On Tue, May 4, 2010 at 12:44 AM, Brent Meeker <meeke...@dslextreme.com> wrote:
> I notice I didn't respond to your first question in this post. So...

I appreciate the response!

>>>> On 5/3/2010 7:41 PM, Rex Allen wrote:
>>>> So, given eternal recurrence, there are an infinite number of Rexs.
>>>> And an infinite number of not-Rexs.  Let's pair the Rexs off in a
>>>> one-to-one correspondence with the not-Rexs.  Then, let's go down the
>>>> list and put an "A" sticker on the Rexs.  And a "B" sticker on the
>>>> not-Rexs.  Then lets randomly arrange them in an infinitely long row
>>>> and select one at random.  What's the probability of selecting a Rex?
>>>> What's the probability of selecting an "A" sticker?
>>> I suppose your intent is to assign equal measure to each position on
>>> the list so, for any finite subsection of the list the measure of As
>>> and Bs will be equal.
>> If that was my intent, what would your response be?
> The usual way of dealing with "infinity" is to use a measure that works for
> finite cases and converges in the limit as the number is arbitrarily
> increased.  Notice that there is no way to "randomly arrange" the infinite
> sets, except by some process that "randomly" selects elements and places
> them on the list.  So you're really back the generating frequency.

Okay, so this is my point.  So let's say we use a process to randomly
distribute our newly-stickered Rexs and not-Rexs so that they are
randomly arranged according to sticker-type.

Even though we have now rearranged them...these are still the same
Rexs and not-Rexs we started with when they were randomly arranged
according to the 6-sided die.

We haven't changed the relative number of Rexs and not-Rexs, we've
just labeled them with an extra property and then rearranged them
according to that additional property.  They retain their original
properties though.

So, we still have a countable infinity of Rexs, and a countable
infinity of not-Rexs.  Who can be placed into one-to-one

SO...what difference does the "measure" make when deciding, as Carroll
put it, "which infinity wins"?

What does winning mean in this context?  Okay, the not-Rexs have a
greater frequency, but so what.  They still don't outnumber the Rexs.
Frequency seems like an arbitrary definition of "winning".

Cardinality seems like the correct measure to decide who won.  At
least in the case of Rexs and not-Rexs, as well as with Boltzmann
Brains and Normal Brains.

The only way for the not-Rexs to "win" is to not allow the "eternal"
part of "eternal recurrence."  To keep it finite, where they win on

At the very least it seems like a defensible position...?

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