↪fishfry I've had more luck with CEX. Just need to verify a bank transfer and then I'm good to go. :) — Michael

Thanks for the tip. I'll check them out.

ps -- CEX has a much higher price for bitcoin than the other exchanges. Also there are some Reddit complaints about it being difficult to get your USD out. Best to check around before committing too much money to these guys. Or any exchange for that matter. All the exchanges are awful at the moment. It's a problem.

Yes the exchanges and wallets are the weak point in the system. The core protocol is cryptographically secure (unless the NSA already has a backdoor, which I regard as highly likely) but the exchanges and wallets are a target for every hacker in the world.

Here's a guy who lost $155k worth of bitcoin (probably worth a lot more at today's price) over the unsecured WiFi in a restaurant. What a brave new world we live in.

Bah. I'm trying to get into this but none of the exchanges are accepting photos of my ID when I'm trying to verify my account. If I miss out on a big profit because of this, I'm going to be super pissed. — Michael

Did you try Gemini? They couldn't verify my ID (same problem you're having) but they emailed me and said I can link a bank account. Coinbase is absolutely hopeless. It's amazing that Coinbase is the supposedly well-funded big dog in the space. No support, the entire system locked up for two days when bitcoin hit 10k. Imagine everyone trying to get their money out during a crash. Coinbase won't last. Bad management IMO.

Plenty of profits to be made in the coming years. This is the Internet in like 1986 when research labs were still developing the underlying protocols. The idea that the public can bet on engineering experiments is nuts. The real money will be made when actual companies provide actual services to the public. Think Amazon, Google, Facebook. Nobody ever heard of them in 1995 when the Internet first hit public awareness.

Yes I'm all over it. Top things you need to know that aren't in the mainstream news:

* Invented by NSA in 1996. Satoshi's 2009 paper is strictly for the tourists. You probably don't believe me. Here's the paper. http://groups.csail.mit.edu/mac/classes/6.805/articles/money/nsamint/nsamint.htm

* Bitcoin security fails if there's centralization of miners, but the top three Chinese miners now control over 50% of the hashing capacity of the network.

* It's a huge pain in the ass to actually fund an account on an exchange and actually buy cryptos and then secure your wallet. There is no way this is going to scale to the public until some of these issues are solved.

* It's not going to bring down the bankers. The bankers will control the space like they've controlled money for thousands of years.

* There's plenty of opportunity to make money. If you missed bitcoin don't worry, there are plenty of altcoins out there. Bitcoin itself has terrible scaling and organizational and centralization problems that will probably kill it at some point.

That said, it's very interesting. Blockchain solves the problem of establishing end-to-end trust on decentralized, distributed ledgers. It's a fantastic technical achievement that will change the world.

After finite time there are a countably infinite number of indistinguishable Hubble Volumes, which, as time progresses, may diverge. — tom

Umm, no. There are uncountably many Hubble Volumes instantiating countably many initial states — tom

[My bolding in both quotes]

In one post you said there are countably many Hubble volumes, and in another post you said there were uncountably many. Can you clarify this?

We would all benefit if you would please lay out a clear version of your argument in one place for reference. Assumptions, argument, conclusions. So we can know we're all talking about the same thing.

I perfectly well admit that I'm ignorant of physics. If someone can explain to me why probability theory doesn't apply to this situation I'd be happy to learn.

My understanding of your argument is that at the moment of the creation of the multiverse every possible state gets instantiated.

As I understand it, the argument for that conclusion is probabilistic. It depends on the behavior of statistical distributions at the limit. You used that phrase as one of your bullet items earlier. You are making a probabilistic argument to demonstrate that it's almost surely the case that every state gets instantiated. And you don't understand that your argument allows for the possibility of a large set of exceptions.

I'm asking for a clear explanation for why the well-known rules of probability don't actually apply to the creation of the multiverse, when these very rules are employed to develop the theory. I'm not saying there's no such explanation. I'm asking what the explanation is. Then I'll learn something.

I've heard that Tegmark himself admits he's no mathematician. I wouldn't be surprised if he's just ignored this point and nobody cares much. It's not that big a deal. As I said earlier, what if the multiverse got started and all but one state were instantiated. It would still be ergodic. Can you tell me the exact reason you claim that can't happen? If it happened, would the universe collapse or would we all have three heads? My intuition doesn't see why it couldn't miss a state yet still satisfy all the relevant statistical properties.

After finite time there are a countably infinite number of indistinguishable Hubble Volumes — tom

I don't understand how you can say this, yet claim you don't understand the idea of countably many coin flips. They're the same mathematical idea.

You have countably many regions, or universes, or coin flips. Each region or universe or flip is assigned one out of finitely many possible states. One out of a zillion in the case of physics, or one out of 2 in the case of coin flips, but the math is exactly the same either way.

You can't mock the idea of coin flips and then come back with the exact same idea in the guise of countably many universes. There's no mathematical difference between a 2-sided coin or a gazillion-sided coin. If the number of states is finite, then probability theory applies. In the large, it is "almost certain" that all states recur infinitely many times, but it is not absolutely certain. The case of coins or universes are exactly the same. It only depends on there being countably many coins or universes or regions, each taking up one out of at most finitely many states.

The coins are a simplification for the purpose of making the probability theory clear. But I can use a gazillion-sided die if you like, where the number of possible arrangements of particles is a gazillion. The math is absolutely the same.

I think maybe you should either accept the physics or try to understand it. — tom

If all you have is insults and you can't coherently explain your point, you've got nothing. You're the one claiming there are infinitely many universes and/or finite regions of an infinite space. My analysis applies.

You earlier claimed that it's been established in this thread that measure zero events can't occur. Two or three people including myself have asked you for the explanation. You have not provided one.

You can go look up the formal definition of ergodicity to learn that the smooth statistical behavior may fail to occur on a set of measure zero. This is a required characteristic of any probabilistic argument, since events on a set of measure zero have no effect on the overall probability.

There are no measure zero events, as explained multiple times in this thread. — tom

Please explain it to me. If there are no measure zero events, then NO distribution of states to universes is possible. Just like if you flip infinitely many coins. Whatever result comes up, that was a measure zero event.

It's only when you aggregate all the possible results as either "Some state only happens once" versus "All states happen infinitely many times," that the latter has probability 1. That's because you're aggregating uncountably many measure zero events, and comparing that to an aggregate of countably many measure zero events.

If you randomly assign one of finitely many possible events to each of infinitely many universes or regions of an infinite universe, the probability of ANY PARTICULAR result is zero. You can just multiply out the probabilities to see that.

I look forward to your refutation of our best Cosological theory, right here, on this forum, by a true expert. — tom

That's a pathetic response.

A non-pathetic response to the question I asked you would be one of the following:

* "I don't quite get this measure zero business, can you explain?" or

* I get measure zero ,but it doesn't apply in this case because _______"; or

* I get measure zero and Tegmark gets measure zero but Tegmark says it doesn't apply in this case because _____."

If you make a statistical argument on an infinite probability space and you don't take measure zero events into account, you have to say why they're not relevant in the particular case under discussion. If you can't formulate a coherent reply, you don't understand the ideas you're promoting.

Are they probabilistic? Perhaps you could show what you mean by that, and which points comply with your meaning?

Infinite realms? Loss of certainty?

Intuitions, infinite probabilities?

Metaphysical claims? — tom

Before I compose a detailed reply to each of the points you raised; can you please tell me which part of my post is giving you trouble? Do you not understand the distinction between almost surely and surely? Or do you understand it but think it doesn't apply in the present discussion? Or think that Tegmark doesn't think it applies? I just want to figure out what level of misunderstanding we're having on this point. There's no question that I'm accurately relaying the math of infinitary probabllity theory. So there must be some disconnect between that and the physics.

All possible states must occur because of:

1. Infinity, which I have mentioned several times previously. Space must be infinite.

2. Mass fluctuations approach zero on large scales.

3. The mass densities at any set of points has a multivatiate Gaussian probability distribution.

Conditions 2 and 3 may be combined into the weaker condition that the correlation functions of all orders vanish in the limit of infinite spatial separation.

Together the above conditions are called ERGODICITY — tom

Thank you for clarifying. I don't feel that you're engaging with my point.

Your bullet items are probabilistic. And you are applying probability theory to an infinite event space.

It is a fact that when you extend finite probability theory to infinitary realms, you lose certainty. Your event space could be extremely well-behaved statistically, and still allow for many anomolies and exceptions. This is the difference between the phrases "surely" and "almost surely." These are technical terms in probability theory. Almost surely means that something happens everywhere except on a set of measure zero. That means that there can be many individual exceptions to the global statistical properties of the event space. And measure zero sets can be quite large and complicated.

I would certainly agree with the statement that given the premises of the discussion, there is almost surely a duplicate earth.

But that does not mean that it must necessarily happen. And this is the point you're not engaging with. The very infinitary probability theory that you're invoking allows for exceptional sets of measure zero.

You are applying without justification your intuitions and beliefs about finite probabilities to infinite ones. You are not giving sufficient consideration to measure zero sets and the loss of absolute certainty when we pass from finite to infinite probability theory.

This being a philosophy forum, it's worth noting that this is a point of epistemology. When you extend finite probability theory to infinite event spaces, you lose certainty. The theory requires the existence of measure zero sets; and sets of measure zero don't have any effect on the limiting behavior of the probability distributions. You have to take this into account when making metaphysical claims based on probabilistic arguments,

Your attributing an agency here, an ‘it’ that ‘does’ something. But the same ‘it’ is what you’re trying to explain, so there’s an issue of recursion involved. — Wayfarer

Only speaking loosely. Ignorant of the physics. Asking if there's an explanation of why all states must occur. Is this all as in all? Or all as in statistically all, but possibly missing one or two with no harm done to the universe?

I don't understand why you think I'm trying to explain anything. Tom said that during inflation all states are created. I'm asking questions about that.

You know, "How does it know how to do that?" seems pretty clear to me. If I said, "How does a bowling ball know how to fall to the earth?" you wouldn't think I am talking about agency, but rather asking how gravity works. So how does the universe know how to generate all the states? That's a pretty common figure of speech I think.

Inflation generates all possible initial conditions. — tom

That's very interesting. How does it know to do that? In the early moments of the universe it's cranking out all these possible configurations, and it's only got one more left. How does it know that? What if it forgets to do one particular configuration? Can it go back and do it later? Can the universe continue to exist or does this one single imbalance make the universe unstable in some way?

This is my preliminary understanding. I no longer think ergodicity absolutely guarantees that there is a duplicate earth. If someone knows better and can walk me through the argument, I'd be grateful.
— fishfry

You also need the Bekenstein Bound and infinity. — tom

So my understanding is correct? I'm gratified. "Almost surely" is not the same as "surely." All these statistical arguments leave the possibility of exceptions, unlikely as they may be.

Now the Beckenstein bound I believe has something to do with relating the energy in a bounded region of space to its information carrying capacity. It takes a certain amount of energy to change the state of a system so that the collection of all possible state changes is finite and can be calculated. Is that about right?

So how do we get from there to duplicate earths?

You know I just don't believe this duplicate earth story. Say there's a universe or a multiverse and it's got every possible state represented infinitely many times ... except there is one state that just happens to only occur once, by incredible amazing luck ... one little blue watery planet with bad politics, third from the sun ... and it's the only one like it in the entire multiverse.

I just don't see why such a universe couldn't exist. It would still have all the required statistical properties. But if your universe is infinite, then "almost surely" is NOT the same as "surely" and individual exceptions to the statistical properties may exist. That's one of the limitations of extending probability theory to infinite spaces. You lose certainty.

But we're making progress if you agree that ergodicity by itself is not sufficient.

You do get your knickers in a twist with great rapidity. — apokrisis

As I attempted to make clear earlier, I'm ignorant of physics and so I have to take on faith a lot of what's written here. But now and then the discussion wanders into areas I'm familiar with; and more often than not what's said is nonsense. At those moments I do my best to straighten out the math.

As far as your snark, it's not necessary.

As you know, that was Michael's terminology. I went along with it for the sake of discussion. — apokrisis

Yes and I corrected it because it's wrong and confuses the subtle probabilistic issues at stake.

But also, 1/infinity is the proper definition of the infinitesimal as far as I'm concerned. — apokrisis

You are entitled to your private definition. The actual definition of an infinitesimal is a quantity x such that 0 < x < 1/n for every natural number n. The gap between your instincts and how math actually works is significant. I understand in general that you do not believe in standard modern math. You made some remark about Cantor earlier along those lines. Philosophically it's perfectly valid that you have alternative ideas about math and don't accept parts of standard modern math. But physics is most definitely based on standard modern math; and to the extent that your outlook diverges from that, you are introducing confusion into the conversation.

Now you will get on your high horse and object no doubt. — apokrisis

Is that really the best you can do? I'm simply doing my best to explain the viewpoint of standard math, which is the math used in modern physics; and you are arguing from your Peircean viewpoint and claiming that you are entitled to make your own definitions for mathematical terms that already have perfectly clear standard definitions.

But I went along with Michael's terminolog — apokrisis

I'm not sure how that supports any point you might be trying to make. I did actually correct his usage first, and I left your first usage alone simply because he was first and you were simply going along. It wasn't till you mentioned it a second time. If that bothers you, I'll apologize if it makes you happy.

For the record I do understand that you have your own private notation and that you reject the standard notation of modern math, on which physics is based. Is that a fair assessment? Since you admit you have your own definition of an infinitesimal, and you assign meaning to the symbol 1/∞. And that I was a terrible person for calling your private ideas, which you are perfectly well entitled to hold, nonsense. They're not nonsense. They're merely your personal ideas and notation, totally at odds with modern math.

But I went along with Michael's terminology largely because I also like that sly implication. — apokrisis

Ok. You are trying to express the idea of an infinitesimal probability. We all have these intuitions of infinitesimals, as Leibniz did. I truly get that. But since we're doing physics, it's important to make sure we get the math right. Else I wouldn't bother to bring it up.

It is another way of getting across that the probability ain't actually zero even if it is almost surely zero when it comes to an infinite spatial universe producing replica earths with replica people doing replica things. — apokrisis

Yes I understand your philosophical point. But your math is wrong. And we're doing physics. If you're using math metaphorically you should say so up front.

If you remember, it was you who introduced the confusion. — apokrisis

This refers to the infinite die. No that's not true. You introduced it, and the moment you did I realized you didn't have any idea what the conversation is about. The duplicate earth argument depends crucially on there being only a finite set of possible states in any bounded region of space. That's fundamental to the argument. When you brought up an infinite state die, I knew you simply had wandered off into some conversation the rest of us aren't having.

I was trying to sort it out for you by pointing out that those would be the kind "coins" you would need to be flipping... — apokrisis

But that's exactly wrong. You need a finite die with a very large number of faces. One for each admissible state of all the particles in some bounded region of space.

It's ironic that just when you are totally losing track of the thread, you think you're helping me. You've done that before and you were wrong then too.

This still reads as nonsense to me. — apokrisis

I'm perfectly willing to stipulate that much of what I write reads as nonsense to you. You have the same effect on me. Your writing seems extremely learned yet you never make a lick of sense.

Why don't we agree not to interact? I was really surprised earlier that you directly replied to something I said. I don't think our interactions are productive. I find your snark annoying, especially since it generally shows up when your degree of wrongness is at a local maximum.

Maybe you agree now as you seem to have discovered ergodicity and moved on to a notion of a universe chopped up into sufficiently large but finite regions - the ensemble of microstates picture that I also have been at pains to criticise. — apokrisis

As it happens I've spent the afternoon chasing down ergodicity. I do know a little about it relative to the irrational rotations of a circle, which are ergodically dense in the circle. What I've learned today is that by the definition of ergodicity, any set that behaves badly must have measure zero. If that's correct, then my NO-duplicate earth possibility is still alive. Ergodicity is a statistical attribute that describes what happens almost surely. But not absolutely surely.

This is my preliminary understanding. I no longer think ergodicity absolutely guarantees that there is a duplicate earth. If someone knows better and can walk me through the argument, I'd be grateful.

Even with finite states does it work? Given an ordinary coin and an infinite series, is it almost certain that there will be an infinite number of heads and an infinite number of tails? — Michael

Yes. Needed to think about that a little. There's only one way there can be zero heads. There are countably many ways there can be one head. (Could be on flip 1, or flip 2, etc.). There are countably many ways there can be exactly two heads: Flips 1-2, 1-3, 1-4, ..., 2-3, 2-4, ..., 3-4, etc., and there are only countably many such combinations. (Countable times countable is countable). Continuing, there are only countably ways to get 3 heads, countably many ways to get 4 heads, etc. Adding up all those possibilities is still countable, since a countable union of countable sets is countable. Each sequence has probability zero, so by countable additivity the probability of finitely many heads is zero. Same analysis for tails.

That's why I said that the duplicate earth theory may fail, but only with probability zero (which can still happen). If you throw in the assumption of ergodicity, you are guaranteed a duplicate earth to the best of my understanding.

I don't see how my argument is any different to your argument here: — Michael

The space of coin flips is uncountable. So countable additivity still allows individual events to have probability zero.

When people are talking about an infinitely-sided die, I assume they mean a countable infinity, which has no uniform probability distribution.

But the talk of an infinitely-sided die is quite incorrect in this context anyway. The entire "duplicate earth" idea depends on their being a FINITE number of possible states in any bounded region of space. If the set of states is infinite, then there is no reason at all that any state should be duplicated.

The introduction of the infinitely-sided die is the moment this thread went completely off the rails. The entire foundation of the duplicate earth theory is that there are only finitely many states possible in a given region of space. That's essential to the argument. Drop that assumption and there is no argument at all.

Perhaps you were talking about infinite sequences of coin flips and not infinite-sided dice. I may have misunderstood you in the general confusion of the last few posts. Once someone mention an infinite-sided die to represent possible states, the entire thread went hopelessly off the rails.

↪fishfry As I said, that was me being lazy. I didn't want to figure out how to do MathJax. — Michael

Doesn't help your argument. The limit is exactly zero. But if the probability of one event in a countable event space is zero, your probabilities don't add up to 1. And if the probability is nonzero, it also doesn't add up to 1. There is no uniform probability distribution on a countable event space. Your notation wasn't lazy, it was mathematically incorrect no matter how you notate it.

ps -- To make this a little more clear, there is a probability distribution on the natural numbers, but it's not uniform. If you assign probability 1/2 to 1, 1/4 to 2, 1/8 to 3, and so forth, the sum of the probabilities is indeed 1 (by countable additivity) so this is a probability distribution. But it is not uniform.

My point is that applying bad math to multiverse theories makes skeptics out of people who know math but don't know physics. Tom's mention of ergodicity makes sense to me because if you assume the distribution of states to regions is ergodic, then there are indeed infinitely many earths. When people say things that make sense to me mathematically, I tend to believe their physics. That's all I'm saying.

Is this the same with infinitesimals? — Michael

There are no infinitesimals in the real number system. Using nonstandard analysis, in which there are infinitesimals, doesn't help your argument.

Which is why I wrote
limn→∞f(1n)=0 — Michael

But that number is exactly zero. There is no such thing as 1/∞.

Isn't this how to describe the probability that an infinite number of coin tosses always lands heads, or always lands tails, or lands once on heads and every other time on tails, or [insert any particular outcome]? — Michael

No. The probability of any particular sequence of infinitely many coin tosses is zero. Exactly zero.

The question at issue with the "infinitely sided die" is that you toss a set of pingpong balls into a hat. The balls are labelled 1, 2, 3, 4, 5, 6, ... and so forth. The question is, what it the probability of pulling out, say, 6. And the answer is that there is no way to assign that event a probability. That's because of a thing called countable additivity. If you say the probability is zero, then the probability of picking any number at all must be zero. But that's false, because you will pick out SOME pingpong ball.

https://en.wikipedia.org/wiki/Sigma_additivity

On the other hand, if the probability is some tiny real number greater than zero, then the total of all the probabilities for individual balls adds up to infinity. That's impossible too, because in a probability space, the probabilities of all the possible events must add up to 1.

The conclusion is that there is no uniform probability distribution on the natural numbers. That's why the notation 1/∞ makes no sense. It's an attempt to express an intuition that turns out to be logically impossible.

Now you may ask, why is the probability of getting some particular sequence of coin tosses exactly zero then? The answer is that there are uncountably many possible sequences, so countable additivity doesn't apply. The probability of any particular sequence of coin tosses is zero, and the probability that you'll get SOME sequence is 1 (since you must get some result), and the math works out.

Here are the mathematical rules for probabilities.

https://en.wikipedia.org/wiki/Probability_axioms

This is a philosophy discussion group. When things seem unarguably right, that's when you know there must be the whole flip-side to the story. Dialectics always rules. So someone's nonsense is always the start of someone else's sense. — apokrisis

When you replied to me earlier spouting buzzwords about the Hubble radius and such, I tended to believe you since I don't know much physics.

But when you write the expression 1/infinity with a straight face; and when I call you on it, and you reply as you did in the quoted paragraph; I have to assume you have no idea what you're talking about. So which is it?

There is no such thing as a probability of 1/infinity. Not in standard analysis and not in nonstandard analysis, which I took the trouble to learn something about a while back.

The problem I have in this thread is that whenever someone lays on the physics jargon, I can't respond or even form an opinion, since I lack the background. But whenever those same individuals (not just you) wander into a domain in which I have technical expertise, they're often spouting bullshit. [In the Frankfurt sense, of course]. And this leads me to think they're doing the same with their physics jargon.

So I ask you again: When you claim an event has probability 1/∞, exactly what do you mean? Is that a claim of standard probability theory? A misunderstanding of nonstandard probability theory? A vague, meaningless intuition that cannot be properly grounded in logic? Or what, exactly?

And if you can't speak sense to me about things I know about, why should I believe you when you speak to me about things I don't know about?

If the odds of earth being the case on any one roll are 1 in infinity, — apokrisis

Why are you doubling down on this mathematical nonsense?

It's exactly as likely as any other outcome: 1/∞. — Michael

Can you explain exactly what you mean by that? Are you using your own private system of mathematics? Making an argument based on a vague misunderstanding of nonstandard analysis? Something else?

Do you know what countable additivity is? Did you know that there's no uniform probability measure on the natural numbers? Does that information affect your use of this notation?

It's called ergodicity. I have mentioned that ergodicity is a requirement for Type 1 multiverse several times. — tom

So you are requiring the assumption that you are then claiming is true? Well by that logic you're certainly right. If you pre-load your desired conclusion into an assumption, your conclusion falls out at the end.

You may be clear in your mind that ergodicity is a requirement, but the other two people arguing for the "duplicate earth" idea certainly never mentioned it. Why do you think that is?

ps -- I had a look at the Wiki page for ergodicity and found no clue to support your argument. Can you be more specific as to how ergodicity rules out the possibility that some state recurs and some state doesn't? https://en.wikipedia.org/wiki/Ergodicity

You obviously can't fit in an infinite number of copies of each, which is what you need for a multiverse. — tom

I'm not the one claiming the universe is infinite. I'm simply pointing out that in an infinite collection of regions, with each region taking on one of a finite number of possible states, there's no reason that any particular state must be necessarily be shared by two regions. It may be that region 1 is in state 1 and all other regions are in some other state. SOME state gets repeated infinitely many times but not necessarily any particular state.

I refer you to NoAxoms' argument. Read it carefully this time. — Michael Ossipoff

The passage you quoted was written by me. I read it very carefully and stand by it.

If there are only finitely many states S1 through Sn, and infinitely many regions R1, R2, ..., there is no reason why some state can't occur only once. That may be statistically unlikely, but it's possible. There is no reason there's necessarily a "duplicate earth."

Yeah. But you get to pick these infinite sequences out of an infinite hat. So you would pick that exact sequence an infinite number of times. — apokrisis

That doesn't make any sense. You're completely misunderstanding the discussion.

The claim (which I disagree with, but accept only for sake of argument in this instance) is that the universe is infinite. So there are infinitely many finite-sized regions of space. Call them R1, R2, R3, and so forth.

Under this circumstance, we assume that each region can only be in one of finitely many possible configurations.

Under that assumption the claim (which I am showing is incorrect) is that there must be two earths.

Now for simplicity I am saying, let's suppose there are only two states, 0 and 1. We might have that R1 is in state 0, and all the other states are in state 1. In that case, state 0 never gets repeated, contradicting the claim that every state must be repeated.

Now in the more realistic case there are zillions of possible state. Still finite, but very large. But then it's still the case that R1 might be in state 0, and every other region is in some other state. It's still the case that it is possible that some state never repeats.

Your maths doesn't give that result as I've argued. You are trying to hardwire in the restriction that Earth or Mars is the binary choice that reality is having to toss a coin on. But this is about a coin with an infinity of faces - one for every possible state of the world. — apokrisis

But this completely undermines the argument. If there are infinitely many possible states then it is possible that R1 is in state 1, R2 is in state 2, etc., so that NO state ever repeats. It's the assumption that there are only finitely many states that makes the idea even plausible that there are possibly two earths. In the infinite state case the argument totally falls apart.

But now you are changing the rules of your own game. Instead of Earth = 0, not-Earth = 1, you are saying reality only has the two options of Earth or Mars. And for some reason, nothing else will get pulled out of the hat. — apokrisis

If there are zillions of states, it's the same argument. The two state example is only a simplified illustration.

So spatial infinity would seem to guarantee that there should be an infinity of Earths — apokrisis

If there are two states and infinitely many universes they could be 0, 1, 1, 1, 1, 1, ...

If 0 is the "earth" state, there is no other earth. 1 is maybe Mars. So Mars exists infinitely many times but not earth. If there are a trillion states, same argument. SOME state recurs infinitely many times, but not necessarily any particular state. Maybe there's only one earth even though there are infinitely many copies of Mars. It's perfectly possible.

I should have quoted more.
I meant to ask if there was no grid in the sky if the space was torrid manifold. — noAxioms

No, there is no grid in the sky on a manifold. There is a distorted grid in the sky at each point. But each point has a different distorted grid, with some regularity conditions to make sure the areas of overlap are consistent.

I then gave the example of the asteroids game which is played on such a manifold, — noAxioms

Asteroids is a torus represented on a 2-D grid. Not an abstract manifold.

and there is very much a noticeable grid to it, despite the lack of lines painted through the space or the fact that the screen happens to line up with it. — noAxioms

Oh I see your point. Yes, the torus is orientable. Some manifolds CAN'T have a consistent coordinate system drawn on them. Famous examples being the Mobius strip and the Klein bottle.

https://en.wikipedia.org/wiki/Orientability

But even in the case of a torus, whose points can be described by rotations on two circles, those are not absolute with respect to the ambient space. They move with the torus, if I'm understanding your point correctly.

Is this true? — noAxioms

That a manifold is a locally Euclidean object that exists on its own and does not live in an ambient Euclidean space? Most definitely.

That doesn't mean you can't have a torus that IS embedded in an ambient Euclidean space. I didn't say that tori can't live in 3-space. I said they don't HAVE to. The latter is accomplished mathematically via the concept of a manifold.

MWI is not an ontological stance. No creation of new universes ('somewhere' as fishfry puts it) occurs — noAxioms

That's pretty mysterious to me. Do you mean that the new universes don't come into existence? Are you saying that they're always there? I should emphasize that my knowledge in this area is limited to Wikipedia. I'm asking on a very simplistic and naive level. My understanding is that MWI says that these universes come into existence at each fork in the road. The cat is dead in one and alive in the other. and those two universes spring out of the one.

If that's not right, can you please explain it to me?

But if the universe is a 3-torus, all three axes have a preferred orientation, — noAxioms

In math, the idea of a manifold is more abstract than that. Naively when we imagine the 2-torus like a big donut in space, we imagine the donut embedded in Euclidean three-space. We can use the equations of the ambient space to describe and manipulate points on the torus.

But the idea of a manifold is that the donut is there ... and the ambient Euclidean 3-space is not. The donut, or sphere, or whatever shape it is, exists on its own. There is no big rectangular grid in the sky.

In the mathematical discipline called differential geometry they note that for each point on the manifold, there some neighborhood that is diffeomorphic to Euclidean n-space for some n. Diffeomorphic means that there's a little neighborhood around each point that can be smoothly deformed into a Euclidean space.

So a manifold does not live in a Euclidean space. All we can say that at each point we can pretend it's pretty much Euclidean. Just like if you're standing in Kansas it seems like you are at the center of a big plane. [Kansas is a very flat state]. Each point on a sphere is locally stretchable into Euclidean 2-space.

Now if you have two points nearby each other, and each point can be deformed into a Euclidean space, what about any points that are in the overlap? Do they get stretched in a consistent manner? That's part of the basic machinery of differential geometry. Then they use these local mappings to do calculus on the surface of the manifold, without regard to any notion of an ambient space.

Einstein basically discovered differential geometry when he developed relativity, then Minkowski came along and recognized it as something mathematicians had already known about since the days of Riemann in the 1840's.

My understanding is that this is how physicists look at things. They don't need an ambient coordinate system. Which is fortunate, since Einstein noted that there is no preferred frame of reference.

I'm no expert but this is what I know about it.

This does not change what I was saying though: like it or not, most of modern physics does use calculus. — SophistiCat

Stating the totally obvious.

You can claim that most of modern physics is misguided — SophistiCat

I have never said any such thing.

And this is why I don't usually engage this person. — SophistiCat

If only.

... I offer you to play a game: give me any number and I can give you an even bigger (but still finite) number. It turns out that that's all we need to get going. — SophistiCat

False.

Consider the Peano axioms. Given a number n, there's a greater number called the successor of n.

That gives us a sequence 0, 1, 2, 3, ...

However, that is NOT ENOUGH to get calculus off the ground. To do that, you need a completed set of numbers {0, 1, 2, 3, ...} which is given by the axiom of infinity.

In other words Peano + Infinity = Calculus.

Without the axiom of infinity, each number has a successor but there is no set of all the numbers; no infinite set; and no calculus.

To say this another way: ZFC is stronger than Peano. Important to keep this in mind when slinging around ideas like infinite numbers of planets.

you would be hard-pressed to find many cosmologists who dismiss the possibility of an infinite universe — SophistiCat

I never dismissed the possibility. I pointed out -- correctly -- that current theory says that the universe is finite. Your own examples support this.

because the level-1 multiverse notion assumes that this universe is infinite. In an infinite amount of space, with an infinite number of solar-systems and planets, there inevitably, somewhere, will be an identical copy of Earth, with, of course, a copy of you. ...an infinite number of exact Earth copies, in fact. — Michael Ossipoff

NoAxioms and I just had a lengthy conversation disproving this very point. Could you please review those posts? What you say is simply not true. At best you have a probabilistic argument that falls short of certainty.

Secondly, the level-1 multiverse only requires a finite universe sufficiently large that light hasn't had time to get from one point to some other point in the age of the universe.

Many Worlds is the only known interpretation of quantum mechanics. — tom

I can only go by the Wiki article. https://en.wikipedia.org/wiki/Many-worlds_interpretation

They say it's an interpretation. You say it's a done deal. I'm not convinced. Where are all these universes supposed to live? But really that's not the point. The point is that Wiki says MW is an intepretation and you say it's the only known interpretation. I"m in no position to evaluate your claim versus Wiki's except that in this case I tend to believe Wiki.

Many Worlds is at least 60 years old, and has not only passed every test, but has led to the discovery of decoherence and the quantum computer. Every quantum interference experiment is a test of Many Worlds, as are interaction free measurements and many other technological examples. — tom

Perhaps I stand corrected. Are we talking about interpretations of QM? Or actually proven multiverses? My understanding is that Many Worlds is an interpretation, not an experimentally verified fact. But if Feynman and Penrose say so, well ok. I don't know.

Well, I hit a different one that cannot be hit, so I'm on thin ice to counter this. But having hit this computable number, I must in addition throw infinite coins and come up heads on them all, or else I just got close to it. I don't need to do that with the 'typical number'. Is there a term for that? I made that up. — noAxioms

By typical you mean random, or noncomputable. Alan Turing described computable numbers as numbers whose decimal (or binary, same concept) expression can be generated by a program. So numbers like 1/3 are computable. Pi is computable since there are many closed-form expressions that could be programmed into a computer.

The noncomputable, or random reals are the ones whose decimal digits can't be expressed or computed by any program. If you throw a dart at the real number line, the chance that you hit a noncomputable number is 1, and the chance that you hit a computable number is 0. But all the speciic numbers we use in math or science are computable.

I do agree with you that if there are infinitely many universes and only finitely many possible state of matter, it would be very unlikely for only one state to occur infinitely many times. But then again, perhaps there are laws of nature that make some configurations of matter more likely than others, so we don't actually know the true probabilities.

ps -- There's another wrinkle to this. If someone believes that the universe is a computation -- this is the "computable universe hypothesis" -- then the noncomputable numbers don't even exist. Only computable numbers exist. I have a hard time with that because then the real line would be full of holes. But some speculative thinkers believe it.

You take a stab at a number line with a pointer and you will hit a 'typical number' as I call it. That number cannot be expressed with any amount of digits. It has zero probability of being hit, and yet it was hit.
Is that more what you're after? — noAxioms

That's exactly right. The set of computable numbers is said to have "measure zero." https://en.wikipedia.org/wiki/Null_set

Edit: I wrote that before reading your last post. Creepy...
No, I would say it is impossible to hit a computable (or expressible) number, except to say 'this one'. — noAxioms

It's not impossible. The computable numbers exist. You might hit one. There are infinitely many of them, after all. In infinite probability spaces, probability zero events may occur; and probability 1 events are not absolutely certain.

Getting back to our original point, this is why even if there are infinitely many universes, SOME state must recur infinitely many times, but not necessarily any particular one. SOME person might have a twin, but probably not you or me.

So at best, the "in an infinite universe there must be two earths" is false in an absolute sense, and is at best a probabilistic argument. Which I think we already agreed on.