I don't know anyone who has said that all others are ignorant. You are ignorant on the subject. That doesn't entail that others are ignorant on it. Indeed, there are people who critique classical set theory who are extremely knowledgeable about it. Critiques of set theory are quite fair game and bring profound insights into the subject. But those are knowledgeable, responsible and thoughtful critiques. And better yet, they are critiques that are followed up with actual mathematical alternatives to classical set theory.
— TonesInDeepFreeze
I didn't say all others are ignorant. I just said there are people who are like this. I did not specify who. — Philosopher19
Did you read anything from the link I gave you? — Philosopher19
It is blatantly contradictory for x to be both x and not x. — Philosopher19
It is blatantly contradictory for a set to be both a member of itself and not a member of itself. — Philosopher19
Yet you want to persist by saying things like the above. — Philosopher19
Once again:
It is blatantly contradictory for x to be both x and not x. It is blatantly contradictory for a set to be both a member of itself and not a member of itself. — Philosopher19
It is blatantly contradictory for x to be both x and not x. It is blatantly contradictory for a set to be both a member of itself and not a member of itself.
Who would reject this but the contradictory/unreasonable/irrational/absurd/insincere? — Philosopher19
is there real math behind the north pole of the riemann sphere? — Mark Nyquist
Says the guy who tried arguing Cardinalities don't have size yet they do, as per the theorem I produced to prove you wrong. Since some Cardinalities are greater than others, we can say that some infinities are larger or even smaller than others. That you got your ass handed to you by someone suffering from "dunning-kruger" — Vaskane
just goes to show you've got a lot to learn, but I'm happy to correct you any time pal. — Vaskane
I'm happy to correct you any time pal. — Vaskane
Here's actually some advice to all non-mathematicians (from a non-mathematician):
If you really can ask an interesting foundational question that isn't illogical or doesn't lacks basic understanding, you actually won't get an answer... because it really is an interesting foundational question!
Yet if the answer is, please start from reading "Elementary Set Theory" or something similar then yes, you do have faulty reasoning. — ssu
The "math boys" here at the forum tend to respond with 'go read some math texts' to anyone who disagrees with them on fundamental principles. — Metaphysician Undercover
That's like telling an atheist to go read some theology, as if this is the way to turn the person around. — Metaphysician Undercover
The "math boys" here at the forum tend to respond with 'go read some math texts' to anyone who disagrees with them on fundamental principles — Metaphysician Undercover
. I am more concerned with what issues you solve with your beliefs. — Lionino
It seems to be a collection of semantic games — Lionino
You have mentioned, for example, that the limit concept is flawed, although it works well most of the time. But I don't recall your argument beyond that point. A more complete knowledge of space and time and points and continuity? Oh yes, something about the Fourier transform and the Uncertainty principle. What are your suggestions to fix that up? Intuitive mathematics? Remind me where doing something specific makes it better. — jgill
Correct! Indeed that is a crucial point that is used in an important proof I gave you. — TonesInDeepFreeze
'There exists a z such that for all y, y is a member of z' contradicts this instance of the axiom schema of separation: For all z, there is a x such that for all y, y is a member of x iff (y is a member of z and yis not a member of y). — TonesInDeepFreeze
That is not "once again". Previously you said that "a set cannot be both a member of itself and a member of other than itself". That is different from "a set cannot be both a member of itself and not a member of itself". — TonesInDeepFreeze
An item in a subset cannot be both a member of the subset and the set. If it is a member of the subset, it is a member of the subset. If it is a member of the set, it is a member of the set. — Philosopher19
Something cannot be both a member of itself and a member of other than itself at the same time. For example, take z to be any set that is not the set of all sets, and take v to be any set. The z of all zs is a member of itself as a z (as in in the z of all zs it is a member of itself). But it is not a member of itself in the v of all vs, precisely because in the v of all vs it is a member of the v of all vs as opposed to a member of itself. If we view the z of all zs as a z, it is a member of itself. If we view the z of all zs as a v, it is a member of the set of all sets. You can't view it as both a member of the z of all zs and a member of the v of all vs at the same time. That will lead to contradictions. In other words, we can't treat two different references as one (as in are we focused on the context of vs or the context of zs?) — Philosopher19
Then you're a dumbass for arguing with me when I was correct that Infinities do indeed have different sizes if that's your stance too. Either way, you're an "egregious" dipshit. "Oh, this guy is arguing the same thing as I am, and he's being comical, I should "correct," him about infinities not having sizes even though that's MY STANCE! Oh, wait, it's my stance, noone else can have it!" Eitherway, the fact is, you're a dumbass when it comes to communication. — Vaskane
An item in a subset cannot be both a member of the subset and the set. — Philosopher19
If it's a member of other than itself, this means that it's not a member of itself. — Philosopher19
I take you to be saying that a set cannot be a member of another set and also a member of itself. — TonesInDeepFreeze
Do you mean: If S is a subset of some set T and x is member of S, then x cannot be a member of T ? — TonesInDeepFreeze
That's incorrect. By the definition of 'subset', if S is a subset of T, and x is a member of S, then x is a member T. — TonesInDeepFreeze
Take z to be any set that is not the set of all sets, and take v to be any set. The z of all zs is a member of itself as a z (as in in the z of all zs it is a member of itself). But it is not a member of itself in the v of all vs, precisely because in the v of all vs it is a member of the v of all vs as opposed to a member of itself. If we view the z of all zs as a z, it is a member of itself. If we view the z of all zs as a v, it is a member of the set of all sets. You can't view it as both a member of the z of all zs and a member of the v of all vs at the same time. That will lead to contradictions. In other words, we can't treat two different references as one (as in are we focused on the context of vs or the context of zs?) — Philosopher19
distinguish between "member of self" and "not member of self" — Philosopher19
I am not sure if this is so much a problem with mathematics though as it is with how it gets applied to the sciences and philosophy. It seems to me that infinite divisibility might be worth investigating even if it doesn't accurately reflect "how things are." — Count Timothy von Icarus
I believe the solution to Russell's paradox is in here:
http://godisallthatmatters.com/2021/05/22/the-solution-to-russells-paradox-and-the-absurdity-of-more-than-one-infinity/ — Philosopher19
I mean that all makes sense, although my understanding was that the question of whether or not space-time is infinitely divisible was an open one. — Count Timothy von Icarus
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