Re: Logarithm of transfinite numbers

Randy Poe said:

Tony Orlow wrote:
Randy Poe said:

Tony Orlow wrote:
Randy Poe said:
I think you mean all the finite substrings have a finite
value, but none of those is the string with a 1 in every
odd position.

If all those 1 bits are in finite positions, then each one has a finite value,
and is greater than the sum of the string to the right of it.

Yes, each of those substrings has a finite value.

None of those is the entire string.

TO logic:

1. Here I have a collection of things which all have
property P.

2. Here I have an object which we all agree is
not a member of the collection.

3. Therefore I conclude my object has property P.

How does not being a member of a collection lead you
to conclude that it has a property of things in the
collection? Why do you think 2 implies 3?

- Randy



Dear god, Randy, haven't I been through this with you enough already?

Is it your theory that there exists M such that M
repetitions will change it from wrong to right?

There is no such M.

Why do
you have to paint my logic to be so flakey?

Gee, I dunno, maybe because... it is so flakey.

I'm PRESENTING your logic unaltered. All I did was abstract
it.

If you start at 1 and go through the inductive set incrementally, at each
point, your max value n IS the size of the set,

And none of those sets is N.

and sits in position n. Every
time you add the next value, the max increments, and the set size increments,
in tandem, so there is no point at which one is ever going to surpass the
other, and no point where the set size will become infinite,

and no point at which one of these sets is N.

but there is no
infinite value in the set. It's that simple.

Right. Therefore, since at no time in this process are you
ever dealing with N, it is fallacious to conclude that you
have ever examined a property of N. You have examined
a sequence of proper subsets, which you agree are not
N, will never be N.

You sure are dense on this one, Randy. I have established a property of the
elements of the set, all of them, that none of them marks any point where the
set acheives infinity. This is a property of every element in the set. For each
and every one of them, the set up to that point is finite.

You all agree that no natural has an infinite number of predecessors, but still
insist that they have an infinite number of successors.

Correct.

But, if x has a
successor y an infinite number of iterations beyond it,

False premise. No x has a successor y "an infinite number
of iterations beyond it".

If all successors are a finite number of iterations beyond any given natural,
then it doesn't have an infinite number of successors. You just think it's an
infinite set because it has no discrete end, but the size of the set is never
equal to an infinite quantitative value. That's what I mean when I say
"infinite". Not, at least as big as any finite, but bigger than every finite.


then y has an infinite
number of predecessors between x and itself.

That would certainly true. The first statement, which does not
hold in the naturals, would imply the second statement which
also does not hold in N.

So what? We're discussing N.

Yes, and you say there are an infinite number of successors to any natural, but
no natural is the infinite successor to any other natural. Doesn't that seem
somewhat contradictory to you? It does to me. But then, I consider infinity to
be a numeric concept, and not just some fluffy "I can't see the end" concept.


So, no natural can have an
infinite NUMBER of successors, despite the fact that there is no last one.

Eh? No, no natural is infinitely far away from another. You
didn't draw a conclusion from "an infinite number of successors",
you drew a conclusion from "a particular successor an infinite
distance away".

What don't you get?

Why you think those two phrases are equivalent.

- Randy


Because I consider infinity to be a quantitative concept, a number larger than
any finite number. I consider that if you add a finite a finite number of times
you get a finite, but if you add it an infinite number of times you get an
infinite. When you say you have an infinite sequence, I want to see an infinite
number of steps, but you cannot identify any two points in the set that are any
more than a finite number of steps from each other.

So, when you say you have an infinite "number" of successions, I think of that
as an infinite quantity, such that there are more than any finite number in
there, and the first is infinite predecessor to some infinite indexed elements.
That's what I mean when I say "infinite", not some really big unboundedly large
finite sequence, like your countably "infinite" naturals.

--
Smiles,

Tony
.

Relevant Pages

  • Re: My investigations into Godels Incompleteness Theorem
    ... Can the natural numbers do something that our string cannot? ... --non-Euclidean goemetries do not exist in NAFL. ... thought that NAFL would outlaw infinite mapping? ...
    (sci.logic)
  • Re: My investigations into Godels Incompleteness Theorem
    ... Can the natural numbers do something that our string cannot? ... --non-Euclidean goemetries do not exist in NAFL. ... thought that NAFL would outlaw infinite mapping? ... a finite upper bound for the naturals, ...
    (sci.logic)
  • Re: An uncountable countable set
    ... Indexing the digit number n is equivalent to covering the ... string up to digit number n. ... Even a finite string of an infinite set of finite ... naturals does exist, and by Cantor, the cardinality is aleph-0. ...
    (sci.math)
  • Re: My investigations into Godels Incompleteness Theorem
    ... Can the natural numbers do something that our string cannot? ... --non-Euclidean goemetries do not exist in NAFL. ... So does this infinite string correspond to an infinitely large ...
    (sci.logic)
  • Re: An uncountable countable set
    ... Tony Orlow wrote: ... positions in the string, it can never achieve any infinite value at any ... very useful model for the entire set of naturals. ...
    (sci.math)
Loading