Re: Logarithm of transfinite numbers

Tony Orlow wrote:
Randy Poe said:

Tony Orlow wrote:
Randy Poe said:


That's the same as saying all strings of finite length, meaning bit positions
in N, which would be, according to your theory, aleph_0 bit positions, and 2
^aleph_0 strings. That's not an additional constraint at all.

It leaves out infinitely many strings. How is something that leaves
out some of the possible strings not a constraint? You are
not allowed the string with all 1's, nor the string with a 1
in every prime position, nor the string with a 1 in every
multiple of 17, nor ....

Those are all infinitely long strings,

They're infinitely-long strings that have infinitely-many 1's.

so you have not reduced the number of
finitely long strings

I didn't say I'm reducing the number of finitely-long strings.
Where did you get that from?

The argument is this: There are 2^aleph_0 infinitely-long
strings, and we're not using all of those infinitely-long strings.
Therefore we're using less than 2^aleph_0 strings.

There are aleph_0 finite naturals, each of which indexes a bit in a string, so
you have aleph_0 bits, each in a finite position. Any combination of bits in
finite positions yields a finite value, and every finite value is a combination
of bits in finite positions. In other words, the set of finite bit strings,
which is the set of all combinations of 0's and 1's in finite positions, is the
set of representations of finite naturals. Given aleph_0 finite bit positions,
each corresponding to a finite natural, a string of this length has 2^aleph_0
possible values, each finite. But, the set of finite naturals has aleph_0
elements, not 2^aleph_0, and you have a contradiction.

Now, you are going to claim that there are an infinite number of finite bit
positions, but even if that were true, it's irrelevant. Consider a 1 in any
finite bit position. The value of this 1 is the sum of the values of all 1's in
previous bit positions, plus 1. If that position is n, the largest value you
can get in n bits is 2^n-1. So, any string up to any finite bit position, which
is what you have, is a finite value.

There are no infinite bit strings under discussion.

I'm quite interested in this stuff, so I'll bite again. By "infinite
bit strings" do you mean bit strings that never terminate? If so, then
isn't the whole point that we *are* talking about bit strings that
never terminate? Each bit string is a string of "length aleph_0"; i.e.
it never terminates. Amongst those strings there are infinitely many
that have "infinitely many 1's" - in other words, if you took out all
the 1's and wrote them one after another then you would never finish
writing them. And none of those strings can map to a natural number?

It's possible I am viewing this at a more naive level than you. There
is also, of course, the possibility that I am just plain wrong.

It's a constraint that eliminates a lot of infinitely-long
strings, and retains only those that have finitely many 1's.
It's a constraint the eliminates a bunch of strings from
the 2^aleph_0 possibilities.


But the 2^aleph_0 possibilities are the set of FINITE strings. If there are
aleph_0 finite naturals, there are aleph_0 bits in finite positions, and 2
^aleph_0 finite-length strings, each corresponding to a finite natural number.



It's not an additonal constraint. If you are including all finite-length
strings, then you are including bits in every finite poisition, which according
to your theory means aleph_0 bits. And yet, that produces an uncountable set of
strings. Any finite number of bits produces a finite set of strings. So, there
is no number of bits in your theory which can produce a countably infinite
number of strings.


I'm constraining you not to include those strings. That's
a constraint.

It's not an additional constraint when you have already restricted yourself to
finite length strings.


You can't cull an
uncountable number of strings from this uncountable set and leave the countable
set of strings that you evision.

And why, pray tell, can't you?

Because your first suggestion removes at most half of the uncountable set, and
this suggestion doesn't remove any of them.


Take the set of all reals. Is that uncountable?
Obviously
Take the set of all reals except 1. Is that uncountable?
Obviously
Remove the second set from the first, leaving {1}. What says
I can't do that?
Go ahead, but that is not what you suggested regarding the binary naturals. You
have failed to identify the countable subset of the uncountable set of strings
which you require to uniquely identify each of the elements in the countably
infinite set.

I did? What about "those strings that have finitely many 1's"
does not uniquely identify the set?

That says nothing different than "the set of all finite-length strings", of
which there are 2^aleph_0. So, you've identified a subset of these 2^aleph_0
strings, but not a proper one.


- Randy



--
Smiles,

Tony

.

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