# Infinite multiverse would contain the ridiculous?

Sorry if this is in the wrong place. Happy for it to be moved.

I've heard it said that a multiverse containing an infinite number of universes, would lead to the ridiculous. The argument is something like this:

1) An infinite number of universes contain an infinite amount of matter.

2) Conceptually, you could keep exploring the multiverse until you encountered something silly like a universe where a chocolate tea pot orbits pluto.

3) Because the multiverse is infinite then this place exists somewhere (like spinning a dice until it lands on 6).

This seems wrong to me. Even if the multiverse contains an infinite set of universes, the set of conceivable universes must also be infinite? My set theory isn't that good, but surely two infinite sets can still overlap rather than contain each other?

I'd like to understand this a little better and I think you guys might be better placed to think it out.

I think I made a mistake, I should have tried the maths forum for sets theory. Or cosmology, for people with thoughts on any multiverse that exists.

Sorry.

EnumaElish
Homework Helper
I have never understood the idea that there must be an infinite number of multiverses. The number of ways in which a finite number of particles can be configured must itself be finite. (If that means I need to assume a granular grid, so be it.) It's a heck lotta configurations, but still a long way from infinity. Starting from here, it would seem that some of the configurations must look very odd indeed. Conversely, I guess, any configuration one might dream up must be included in the great multitude.

mfb
Mentor
The number of ways in which a finite number of particles can be configured must itself be finite.
This is not trivial, but it follows from quantum mechanics.
We don't need it if we don't care about the chocolate tea pot with infinite precision - in every universe it has some non-zero probability to exist (under reasonable assumptions about the initial state and in most interpretations of quantum mechanics).

EnumaElish
hilbert2
Gold Member
Because the multiverse is infinite then this place exists somewhere (like spinning a dice until it lands on 6).

I think there's some misunderstanding of the concept of infinity here... A set can be infinite without containing everything you can imagine. The set of even integers is infinite, but it still lacks one half of all integers.

Saying that the infinite number of universes implies that "there must be a universe where Santa Claus exists" or "where a chocolate tea pot orbits Pluto", is like the argument made by Ancient Greeks to "prove" that there is a finite number of different kinds of atoms: "if there were infinitely many varieties of atoms, there would have to be atoms of all sizes, including those large enough to be visible. Because we have never seen a visible atom, there must be a finite number of atoms".

Tazerfish, DrClaude, Grinkle and 1 other person
Alan Guth has said something like in a multiverse "Anything that can happen will happen and it will happen an infinite number of times". Of course one might argue Sanat Claus can't happen as he breaks some law of physics. But i don't see how atoms spontaneously forming a teapot breaks the law of physics. Its amazingly unlikely but not impossible. So why shouldn't there be an infinite number of such tea pots? Of course there will also be an infinite number of Plutos without tea pots. As I understand it some infinites can be bigger than others. Perhaps thats the answer.

jbriggs444
Homework Helper
Alan Guth has said something like in a multiverse "Anything that can happen will happen and it will happen an infinite number of times". Of course one might argue Sanat Claus can't happen as he breaks some law of physics. But i don't see how atoms spontaneously forming a teapot breaks the law of physics. Its amazingly unlikely but not impossible. So why shouldn't there be an infinite number of such tea pots? Of course there will also be an infinite number of Plutos without tea pots. As I understand it some infinites can be bigger than others. Perhaps thats the answer.
One simple approach is to compute the probability of a teapot sitting on your coffee table compared to the probability of a bunch of disassembled teapot fragments. The fact that on an infinite number of coffee tables, some are likely to randomly have a teapot has no bearing on the probability you determine for the one in front of you.

Cardinality is one way of comparing the sizes of infinite sets. Measure theory is another way of comparing the sizes of infinite sets if you want to go down that road.

Alan Guth has said something like in a multiverse "Anything that can happen will happen and it will happen an infinite number of times". Of course one might argue Sanat Claus can't happen as he breaks some law of physics. But i don't see how atoms spontaneously forming a teapot breaks the law of physics. Its amazingly unlikely but not impossible. So why shouldn't there be an infinite number of such tea pots? Of course there will also be an infinite number of Plutos without tea pots. As I understand it some infinites can be bigger than others. Perhaps thats the answer.

I've said it before and I'll say it again. To deal with probabilities in infinite sets you have to use measure theory. Trying to use the methods of finite probability on an infinite set doesn't work.

For example, the probability that a natural number is prime is zero. The probability is smaller than any positive number, so it must be zero.

mfb
Mentor
For example, the probability that a natural number is prime is zero.
That depends on your chosen distribution, or way to determine that probability as limit.

By the way, the most likely process for the teapot is probably intelligent life sending it there.

EnumaElish
That depends on your chosen distribution, or way to determine that probability as limit.

By the way, the most likely process for the teapot is probably intelligent life sending it there.

I'm talking about the probability that a natural number is prime. So you are dealing with an infinite set and have to use measure theory.

Measure theory is quite simple. An hour or less of study would give you the basic ideas, which are enough for questions like this. You just have to get used to probability zero not meaning "impossible" and probability one not meaning certainty.

I have read that someone long ago proved the Reimann Hypothesis with probability one.

You won't get anywhere using the methods of finite probability on infinite sets.

EnumaElish
Homework Helper
If we think of history as a long series of bifurcations then the resulting multiverse is finite.

mfb
Mentor
I'm talking about the probability that a natural number is prime.
That is not well-defined on its own. There are ways to define such a probability that make sense, but you have to be careful with those definitions because the result can depend on them.

{2,3,1,5,7,4,11,13,6,17,19,8,23,29,...}
Pick a number at random. Chances are good you'll hit a prime, and this set can be continued to infinity and contains all natural numbers.

EnumaElish and jbriggs444
jbriggs444
Homework Helper
If we think of history as a long series of bifurcations then the resulting multiverse is finite.
Finite if we look at the set of universes now (set of paths of some predetermined length), countably infinite if we look at the set of all possible future universes (set of paths of all finite lengths), uncountably infinite if we look at all all possible futures (set of all possible unending paths).

EnumaElish
Homework Helper
That is not well-defined on its own. There are ways to define such a probability that make sense, but you have to be careful with those definitions because the result can depend on them.

{2,3,1,5,7,4,11,13,6,17,19,8,23,29,...}
Pick a number at random. Chances are good you'll hit a prime, and this set can be continued to infinity and contains all natural numbers.
How does this square with the primes within the reals having a measure of zero?

But what if I approach it sequentially? I know 2 is prime, and a version of the Bertrand-Chebychev theorem tells me that the next prime has to be less than twice the previous prime, i.e. 4. There is one natural number which satisfies this, and it is 3. So 3 is prime with a 100% probability. Between 3 and 6 there are two naturals, so each is a prime with at least a 50% probability. And so on. The probability of each natural number gradually converges to zero as the primes become larger and the interval (p, 2p) expands toward infinity.

jbriggs444
Homework Helper
How does this square with the primes within the reals having a measure of zero?
The above is a demonstration that the primes within the naturals have an asymptotic density that depends on the ordering that you choose for the naturals. It has little to do with the fact that a countable set will always be of measure zero.
But what if I approach it sequentially? I know 2 is prime, and a version of the Bertrand-Chebychev theorem tells me that the next prime has to be less than twice the previous prime, i.e. 4. There is one natural number which satisfies this, and it is 3. So 3 is prime with a 100% probability. Between 3 and 6 there are two naturals,...
Between 3 and 6 there are as many or as few naturals as you please, depending on the order you choose. In the case of the order suggested by mfb, there are six natural numbers strictly between 3 and 6.

If you use Lebesgue Measure, it is possible to do the same trick on the reals. For instance, consider Cantor's ternary set. It has measure zero. But it is uncountable. For any positive measure, there is a re-ordering of the reals in which Cantor's ternary set has that measure. There is even a re-ordering in which its complement has measure zero. There are also orderings where it is not measurable at all.

For proof, consider a bijection between the Cantor set and the set of binary sequences and a further bijection between the set of binary sequences and an arbitrary interval within the reals.

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EnumaElish and mfb
Measure theory is quite simple. An hour or less of study would give you the basic ideas, which are enough for questions like this. You just have to get used to probability zero not meaning "impossible" and probability one not meaning certainty.

Bearing in mind I am a novice, does this mean that an infinite multiverse could mathematically have a probability value 1 of containing a silly universe and still not actually contain such a universe?

The above is a demonstration that the primes within the naturals have an asymptotic density that depends on the ordering that you choose for the naturals. It has little to do with the fact that a countable set will always be of measure zero.

Again I am a novice. Could an infinite multiverse contain an 'asymptotic density' of silly universes and what would that actually mean?

Bearing in mind I am a novice, does this mean that an infinite multiverse could mathematically have a probability value 1 of containing a silly universe and still not actually contain such a universe?

Good question!

We are now talking about the measure of sets. There is the set of all possible events. What are the measures of subsets? (As we will see, the frequentist approach doesn't always work sensibly infinite subsets. So we don't have much choice other than to identify the measures of subsets with probabilities.)

Is the set of silly universes empty? If so, then there aren't any. If not, then there are silly universes. But even if the set of silly universes is not empty, then that does not mean that the measure is greater than zero. The measure might be zero. In that case, you could in theory look at an infinite number of universes and not find a silly one. But that tells you nothing about whether silly universes exist. It just tells you that the measure of the set of silly universes is zero with probability one. Not that you could perform such an experiment anyway.

If you did find a finite number of silly universes in your infinite subset, neither would that prove that the measure of silly universes was not zero. Though you would have proved that silly universes exist.

As you can see, the frequentist approach does not work with infinite sets.

If you can prove that the probability of any given silly universe is greater than zero, then the probability that a silly universe exists is one. That doesn't prove that one exists, but it's close enough for all practical purposes.

But it is possible that the probability of any given universe is zero, including ours. The fact that our universe exists does not at all indicate that the measure of the set of all universes identical to it is greater than zero. This observation tells us nothing about the measure of such a subset.

This stuff you read is based on the assumption that the number of distinct universes is finite, and that each distinct universe has probability greater than zero. I do not believe these assumptions. I'm not saying they aren't true. They could be. I'm saying I don't believe it. Since this is science, it is their job to convince me.

If the number of distinct universes is finite then there must be subsets of identical universes with measures greater than zero, and these measures sum to one. But our universe might yet be in a subset with measure zero, and there could be a great many such subsets.

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Woah! I need a stiff coffee.

1oldman2 and EnumaElish
EnumaElish
Homework Helper
Finite if we look at the set of universes now (set of paths of some predetermined length), countably infinite if we look at the set of all possible future universes (set of paths of all finite lengths), uncountably infinite if we look at all all possible futures (set of all possible unending paths).
For the last point, don't you need infinite time (or infinite speed)?

If the bifurcation analogy is accepted, the limiting set of universes might be thought of as a Cantor set, to the extent that a Cantor set is similar (not sure whether the right concept here is a homeomorphy or an isomorphy) to a series of coin tosses (or double-slit experiments with recorded outcomes).

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jbriggs444
Homework Helper
For the last point, don't you need infinite time (or infinite speed)?
Yes, agreed. Infinite time, infinite "speed" or at least some way of invoking infinitely many bifurcations.

EnumaElish
PeroK
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Gold Member
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For example, the probability that a natural number is prime is zero. The probability is smaller than any positive number, so it must be zero.

At the moment, as I look at this page, this thread has 23 replies. The similar discussions below have 19, 5, 7, 7 and 17 replies respectively.

That's 6/6 prime numbers. Quite a hit rate for something with zero probability!

EnumaElish
hilbert2
Gold Member
At the moment, as I look at this page, this thread has 23 replies. The similar discussions below have 19, 5, 7, 7 and 17 replies respectively.

That's 6/6 prime numbers. Quite a hit rate for something with zero probability!

Nice observation. :)

What if we choose a random natural number in such a way that any number ##n## is given a relative statistical weight of ##e^{-an}##, where ##a## is a positive real number? How about the case where the statistical weight is ##\frac{1}{n}##?

EnumaElish
Homework Helper
How about the case where the statistical weight is ##\frac{1}{n}##?
The sum of those "probabilities" would be infinite.

hilbert2
Gold Member
The sum of those "probabilities" would be infinite.

We could form a sequence of sets of natural numbers ##A_k## so that in the set number ##k## the natural number ##n## appears ##\lfloor\frac{k}{n}\rfloor## times (note the floor function that rounds to nearest smaller integer, sufficiently large numbers ##n## don't appear even once), and find the limit of the fraction of primes in those sets when ##k## approaches infinity. I know that technically, the same element can't appear many times in a set, maybe ##A_k## should be called a vector of natural numbers.

mfb
Mentor
What if we choose a random natural number in such a way that any number ##n## is given a relative statistical weight of ##e^{-an}##, where ##a## is a positive real number?
Then we get a positive sum, as every prime contributes a positive value.
How about the case where the statistical weight is ##\frac{1}{n}##?
The sum of weights would diverge.

##\displaystyle \lim_{n \to \infty} \frac {\pi(n)}{n} = 0## where ##\pi(n)## is the number of primes up to n. That is a useful and quite natural way to talk about "the fraction of natural numbers being X".

PeroK
hilbert2
Gold Member
##\displaystyle \lim_{n \to \infty} \frac {\pi(n)}{n} = 0## where ##\pi(n)## is the number of primes up to n. That is a useful and quite natural way to talk about "the fraction of natural numbers being X".

That's not the same limit that I was talking about. If you define a Wolfram Mathematica sequence equivalent to the ##A_k## in my previous post, and plot the proportion of primes in those sets up to large values of ##k##:

Code:
DiscretePlot[
Sum[Boole[PrimeQ[n]] Floor[k/n], {n, 1, k}]/
Sum[Floor[k/n], {n, 1, k}], {k, 1, 1000}]

you get this kind of a graph:

It's impossible to be sure about whether this converges to zero or some positive real number, but it definitely doesn't decrease as fast as ##\frac {\pi(k)}{k}##.

EnumaElish