying that, even though 2^n is always bigger than n for any n>=0, that 2^n=n for n=oo. It's simply untrue, stupid, and ridiculous.


There is no natural number n such that n=oo. (Again, assuming the Axiom of
Choice.) Your second equality is empty of content.

I choose to dismiss the axiom of choice. Answer about the limit above stated twice. 

I think you're asking me to evaluate the limit, as n increases without bound, of
2^n/n. That's simple. The limit does not exist, because the sequence {2^n/n}
diverges for increasing n.

Again, if you dismiss the Axiom of Choice, I'm not sure that we can talk
further without a recognition that you are working with non-standard math.
(Standard math includes the Axiom of Choice.)

Matt 


Bull***, as you have just had your nose rubbed in. What kind of
mathematician contends that three digits can only represent htree
values in a digital number?  Sounds like chicken-scratch math to me.


Your problem is still that you are confusing "three" with "cardinality
of the naturals".  They don't obey the same laws.


Pay attention. I used 3 as an example of a number of bits, to show
you concretely that three digits does NOT mean three possible
strings. If that example wasn't concrete enough for you then you're
hopeless.

You are still thinking that 3 is governed by the same laws as infinity.

I am thinking that symbolic systems are governed by the laws of symbolic systems, and showing you that those laws apply for infinite sets as well as finite sets, since the laws governing finiteness apply to the formulas that govern strings. The number of strings of length L is S^L, so if that number of strings is infinite, then either S or L is infinite, since S and L finite implies S^L finite. It's pretty sad that you're so indoctrinated in this hare krishna-style math that you can't think straight any more.


The laws governing finiteness apply to the formulas that govern strings of
finite length. If they applied to strings of infinite length, we would have
no way of distinguishing finite from infinite strings.

The laws governing finiteness are what distinguishes between finite and infinite values for formulas, given the finiteness or infinitude of the terms it contains. For instance, for a^b to be finite, a and b must be finite.

The number of strings of length L (L a natural number), drawn from an alphabet of S characters (S a natural number), is indeed S^L. But if L is infinite,
the expression "S^L" has no meaning as a natural number.

It assumes an infinite value, which represents the sze of the set of strings. Either you need infinite S or infinite L to get infinite S^L, which is my point about the infinite set of naturals requiring either infinite strings, or an infinite number base.

Matt



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