to infinity, and then some...

[nathan]

Because physically possible arrangements of particles are not unique, unlike numbers. From there, I suppose it's just a matter of what is possible. It may be that the existence of our planet in it's current state depends on the way our solar system developed, and so it would need a very similar history in a very similar solar system. The idea is that if it has happened once, it has a non-zero probability, so in an infinite universe you would expect it to happen infinitely often. I suppose this assumes that the local conditions in our neck of the galaxy are infinitely common in the universe (does this imply homogeneity?).

I think it implies quantization of space-time.

IIUC an unbounded infinite universe would be a countably infinite set of repeated regions. If space-time is not quantized, then the positions (eg) of particles are on a continuum. Thus there would be infinitely more configurations than there are regions -- in the same way there are more real numbers than integers.

 

[Michael C]

I'm not sure, but I'll try.

Compare "even numbers" with "even numbers and the numbers 3 and 5". Logically, by including two more numbers, the size should be two more, so they're not the same size. But because the cardinality remains unchanged, they're mathematically equivalent, and can be used interchangeably.

(Probably not the clearest explanation, but the best I can do at the time of night it is over here.)

In as much as we can talk about "size", they are the same size. Here's another argument I saw recently that neatly demonstrates how an infinity can be divided by 2 and remain the same size:

Imagine we make a book containing a list of all the finite "words" (i.e. permutations) that can be created using only the letters "a" and "b". It might start like this:

a
b
aa
ab
ba
bb
aaa
aab
aba
abb
---
Of course the list will be infinitely long: it will be a big book!

For the second edition, we decide to split the book into two volumes. Volume 1 contains all the words that start with "a" and volume 2 contains all the words that start with "b". Clearly each of these volumes is half the size of the original first edition.

For the third edition, to save space, since we know that all the words in volume 1 start with an "a", we omit this initial "a" for every word. Similarly for volume 2 we omit the initial "b" for every word. Now of course volume 1 and volume 2 will contain the same list of words. But this list will also contain all the words of the list in the original one-volume version. We have decomposed the list into two halves, each of which is identical with the original list.

 

[Modified]

...We have decomposed the list into two halves, each of which is identical with the original list.

Except that each contains a zero-length word, which the original did not.

 

[Michael C]

Except that each contains a zero-length word, which the original did not.

Yes, you're right. So in fact each of the two "halves" has one more element than the original list

 

[Myriad]

I think the zero-length word has to be in the original first edition list too, given how that list is defined.

Since it does not start with a or b, it doesn't appear in either volume of the second edition. But it reappears in both volumes of the third edition.

Those are the kind of details that make different editions so sought after by collectors.

I'd settle for a copy of any edition, seeing how (by interpreting a and b as 0 and 1) it must contain every text passage, book, software, audio track, and digital image that ever has been or ever could be created. (The only problem is it needs a really good index... and somewhere in there must be the best possible index, so if you can find that you've got it made...)

(Yeah, yeah, Borges said it first. Don't remind me.)

Respectfully,
Myriad