Michael

Yes, and our current scientific theories suggest that the "if" is true. And it's still always the case that the probability that I am a Boltzmann brai...

You misunderstand the paradox. Naive set theory allows the Russell set. The Russell set is a contradiction. Therefore, naive set theory is inconsisten...

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 ...

See here.

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 b...

But never as likely that I am a Boltzmann brain.

From the Wikipedia article: In no case is there an infinity of non-Boltzmann brains. In some cases there are an infinity of Boltzmann brains. To avoid...

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 ti...

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 ...

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 argumen...

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 ...

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...

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...

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.

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 s...

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?

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 natura...

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...

Both a member of itself and a member of B. Yes it is.

Both

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.

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 s...

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. Le...

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...

@"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? Eith...

@"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) ...

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

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...

No

https://www.bbc.co.uk/news/world-us-canada-68266447 The ironic thing is that article 5 has only ever been invoked once; when the USA was attacked on 9...

I can’t speak on the more scientific aspects of that paper, but on that final section, although it’s the case that any randomly selected brain is most...

Sure. Same with brains in a vat. I’m not saying they’re true, only that if our understanding of physics is correct then it’s most likely.

“I am Banno” and “I am a Boltzmann brain” are not in conflict. You are Banno, and if our physics is correct then you are also most likely a Boltzmann ...

The conclusion is that there is an external world that behaves according to the laws of physics but that we are most likely brains floating in a vacuu...

I’m not sure if I would. I just accept the existence of a material world and that my everyday experiences are of that material world as a matter of fa...

Certainly more than my thoughts but possibly not more than my thoughts and experiences. I can be surprised when I dream but it doesn’t follow that the...

Perhaps check out this article: Epistemological Problems of Perception

Perhaps at the very least it presupposes that solipsism is false. It need not presuppose the existence of a material world (e.g. it allows for idealis...

Extended real number line

The full quote, in the body, continues with: Sure, after Trump.

https://thephilosophyforum.com/discussion/5224/mathjax-tutorial-typeset-logic-neatly-so-that-people-read-your-posts/p1

A question for mathematicians: looking at what I've done above, can this be written as a matrix like this? \begin{bmatrix}0.1 & 0.01 & 0.001 & \cdots ...

Yes, there are an "infinite number" of infinite sets: \{0.1,\text{ }0.01,\text{ }0.001,\text{ }...\}\\\{0.2,\text{ }0.02,\text{ }0.002,\text{ }...\}\\...

The epistemological problem of perception seeks to understand the relationship between visual experiences and the external world objects that such exp...

On that I agree.

I've tidied up my comment. Perhaps you could explain where you think I've gone wrong?

If all it proves is that every T has the true and unprovable sentence "this sentence is true and unprovable" then it seems vacuous. Or does it prove t...