Finite number divided by infinity

[Dale]

I think you are confusing “computable” with “physical”. Physical locations are isomorphic to the reals (or at least we have no evidence to the contrary). So picking a physical position is isomorphic to picking a real number, even if you cannot write it down or enter it into a computer exactly.

 

[PeroK]

I think you are confusing “computable” with “physical”. Physical locations are isomorphic to the reals (or at least we have no evidence to the contrary). So picking a physical position is isomorphic to picking a real number, even if you cannot write it down or enter it into a computer exactly.

How do you choose a point in the continuum? QM explicitly avoids localising a point particle precisely (even if a particle can be a point). You cannot wave your arms and say: that particle is precisely at location $(x, y, z)$.

The mathematical model involves a continuum, but you cannot choose a point in the continuum, except mathematically.

 

[A.T.]

You can't "choose" one in the sense that the zero probability of doing so has any real meaning.

Going back to the original topic, one can choose a planet, and there might be infinitely many planets.

 

[kuruman]

Y'all missed the point of "Hitchhiker's Guide." Clearly, $$\text{lim}_{x\rightarrow\infty} \frac{1}{x}=42.$$

 

[PeroK]

Going back to the original topic, one can choose a planet, and there might be infinitely many planets.

Yes, there is a distribution on the natural numbers where, for example: $p(1) = 1$ and $p(n) = 0, \ n \ne 1$.

Or, where: $\forall n: \ p(n) = \dfrac{1}{2^n}$.

 

[PeroK]

I think you are confusing “computable” with “physical”. Physical locations are isomorphic to the reals (or at least we have no evidence to the contrary). So picking a physical position is isomorphic to picking a real number, even if you cannot write it down or enter it into a computer exactly.

PS if you can do this, then any theory that is limited to a finite number of spacetime events, say, has been disproven. If you can genuinely map spacetime isomorphically to a continuum, then discrete spacetime theories are, as fundamental theories, impossible.

 

[DaveC426913]

Mathematically, we can have a uniform distribution on the interval $[0, 1]$, where the probability density is uniform. In reality, however, it's not possible to select from more than a finite number of outcomes with equal probability.

The paradox "something with zero probability can happen" occurs by misapplying mathematics. Or, at least, imagining that there is a scenario where you can apply that mathematics. It's impossible to select real numbers uniformly. More explicitly, the positive integers cannot be selected uniformly.

Another example is Hilbert's Hotel. That's valid mathematics and a good example of the behaviour of infinite sets; but, we cannot build a hotel with an infinite number of rooms.

The Internet is awash with the fallacy that these things could be physically realised, rather than being purely mathematical constructions.

Sure but who said anything about physically realizing them as a requirement ?

Can any scenario that contains infinities be physically realized?

 

[Dale]

QM explicitly avoids localising a point particle precisely

QM says that the probability distribution of a point particle has a non-zero standard deviation. It does not say that the sample space is discrete. In fact, in QM the position of a point particle is explicitly modeled as a continuous probability density function on a real-valued sample space.

You cannot wave your arms and say: that particle is precisely at location (x,y,z).

Sure. But again that is a statement about a probability density function. It does not imply that the sample space is not real valued.

The mathematical model involves a continuum

To me, this is the actual key point. The mathematical model involves a continuum, and the model makes good physical predictions. As far as we can tell the continuum is a valid description of the world.

I am not aware of any experimentally confirmed models that avoid the continuum assumption. So I think that it is worth taking the assumption seriously.

 

[Matterwave]

There are many many theorems dealing with $\mathbb{R}$, $\mathbb{C}$ and other complete spaces that we (implicitly or explicitly) make use of in physics.

For example, a key property of Hilbert spaces is that they are complete. Manifolds are defined via charts mapping open subsets to $\mathbb{R}^n$ or $\mathbb{C}^n$. Etc. etc.

Even spacetime quantization methods like Loop Quantum Gravity make use of $\mathbb{R}$ and $\mathbb{C}$.

These are just some thoughts I had while thinking on uncountable vs countable infinities. It's not meant as an argument to support or refute others in this thread.

 

[PeroK]

Sure but who said anything about physically realizing them as a requirement ?

Can any scenario that contains infinities be physically realized?

It's a good question. I think the difference is in the conclusions you draw. We can talk about uniform circular motion, say, and even though that is never a totally exact description, it's essentially correct. The problem in your scenario is that you infer that you really can do something with probability zero. If you say simply that the rational numbers have measure zero, then no one can argue. But, if you use that to infer a highly dubious paradox in the real world - and emphasise with your use of language that you really can do something that has zero probability - then that is essentially incorrect (IMO).

In other words, the mathematics of the infinite is extremely useful and many conclusions map closely to what we can do physically. But, there are cases where the mathematics of the infinite leads to conclusions that are not physically realisable. Circular motion is physically realistic; Hilbert's Hotel is not.

 

[PeroK]

QM says that the probability distribution of a point particle has a non-zero standard deviation. It does not say that the sample space is discrete. In fact, in QM the position of a point particle is explicitly modeled as a continuous probability density function on a real-valued sample space.

The point is simply that you cannot, even theoretically, use the position of a point particle to identify a point in spacetime. And, by implication, you cannot use any physical object to identify a point in spacetime. It's not clear, therefore, how you can choose a point in the spacetime continuum.

If you say "just by thinking of one", then that's valid up to a point. We can imagine the origin and the point $(t, x, y, z)$ for some specific values of the coordinates. But, a typical point in the continuum requires an infinite amount of information. It's not clear to me how you can hold an infinite amount of information in your head?

I don't see how your brain could possibly choose a non-computable real number. There is explicitly no algorithm you could use. You certainly couldn't do it using a computer.

I just don't see any way to choose a real number, other than from a finite subset.

You say you can do it? Perhaps I can't argue with that. But, I certainly can't do it with my finite brain.

 

[A.T.]

But, if you use that to infer a highly dubious paradox in the real world - and emphasise with your use of language that you really can do something that has zero probability - then that is essentially incorrect (IMO).

I have no problem with things happening in the real world that have zero probability, if they are one of infinitely many equally likely outcomes, from which one had to happen. Every outcome having zero probability doesn't mean that every outcome is impossible, because one of them has to happen.

Here, zero probability for me just means that the outcome is impossible to guess, or impossible to repeat with finite attempts.

 

[PeroK]

I have no problem with things happening in the real world that have zero probability, if they are one of infinitely many equally likely outcomes, from which one had to happen. Every outcome having zero probability doesn't mean that every outcome is impossible, because one of them has to happen.

Here, zero probability for me just means that the outcome is impossible to guess, or impossible to repeat with finite attempts.

Give me an example of something that is possible with zero probability. Perhaps we could organise a bet on it?

 

[A.T.]

Give me an example of something that is possible with zero probability.

The key condition is:

... if they are one of infinitely many equally likely outcomes, from which one had to happen.

Since we don't know if infinite sets even exist in the real world, it's not possible to give a specific example.

 

[PeroK]

The key condition is:

Since we don't know if infinite sets even exist in the real world, it's not possible to give a specific example.

In other words, you're bs-ing me!

 

[A.T.]

In other words, you're bs-ing me!

No, you just have to read the whole sentence.

 

[Hornbein]

Do we know for sure that infinite sets (e.g. all planets) are not physically realizable?

It is believed that our Universe is infinite. This is consistent with the best measurements.

 

[Hornbein]

Give me an example of something that is possible with zero probability. Perhaps we could organise a bet on it?

In my opinion a reasonable way to apply measure theory is to have all events have zero probability. We are interested mainly in infinite subsets of the space. For example, coin flips. If one includes the exact real number of the time and location of the flip then all events have zero probability.

If the Universe is infinite then each planet has zero probability. Some physicists deny this, my stance is that no one really knows and I'm not interested in debating it.

Mostly my point is that in infinite sets the "it happened so the probability must have been greater than zero" is not correct. Measure zero is not the same as impossible.

I have long found it amusing that the subset of prime numbers has measure zero.

 

[Herman Trivilino]

TL;DR: How does this make sense - ''Any finite number divided by infinity is to near as to nothing as makes no odds, so the average population of the Universe can be said to be zero.''

I was reading Hitchhiker's Guide to the Galaxy

You're trying to make sense out of whimsy. The author is having fun with you!

From a google search:

"Radio origin: March 1978 (broadcast as a BBC radio comedy series before it became a book)."

Emphasis on comedy added.

Plus, if you want to get serious, and I can't fault you for that, our understanding of cosmology has progressed since 1978. The descriptions of the universe in 1978 compared to 2026 are entirely different.

 

[PeroK]

In my opinion a reasonable way to apply measure theory is to have all events have zero probability. We are interested mainly in infinite subsets of the space. For example, coin flips. If one includes the exact real number of the time and location of the flip then all events have zero probability.

I guess you are entitled to your opinion. But, that is not the way probability theory is taught or studied as an academic discipline.

 

[Dale]

The point is simply that you cannot, even theoretically, use the position of a point particle to identify a point in spacetime.

Why not? If your claim is based on technological limitations then that isn’t an “even theoretically” argument. If your claim is based on the uncertainty principle then you are arguing about the probability density rather than the sample space. Then theoretically all you have to do is physically sample from the distribution.

I think your “even theoretically” goes too far and undermines your claim. “Practically” I would agree.

a typical point in the continuum requires an infinite amount of information.

That isn’t correct. Writing down a non-computable number requires an infinite amount of information. But I am explicitly not talking about writing or storing the number. I am talking about the physical existence of the continuum itself (and points in it) as part of the physical state itself.

I don't see how your brain could possibly choose a non-computable real number. …

I just don't see any way to choose a real number, other than from a finite subset.

These are two very different claims. The computable real numbers is not a finite subset.

The issue is that our experimentally verified models rely heavily on the continuum hypothesis. So I think that we have to take that assumption seriously. Despite our inability to store non-computable real numbers, despite our technological limitations, all of our evidence indicates that many physical quantities (such as position) are continuous.

 

[Hornbein]

I guess you are entitled to your opinion. But, that is not the way probability theory is taught or studied as an academic discipline.

I have a masters degree in probability and statistics. I guess you are entitled to your opinion.

 

[PeroK]

Why not? If your claim is based on technological limitations then that isn’t an “even theoretically” argument. If your claim is based on the uncertainty principle then you are arguing about the probability density rather than the sample space. Then theoretically all you have to do is physically sample from the distribution.

In QM, a particle does not have a position. The wave-function (technically its modulus squared) is not a probability density function for where the particle is. It isn't anywhere in a classical sense.

Even if you make a measurement of position (although even that is a slighty tricky concept in QM), all you do is localise the wave-function somewhat. It's still not in a definite position. It's just more likely to be found in a finite volume.

Despite our inability to store non-computable real numbers, despite our technological limitations, all of our evidence indicates that many physical quantities (such as position) are continuous.

States of definite position are part of the mathematical model (Dira Delta functions) but explicitly not physically realisable states. See, for example, Griffiths section 2.4.

 

[PeroK]

I have a masters degree in probability and statistics. I guess you are entitled to your opinion.

Show me a reference and I'll admit I'm wrong. I.e. the theory of probability where all events have probability zero.

 

[Hornbein]

Show me a reference and I'll admit I'm wrong. I.e. the theory of probability where all events have probability zero.

The real numbers [0,1] with a uniform distribution.

 

[Dale]

In QM, a particle does not have a position.

But it does have a probability distribution of position, and the sample space for that probability distribution is the reals.

The wave-function (technically its modulus squared) is not a probability density function for where the particle is.

But the wavefunction is a function over the continuum.

 

[PeroK]

But it does have a probability distribution of position,

Not in QM. Only in classical statistical mechanics.

and the sample space for that probability distribution is the reals.

In QM, the sample space is determined by the measurement apparatus. You can't, even theoretically, get a single real number as the position of a particle. That is not a valid outcome in QM.

But the wavefunction is a function over the continuum.

The wavefunction is part of the mathematical model - it is defined for all points in space. You claim you can get an arbitrary real number from identifying the position of a particle. The wave-function is defined everywhere - although it may negligible outside a small volume.

I'm not claiming that the real numbers don't exist as a mathematical set. I claim that you cannot produce arbitrary real numbers from events in spacetime. If you could, then any potential theories where spacetime is discrete could immediately be dismissed - as you would already have disproved them. It's not a question of the continuum being a working model. It's a question of the continuum being absolutely necessary, with no possibility of a theory that avoids it.

 

[A.T.]

Yes, there is a distribution on the natural numbers where, for example: $p(1) = 1$ and $p(n) = 0, \ n \ne 1$.

Or, where: $\forall n: \ p(n) = \dfrac{1}{2^n}$.

What if all natural numbers are equally likely?

 

[Matterwave]

What if all natural numbers are equally likely?

By this I take you to mean there is a valid (satisfies the Kolmogorov axioms) probability distribution over $\mathbb{N}$ such that every element $n\in\mathbb{N}$ is assigned an equal probability in $[0,1]$.

We can show that this is impossible.

Given any $c\in(0,1]$ ($c\neq0$) pick a number $n>\lfloor 1/c \rfloor + 1$ and you'll see that the probability over the first $n$ numbers exceeds 1 so this is not a valid probability distribution.

If you take $c=0$, by sigma additivity you will find $P(\mathbb{N})=0$ which is also not a valid probability distribution.

Therefore, no such distribution exists.

 

[Matterwave]

In my opinion a reasonable way to apply measure theory is to have all events have zero probability. We are interested mainly in infinite subsets of the space.

Sounds reasonable. Your example in post #55 seems like a reasonable application of this. But the infinity you needed there is uncountable.

For example, coin flips. If one includes the exact real number of the time and location of the flip then all events have zero probability.

Aren't coin flips usually given as an example of a discrete event space? What's your event space? It's not {heads, tails} as usual?

If the Universe is infinite then each planet has zero probability.

Zero probability of what? What events are you measuring over the planets?

Some physicists deny this, my stance is that no one really knows and I'm not interested in debating it.

Debating what?

Mostly my point is that in infinite sets the "it happened so the probability must have been greater than zero" is not correct. Measure zero is not the same as impossible.

It seems to me that you need an uncountable infinity for this argument. I would (genuinely) appreciate an example where I'm wrong.

I have long found it amusing that the subset of prime numbers has measure zero.

What's this mean? Measure 0 relative to what (in what sigma algebra)?

Do you perhaps mean relative to the Lebesgue measure over $\mathbb{R}$? But note that the Lebesgue measure is not a probability measure since it's not normalized to 1.