Re: Logarithm of transfinite numbers

Randy Poe said:

Tony Orlow wrote:
Virgil said:
In article <MPG.1e84b7c66271a2f898ab0f@xxxxxxxxxxxxxxxxxxxxxxxxx>,
Tony Orlow <aeo6@xxxxxxxxxxx> wrote:

Randy Poe said:

Tony uses words vaguely and in his own way, so I don't know what
*he* means. The rest of us in this discussion mean that the natural
numbers N form an index to the string: there's a bit in the n-th
position for every n in N.

Are all of those bits, then, in finite positions?

YES!


Can any of them
represent an infinite value

NO!

if the interpretation of the nth bit is
2^n? Can all 1's up to any finite position represent an infinite
value? Nope.

Your countably infinite bit string can only represent
finite values. There is no infinite position in the string.

What is the allegedly finite value that TO assigns to the string with
1's in all infinitely many postions?

There aren't infinitely many positions. There is a finite but unbounded
sequence of bits, which you call countably infinite. That's no matter. The
string you talk about, with 1's in all finite positions, in binary, would be
your largest finite. So, YOU tell me what that is.


Funny, you think he's following you, and yet, he asked me the
question, as far as I can tell, and hasn't objected to what I've
said. He says, "the penny dropped", and you declare victory. Whatever
shakes your tree, Randy. Try responding directly to my statement
above. Where did I go wrong?

That appears to have happened well before TO's appearance in sci.math,
and is apparently uncurable.

You didn't answer my question above. How can you have an infinite value in a
string with only finite positions?

Consider the string with a 1 in all odd positions.

Its value is 2^1 + 2^3 + 2^5 + ....

That is larger than any finite value. I think you agree that
something larger than any finite value is not given the name
"infinite".

sum(n=1->N: 2^(2n-1)) diverges to an infinite value, for sure, if N is
infinite, but is finite for any finite value of N. Do any of your naturals
achieve any infinite values? Is there any element which has an infinite index
in the set? Until you reach infinite n, you do not have an infinite sum, but
only the "initial segments" of the sum, as you like to say. They're all finite.


You can't. Up to any finite position, the
string still has a finite value.

Is this agreeing or not?


But 2^1 + 2^3 + 2^5 + ... is not "up to any finite position". It
goes beyond any finite position.

Except that you don't HAVE any infinite positions in the set. n never reaches
any infinite value. So, the sum never reaches any infinite value either, and
you have no infinite values in the set of all possible natural values for a
countable binary string.


- Randy



--
Smiles,

Tony
.

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