Re: Logarithm of transfinite numbers

Randy Poe said:

Tony Orlow wrote:
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.

Hint: My unwillingness to agree with your gobbledegook
version of "logic" has nothing to do with denseness.

I have established a property of the
elements of the set, all of them,

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

that none of them marks any point where the
set acheives infinity.

Right. No set of the form {1,...,n} is the set N of
interest. The "collection of things" is the sets of the
form {1,...,n} and we all agree that N is none of these,
i.e., that

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

Right? Do we agree that N is not one of the set of
the form {1,...,n}?


The set of consecutive naturals starting from 1 and having size N is of the
form {1, 2, 3 ... N-1, N}, so if the size of the set of finite n in N is aleph_
0, then that set is of the form {1, 2 ... aleph_0-1, aleph_0). Your theory
doesn't allow this, but that's the way it is.



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.

Yes. That's #1. Every member of the collection has that
property. But remember #2: N is not in the collection.

All of your arguments go around and around this kind
of logic. Over and over you discuss the properties
of the collection, over and over you agree N is not
in the collection, and then over and over you do
this.

The point is that there is no finite n in N for which the set achieves
infinity. This is true for every finite natural, that it has this relation to
the set size. So, this inductively proven property of the element in relation
to the set size has implications for the set size as a whole. As long as the
elements are all finite, so is the set size.


"3. Therefore I conclude my object has property P."

N is not in the collection. It doesn't matter that
every element of n can be used to form a set of
the form {1,...,n}. When you form that collection of
sets, you have a collection which DOES NOT
INCLUDE N, AND WE ALL AGREE ON THAT.

Nevertheless, you think N shares a property with
every member of a collection OF WHICH WE ALL
AGREE IT IS NOT A MEMBER.


Calm down. You're flailing. Pay attention. it's very simple:

If at every point in the set the value of the element at that point is equal to
the count of elements up to that point, and if at every point in the set the
value of the element at that point is finite, then at every point in the set
the number of elements up to that point is finite, so there is NO point in the
set where the number of elements up to that point is infinite.

Explain exactly where this argument fails.


Despite your belief that some axiom or other ought
to get you there, there isn't. N is not a member of
the collection whose properties you insist on
focusing on.

Indeed, it is the limit of {1...n} as n approaches aleph_0, or alpha, or
whatever largest finite you care to envision. There isn't any largest finite,
and so your set's not well defined, but if you insist on talking about
finiteness as part of the definition of a set, then you're going to have these
problems.


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.

Both of those are correct.

Doesn't that seem somewhat contradictory to you?

No, since it's so trivially easy to get one's intuition
around both of them.

It does to me.

Your inability to understand does not mean the concepts
are incomprehensible. It is something lacking in your
own brain.

- Randy


It is the presence of a clear vision for how this stuff fits together that is
lacking in your brain, but not mine.

--
Smiles,

Tony
.

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