Boltzmann brains are a matter of speculation and not observation. — wonderer1

You're right that it's not a matter of observation (and perhaps that my binary distinction is imprecise), but it's wrong to suggest that it's as simple as speculation. Rather it's a consequence of our best understandings of quantum mechanics and thermodynamics.

So either we are most likely Boltzmann brains or our best understandings of quantum mechanics and thermodynamics is mistaken.

Given that the scientific evidence supports our best understandings of quantum mechanics and thermodynamics, the scientific evidence doesn't support the claim that we are not likely to be Boltzmann brains.

So what justifies your claim that we are not likely to be Boltzmann brains? Is it just "common sense" or "intuition"? Are "common sense" and "intuition" more reasonable than scientific evidence?

Are not all theories interpretations? Are our memories of what we have learned of science and mathematics also illusions? If so, then how can we justifiably use them to support any conclusions at all? — Janus

See here.

It pays to remember that scientific theories, and science generally, only tell us how to make sense of how things appear to be to ordinary humans. — Janus

They tell us how to make sense of how things appear to us. Whether or not we are ordinary humans or Boltzmann brains is the very question being considered.

it is a fact that some interpretations of our current scientific theories entail that we are more likely to be Boltzmann brains than ordinary humans. — Janus

Not just some, but the leading theories.

And given that there is a brain, the longer it persists the less likely it is to be merely a quantum fluctuation. — Banno

But still more likely than not being a Boltzmann brain.

It's odd to me when one exclaims that they are more likely to be a philosophical tool of thinking than a human. — creativesoul

It is a fact that our current scientific theories entail that we are more likely to be Boltzmann brains than ordinary humans.

It's certainly counter-intuitive, but then so is much of science. I won't claim that my intuitions ought take precedence over scientific evidence.

It is impossible for a human to not be a human. — creativesoul

And it's impossible for a Boltzmann brain to not be a Boltzmann brain, or for a horse to not be a horse.

What of it?

As if basing one's beliefs on empirical evidence were not an act of faith... If you are a Boltzmann brain, what are the chances of your having just happened to have imagined into being a world that exactly corresponds to the actual world? You happened to drop into existence in a way that allows you to realise you are a Boltzmann brain... — Banno

There are, broadly speaking, four possibilities:

1. We are Boltzmann brains and our scientific theories are mostly correct
2. We are Boltzmann brains and our scientific theories are mostly incorrect
3. We are not Boltzmann brains and our scientific theories are mostly correct
4. We are not Boltzmann brains and our scientific theories are mostly incorrect

If our scientific theories are mostly correct then either (1) or (3) is the case, with (1) being most likely (as per those very scientific theories).

So one of these is true:

a. We are most likely Boltzmann brains (1 or 3)
b. Our scientific theories are mostly incorrect (2 or 4)

You seem to be arguing for the paradox after the paradox has been dismissed. — Mark Nyquist

I'm arguing that Philosopher19 doesn't understand set theory, and that his attempted "solution" to the Russell paradox makes no sense.

The Russell paradox has already been "solved": see ZFC, for example.

I will continue to take the world as being pretty much as it appears. — Banno

And you're welcome to do so. But it's a faith, not something supported by empirical evidence.

If. — Banno

Yes, and our current scientific theories suggest that the "if" is true.

And each time you reply, that chance shrinks, and not just a little bit, but by a truely extraordinary quantity. — Banno

And it's still always the case that the probability that I am a Boltzmann brain is greater than the probability that I am not a Boltzmann brain.

And you can't have a paradox if the defined mathematical object does not exist. — Mark Nyquist

You misunderstand the paradox.

Naive set theory allows the Russell set. The Russell set is a contradiction. Therefore, naive set theory is inconsistent.

That's all it is.

We don't need to bring up mathematical realism.

But we are not talking about whether there are any Boltzmann brains, so much as whether you are a Boltzmann brain.

And the chances of that continue to shrink. — Banno

And it's always the case that the probability that I am a Boltzmann brain is greater than the probability that I am not a Boltzmann brain. Even if we were to continue this discussion for 1,000 years.

It seems to me there is a problem with you being a Boltzmann brain and yet so predictable. Should we expect that if you are a BB? Where's the batty? — wonderer1

See here.

Or maybe we will reach agreement that there is something quite specious about this argument. — Banno

The argument is valid:

1. There are far more long-lived Boltzmann brains than long-lived humans
2. I am long-lived
3. Therefore, I am more likely to be a Boltzmann brain than a human

Our current scientific theories suggest that (1) is true.

It would be strange to suggest that our current scientific theories are probably wrong simply because you don't like the conclusion.

Unless you have some actual evidence against either (1) or (3), your rejection of the argument is simply a matter of faith (as I said before).

Again, the longer you persist, the more likely that you are an ordinary brain. — Banno

But never as likely that I am a Boltzmann brain.

In an infinite space of infinite possibilities, there are presumably an infinity of non-Boltzmann brains, so I dont; see that you have grounds for asserting that they are less common than ordinary brains... — Banno

From the Wikipedia article:

In a single de Sitter universe with a cosmological constant, and starting from any finite spatial slice, the number of "normal" observers is finite and bounded by the heat death of the universe. If the universe lasts forever, the number of nucleated Boltzmann brains is, in most models, infinite; cosmologists such as Alan Guth worry that this would make it seem "infinitely unlikely for us to be normal brains". One caveat is that if the universe is a false vacuum that locally decays into a Minkowski or a Big Crunch-bound anti-de Sitter space in less than 20 billion years, then infinite Boltzmann nucleation is avoided. (If the average local false vacuum decay rate is over 20 billion years, Boltzmann brain nucleation is still infinite, as the universe increases in size faster than local vacuum collapses destroy the portions of the universe within the collapses' future light cones). Proposed hypothetical mechanisms to destroy the universe within that timeframe range from superheavy gravitinos to a heavier-than-observed top quark triggering "death by Higgs".

If no cosmological constant exists, and if the presently observed vacuum energy is from quintessence that will eventually completely dissipate, then infinite Boltzmann nucleation is also avoided.

In no case is there an infinity of non-Boltzmann brains. In some cases there are an infinity of Boltzmann brains.

To avoid the Boltzmann brain hypothesis you need to hope that either there is no cosmological constant or that the universe is a false vacuum.

But it hasn't disappeared yet, and the longer it doesn't disappear the less likely that it is a quantum fluctuation. — Banno

It doesn't follow that I am most likely not a Boltzmann brain. It only follows that the probability that I am a Boltzmann brain gets smaller as the time increases. But due to the sheer number of Boltzmann brains, it is always the case that the probability that I am a Boltzmann brain is greater than the probability that I am not a Boltzmann brain.

Yes, it is. You claimed that:

1. Because most Boltzmann brains are short-lived then if I am long-lived then I am probably not a Boltzmann brain.

This can be generalised as:

2. Because most X are Y then if not Y then probably not X

Substituting in something else for X and Y:

3. Because most red balls have no stripe then if the ball has a stripe then it is probably not a red ball

My example above shows why (3) is false, and so why (2) is false, and so why (1) is false.

You've argued that there is a set of all sets, U.

If A is the set {A} then A is a member of both A and U.

There are 1,000 red balls with no green stripe.
There are 100 red balls with a green stripe.
There are 10 blue balls with a green stripe.

Your argument is that because most red balls have no green stripe then if my ball has a green stripe then it is most likely not a red ball. That is wrong. If my ball has a green stripe then it is most likely a red ball.

So:

There are 1,000 short-lived Boltzmann brains.
There are 100 long-lived Boltzmann brains.
There are 10 long-lived human brains.

Most Boltzmann brains are short-lived brains, but most long-lived brains are Boltzmann brains. Therefore if I am a long-lived brain then I am most likely a Boltzmann brain.

Yep, I think that's right.

But there is a further step. There are far more batty brains than Boltzmann brain. But there is a further step. Supose you are a quantum fluctuation, having just popped into existence last Tuesday. The chances of you persisting into the next few seconds are vanishingly small. Chances are the world around you is ephemeral, and will disappear, or at the least not continue in a coherent fashion.

And yet for us, the world continues on in a regular and predictable fashion. Well, at least outside of dormitory kitchens.

And that is the argument from Batty Brains - that the world persists shows that it is very unlikely that you are a Boltzmann brain.

That seems to be how the argument goes. — Banno

That would be an invalid argument.

Assume that there are 1,000 short-lived Boltzmann brains, 100 long-lived Boltzmann brains, and 10 long-lived human brains. Most Boltzmann brains are short-lived, but most long-lived brains are Boltzmann brains.

From the Wikipedia article:

In Boltzmann brain scenarios, the ratio of Boltzmann brains to "normal observers" is astronomically large. Almost any relevant subset of Boltzmann brains, such as "brains embedded within functioning bodies", "observers who believe they are perceiving 3 K microwave background radiation through telescopes", "observers who have a memory of coherent experiences", or "observers who have the same series of experiences as me", also vastly outnumber "normal observers". Therefore, under most models of consciousness, it is unclear that one can reliably conclude that oneself is not such a "Boltzmann observer", in a case where Boltzmann brains dominate the universe. Even under "content externalism" models of consciousness, Boltzmann observers living in a consistent Earth-sized fluctuation over the course of the past several years outnumber the "normal observers" spawned before a universe's "heat death".

As stated earlier, most Boltzmann brains have "abnormal" experiences; Feynman has pointed out that, if one knows oneself to be a typical Boltzmann brain, one does not expect "normal" observations to continue in the future. In other words, in a Boltzmann-dominated universe, most Boltzmann brains have "abnormal" experiences, but most observers with only "normal" experiences are Boltzmann brains, due to the overwhelming vastness of the population of Boltzmann brains in such a universe.

As you seem unwilling to accept facts about maths, let's use your own reasoning against you.

1. x is a member of A if and only if x is a member of x
2. Let x = B
3. B is a member of A if and only if B is a member of B

But according to your reasoning, (3) is a contradiction. Therefore (1) is a contradiction.

Your axioms lead to an inverse Russell paradox.

We can resolve this either by allowing that B is a member of both A and B, or by not allowing a set to be a member of itself.

But none of this addresses the fundamental problem with this discussion, and that is that this is Russell’s paradox:

1. x is a member of R if and only if x is not a member of x
2. Let x = R
3. R is a member of R if and only if R is not a member of R

You haven’t proved that (3) is not a contradiction. And you can’t because it is.

The solution to the paradox is known: construct a set theory with axioms that do not entail (1).

ZFC does this by not allowing a set to be a member of itself. New Foundations does this by restricting which sorts of sets can be members of themselves.

Am I the one that's confused? So it's not the case that in A it's a member of one thing and in B it's a member of another thing? So it's not the case that in A it's a member of itself and in B it's not a member of itself? — Philosopher19

Yes, you’re confused. A is a member of A and B. 1 is a member of N and R. That’s all there is to it.

I have not disagreed with scenario 2. I have said that in B, A is not a member of itself precisely because it is a member B (as opposed to itself), and in A, A is a member of itself. — Philosopher19

And you’re confused. It’s not the case that “in A” it’s a member of one thing and “in B” it’s a member only of something else.

It’s the case that in scenario 2, A is a member of A and B.

Explain to me the difference here, and why you disagree with scenario 2:

Scenario 1
B = {0, A}, where A = {1}

Scenario 2
B = {0, A}, where A = {A}

So why is it that A can be both a member of B and C but not a member of both A and B?

And in B, A is not a member of itself. — Philosopher19

N is the set of natural numbers.
R is the set of real numbers.

Every natural number is a member of both N and R (every natural number is both a natural and a real number).

We don't say "in R, 1 is not a member of N".

Just take a math lesson or two.

There's no such thing as "in A" and "in B".

It is just the case that the symbol "A" is defined recursively as "{A}" and that the symbol "B" is defined as "{A, 0}", which is the same as "{{A}, 0}" given the recursive definition of "A".

$A=\{A\}\\B=\{A,0\}=\{\{A\},0\}$

So once again, in B, is A a member of itself or not a member of itself? — Philosopher19

Both a member of itself and a member of B.

In B, A is not a member of both A and B. — Philosopher19

Yes it is.

In the case of B = {A, 0}, is A a member of A/itself, or is A a member of B/non-itself? — Philosopher19

Both

When a set is a member of itself, it is not a member of another set — Philosopher19

And this is a fundamental misunderstanding of set theory.

If A = {A} and if B = {A, 0} then A is a member of A and a member of B.

But a set can either be a member of itself or a member of other than itself. — Philosopher19

A set can be a member of more than one set. You just don't understand the basics of set theory.

You should really take a few math lessons before you start telling mathematicians that they're wrong about maths.

What you say in response doesn't prove that Russell's paradox isn't a contradiction.

1. x is a member of R if and only if x is not a member of x
2. Let x = R
3. R is a member of R if and only if R is not a member of R

(3) is a very obvious contradiction. You don't even need to know maths to see that.

A is a member of both A and B.

I'll explain it to you in non-math terms:

I am a member of the football team and a member of the tennis team.

These are two different claims:

1. I am not a member of the football team
2. I am a member of a non-football team

(1) is false and (2) is true.

@Philosopher19

Regarding Russell's paradox, it is simply this:

1. $x$ is a member of $R$ if and only if $x$ is not a member of $x$.

Is $R$ a member of $R$?

Either answer entails a contradiction, and so (1) is a contradiction. Given that naive set theory entails (1), naive set theory is shown to be inconsistent.

@Philosopher19

These are two different claims:

1. A is not a member of itself
2. A is a member of some other set

Given this:

$A=\{A\}\\B=\{A,0\}$

(1) is false and (2) is true.

The Moving Spotlight Theory? Seems to be a hybrid view that allows for both eternalism and a dynamic time.

That would assume an eternalist view of time, in which time is treated much like a length, or as another spatial dimension. Whereas - prior to the untimely demise of this discussion - I was seeking to explore the limitations of eternalism, such as its logical omission of progress, happening or motion; characteristics that I consider to be absent from eternalism but logically aligned with the opposing view of presentism. However, many eternalists disagree. — Luke

I assume you're also against the growing block theory of time?

If you're arguing for presentism then this might be interesting:

Presentists Should Not Believe in Time Travel

The general gist being that the very concept of traveling to the past depends on the past existing in some sense as a location to travel to, and so requires either the growing block universe or eternalism.

If presentism is correct then any supposed time machine would work by rebuilding the universe into a facsimile of one of its past states, which isn't really time travel.

there exists a history — Luke

If there exists a history then presentism is false.