There is no object called 'Infinity' in the sense you have been using it.
Here is a way to say what you want to say:
In mathematics, there are sets that are infinite but that have different cardinality from one another.
Better yet:
If x is infinite then there is a y that is infinite and y has greater cardinality than x. — TonesInDeepFreeze
There is no object called 'Infinity' in the sense you have been using it. — TonesInDeepFreeze
There is no x such that for all y, y is a member of x iff y is not a member of y. Proof: — TonesInDeepFreeze
the axiom schema of separation — 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
We do NOT claim that from "after each natural number there is a next number" and "there is no greatest natural number" that we can infer that there is a set of all the natural numbers. Indeed such an inference IS a non sequitur. And every mathematician and logician knows it is a non sequitur. So, we recognize that to have a set with all the natural numbers we need an AXIOM for that, which is NOT an inference. — TonesInDeepFreeze
Something cannot be both a member of itself and a member of other than itself at the same time. — Philosopher19
I'm not sure what you mean by "So, we recognize that to have a set with all the natural numbers we need an AXIOM for that, which is NOT an inference." — Philosopher19
So what semantic are mathematicians using when they use the world/label "infinite"? — Philosopher19
So what semantic are mathematicians using when they use the world/label "infinite"? — Philosopher19
I have an ability to understand concepts without even knowing of them — Vaskane
Great post, thanks. How do you prove then N is different size to P?Here is a finite definition of an infinite set: "A given set S is infinite iff there exists a bijective function between S and a proper subset of S." Furthermore, such a bijective function can be stated finitely.
Here is an example. Take the set of natural numbers ℕ = { 0, 1, ··· }. Now take a proper subset of ℕ containing only even the numbers, ℙ = { 0 , 2 , ··· }. These two are equinumerous because there is a bijective function f : ℕ → ℙ, given by f(n) = 2n.
The proof that "f" is bijective is finite. So is the proof that ℙ is a proper subset of ℕ. — DanCoimbra
In certain alternative set theories, there are sets that both members of themselves and of other sets. — TonesInDeepFreeze
We go in a circles, as it is with cranks. The crank makes false claims and terrible misunderstandings. Then the crank is corrected and their error is explained. Then the crank ignores all the corrections and just posts the false claims and misunderstanding again as if the corrections and explanations never existed. — TonesInDeepFreeze
Here's actually some advice to all non-mathematicians (from a non-mathematician):This isn't really the place to come to get people to agree with you. I think the math boys really did give you a good amount of feedback that would be hard to get anywhere else. So if you want to run something past us we'll tell you what we think and you can react accordingly. Most of what you say really irks a formally trained mathematician. — Mark Nyquist
How do you prove then N is different size to P? — Corvus
We go in a circles, as it is with cranks. The crank makes false claims and terrible misunderstandings. Then the crank is corrected and their error is explained. Then the crank ignores all the corrections and just posts the false claims and misunderstanding again as if the corrections and explanations never existed.
— TonesInDeepFreeze
Evidently, there's no point in continuing this discussion. — Philosopher19
If you believe your mathematics is free from contradictions or paradoxes — Philosopher19
I see no paradoxes or contradictions or foundational incompleteness in the beliefs that I uphold (mathematical or otherwise). — Philosopher19
I see no paradoxes or contradictions or foundational incompleteness in the beliefs that I uphold (mathematical or otherwise). — Philosopher19
That a set cannot be both a member of itself and a member of other than itself is the equivalent of saying that a shape cannot be both a square and a triangle — Philosopher19
From the point of the set N, it looks like it is. But from the point of the set P, it looks like it is only a half set to N. What's going on?We don't. He proved that they are the same size. — TonesInDeepFreeze
He did it again! He completely skipped recognizing the refutation given him. — TonesInDeepFreeze
By the way, we don't need to use temporal phrases such as "at the same time". Set theory does not mention temporality.
Then the rest of your z's and v's is irrelevant if it is supposed to refute the proofs I gave. — TonesInDeepFreeze
I will just say this. That a set cannot be both a member of itself and a member of other than itself is the equivalent of saying that a shape cannot be both a square and a triangle (I have taken out the "at the same time" and the effect is still the same). — Philosopher19
I don't think I'm the one that has been showing the disrespect — Philosopher19
I don't think I entered the discussion closed-minded or dogmatic. — Philosopher19
"expert" in the field — Philosopher19
They want to hold on to their paradoxical or contradictory theory — Philosopher19
incomplete — Philosopher19
act as though they are the knowledgeable ones whilst all others are ignorant — Philosopher19
What good is an expert in multishapism geometry that deals with the study of shapes such as round triangles and circular pentagons? — Philosopher19
Is it logically possible for a set to be both a member of itself and a member of other than itself? If it is a member of other than itself, then it is not a member of itself, is it? And if it is a member of itself, it is not a member of other than itself is it? — Philosopher19
All of them are here:
godisallthatmatters.com — Philosopher19
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 gave you a refutation. You started with insults — Philosopher19
You started with insults — Philosopher19
emotional or biased — Philosopher19
You have not yet answered:
Is it logically possible for a set to be both a member of itself and a member of other than itself? If it is a member of other than itself, then it is not a member of itself, is it? And if it is a member of itself, it is not a member of other than itself is it? — Philosopher19
And don't say to me something like "some set theories allow for this or that". — Philosopher19
There is no need to dance around anything. — Philosopher19
Is it logically possible for a set to be both a member of itself and a member of other than itself? If it is a member of other than itself, then it is not a member of itself, is it? And if it is a member of itself, it is not a member of other than itself is it?
— Philosopher19
Where is my response? Is it me who ignores you or you who ignores me? — Philosopher19
The relative consistency of those theories indicates that it is not contradictory that a set is a member of itself and also a member of other sets. — TonesInDeepFreeze
Is it logically possible for a set to be both a member of itself and a member of other than itself? If it is a member of other than itself, then it is not a member of itself, is it? And if it is a member of itself, it is not a member of other than itself is it? — Philosopher19
And don't say to me something like "some set theories allow for this or that". I'm asking a basic logical question that has a basic and straight forward answer. There is no need to dance around anything. Just deal with the main issue at hand. — Philosopher19
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