The set of all sets exists.
The set of all sets exists.
Axiom 1: S is the set of all sets
Axiom 2: The partial function ∈(x,y) is undefined for (x:=S,y) for similar reasons why division is undefined for (x,y:=0).
Axiom 3: The powerset of S is a fixed point. P(S) = S.
Bye, Cantor.
Edit: Add bindings " :="
Axiom 2: The partial function ∈(x,y) is undefined for (x:=S,y) for similar reasons why division is undefined for (x,y:=0).
Axiom 3: The powerset of S is a fixed point. P(S) = S.
Bye, Cantor.
Edit: Add bindings " :="
Last edited by Skepdick on Thu Jun 19, 2025 11:40 am, edited 2 times in total.
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Impenitent
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Re: The set of all sets exists.
i is the set of the temporarily unseen
join with set(i) and we'll find the aliens
-Imp
join with set(i) and we'll find the aliens
-Imp
Re: The set of all sets exists.
The set of all sets would effectively be a set that occurs through all sets as all sets for the most central thing in a hierarchy must occur through all things, thus leaving it a meaningless set as it is self similar and without contrast.
The metaphysical observation of this would be the totality argument:
There is only the totality and this totality has no comparison or it would not be the totality as something would be beyond it. Without comparison a thing cannot have the distinction that would allow it to exist as a thing, thus the totality is nothing.
Re: The set of all sets exists.
If you remove and deny the axiom of infinity (ZF-INF), and in this way render your custom axiomatization bi-interpretable with arithmetic, this particular custom axiom insist that there is a solution for n=2^n for some n in the natural numbers.
Such solution does not exist, however, purely for arithmetical reasons independent of set theory. This is provable by simple induction over the natural numbers. Hence, your custom axiom creates havoc in arithmetic.
Given the enormous size of S, the collision will occur in a nonstandard model of arithmetic, but that is still enough to invalidate the proof in the (standard) natural numbers as well. Arithmetic does not tolerate a solution for n=2^n anywhere at all.
So, you would really need to further hack the natural numbers for this problem, because otherwise you have no model in arithmetic and therefore also no model in ZF-INF either, and therefore no model for your custom axiomatization. If a fragment of your custom axiomatization is inconsistent, then the entire axiomatization has no model.
In my impression, the arithmetic theory associated to a fragment of your custom set axiomatization will go up in smoke, if you add this custom axiom.
Re: The set of all sets exists.
That's not true. The natural numbers have a least element. This doesn't. It has a maximal element.
Re: The set of all sets exists.
S has a cardinality, card(S). So, does P(S). In your custom axiomatization, card(S) = card(P(S)) = 2^card(S). How do you make that work in the corresponding fragment of arithmetic theory?Skepdick wrote: ↑Thu Jun 19, 2025 6:55 amThat's not true. The natural numbers have a least element. This doesn't. It has a maximal element.
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Magnus Anderson
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Re: The set of all sets exists.
False premise. No such set exists.
This redefines the membership relation in order to hide the fact that S does not actually exist.Axiom 2: The partial function ∈(x,y) is undefined for (S,x) for similar reasons why division is undefined for (x,0).
This redefines the word "powerset" for the same exact reason.Axiom 3: The powerset of S is a fixed point. P(S) = S.
The art you're practicing is called "sophistry"
Re: The set of all sets exists.
OK... by the convention you've just established I reject Peano's axiom
False premise. No such natural number exists.0 is a natural number.
The same way we redefine the relation x / 0 in order to hide the fact that 0 does not actually exist.Magnus Anderson wrote: ↑Thu Jun 19, 2025 7:35 am This redefines the membership relation in order to hide the fact that S does not actually exist.
OK, so given that 0 doesn't exist; what is 1 the powerset of?Magnus Anderson wrote: ↑Thu Jun 19, 2025 7:35 am This redefines the word "powerset" for the same exact reason.
Oh, I don't know, buddy.
It's just logic.
The set of all sets is the unit.
It's 1
Last edited by Skepdick on Thu Jun 19, 2025 10:16 am, edited 1 time in total.
Re: The set of all sets exists.
He axiomitizes S into existence. I am actually not particularly critical about this particular custom axiom.
He just says that S is a proper class:Magnus Anderson wrote: ↑Thu Jun 19, 2025 7:35 amThis redefines the membership relation in order to hide the fact that S does not actually exist.
https://en.m.wikipedia.org/wiki/Class_(set_theory)
This is the axiom where I also see serious issues.Magnus Anderson wrote: ↑Thu Jun 19, 2025 7:35 amThis redefines the word "powerset" for the same exact reason.
It implies that the cardinality of S, say n, is equal to the cardinality of the power set of n, that is 2^n.
So, it says that there exists a natural number n for which n=2^n. You can prove by induction over the natural numbers that such n cannot exist.
This axiom in his customized set theory leads to breakage in arithmetic theory.
Re: The set of all sets exists.
I hope and expect any Cantorian to have issues. Fuck Cantor. Fuck that fucking relativist.
Nothing of such sorts is implied. It implies The powerset function has a fixed point.
Which implies that S = P(S) = P(P(S)) = P(P(P(S)) ...
You can give them "friendly labels" if you want 0 = 1 = 2 = 3 = 4 ...
You are welcome to tell me what the inverse of the powerset operation is.
If you want consistent AND complete Mathematics you can't have irreversible operations!
Irreversibility is information loss.
Last edited by Skepdick on Thu Jun 19, 2025 11:51 am, edited 1 time in total.
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Magnus Anderson
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Re: The set of all sets exists.
Oh really? Does the axiom of identity exist?Magnus Anderson wrote: ↑Thu Jun 19, 2025 10:33 am You can't "axiomitize" things into existence.
If something does not exist, it does not exist. You can't bring it into existence.
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Magnus Anderson
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Re: The set of all sets exists.
M̶a̶n̶g̶n̶u̶s̶ ̶A̶n̶d̶e̶r̶s̶o̶n̶ Skepdick wrote: ↑Thu Jun 19, 2025 10:38 am That's not really true. But in order to actually see it, one has to stop being a literalist for a bit.
Last edited by Skepdick on Thu Jun 19, 2025 10:42 am, edited 1 time in total.
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Magnus Anderson
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Re: The set of all sets exists.
You can't see something that does not exist. And if you're talking about something that does exist, then you're merely calling something else the same name as something that does not exist.