Philosophy Now Forum

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 " :="

i is the set of the temporarily unseen

join with set(i) and we'll find the aliens

-Imp

Skepdick wrote: ↑Wed Jun 18, 2025 3:35 pm Axiom 1: S is the set of all sets
Axiom 2: The partial function ∈(x,y) is undefined for (S,x) for similar reasons why division is undefined for (x,0).
Axiom 3: The powerset of S is a fixed point. P(S) = S.

Bye, Cantor.

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.

Skepdick wrote: ↑Wed Jun 18, 2025 3:35 pm Axiom 3: The powerset of S is a fixed point. P(S) = S.

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.

godelian wrote: ↑Thu Jun 19, 2025 2:28 am

Skepdick wrote: ↑Wed Jun 18, 2025 3:35 pm Axiom 3: The powerset of S is a fixed point. P(S) = S.

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.

That's not true. The natural numbers have a least element. This doesn't. It has a maximal element.

Skepdick wrote: ↑Thu Jun 19, 2025 6:55 am

godelian wrote: ↑Thu Jun 19, 2025 2:28 am

Skepdick wrote: ↑Wed Jun 18, 2025 3:35 pm Axiom 3: The powerset of S is a fixed point. P(S) = S.

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.

That's not true. The natural numbers have a least element. This doesn't. It has a maximal element.

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: ↑Wed Jun 18, 2025 3:35 pm Axiom 1: S is the set of all sets

False premise. No such set exists.

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 membership relation in order to hide the fact that S does not actually exist.

Axiom 3: The powerset of S is a fixed point. P(S) = S.

This redefines the word "powerset" for the same exact reason.

The art you're practicing is called "sophistry"

Magnus Anderson wrote: ↑Thu Jun 19, 2025 7:35 am

Skepdick wrote: ↑Wed Jun 18, 2025 3:35 pm Axiom 1: S is the set of all sets

False premise. No such set exists.

OK... by the convention you've just established I reject Peano's axiom

0 is a natural number.

False premise. No such natural number exists.

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.

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 word "powerset" for the same exact reason.

OK, so given that 0 doesn't exist; what is 1 the powerset of?

Magnus Anderson wrote: ↑Thu Jun 19, 2025 7:35 am The art you're practicing is called "sophistry"

Oh, I don't know, buddy.

It's just logic.

The set of all sets is the unit.

It's 1

Magnus Anderson wrote: ↑Thu Jun 19, 2025 7:35 am

Skepdick wrote: ↑Wed Jun 18, 2025 3:35 pm Axiom 1: S is the set of all sets

False premise. No such set exists.

He axiomitizes S into existence. I am actually not particularly critical about this particular custom axiom.

Magnus Anderson wrote: ↑Thu Jun 19, 2025 7:35 am

Skepdick wrote: ↑Wed Jun 18, 2025 3:35 pm 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 membership relation in order to hide the fact that S does not actually exist.

He just says that S is a proper class:

https://en.m.wikipedia.org/wiki/Class_(set_theory)

Magnus Anderson wrote: ↑Thu Jun 19, 2025 7:35 am

Skepdick wrote: ↑Wed Jun 18, 2025 3:35 pm Axiom 3: The powerset of S is a fixed point. P(S) = S.

This redefines the word "powerset" for the same exact reason.

This is the axiom where I also see serious issues.

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.

godelian wrote: ↑Thu Jun 19, 2025 10:16 am This is the axiom where I also see serious issues.

I hope and expect any Cantorian to have issues. Fuck Cantor. Fuck that fucking relativist.

godelian wrote: ↑Thu Jun 19, 2025 10:16 am It implies that the cardinality of S, n, is equal to the cardinality of the power set of n, 2^n.

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.

godelian wrote: ↑Thu Jun 19, 2025 10:16 am He axiomitizes S into existence. I am actually not particularly critical about this particular custom axiom.

You can't "axiomitize" things into existence.

If something does not exist, it does not exist. You can't bring it into existence.

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.

Oh really? Does the axiom of identity exist?

Skepdick wrote: ↑Thu Jun 19, 2025 10:10 am 0 does not actually exist.

That's not really true. But in order to actually see it, one has to stop being a literalist for a bit.

Magnus Anderson wrote: ↑Thu Jun 19, 2025 7:35 am

Skepdick wrote: ↑Wed Jun 18, 2025 3:35 pm Axiom 1: S is the set of all sets

False premise. No such set 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.

Skepdick wrote: ↑Thu Jun 19, 2025 10:41 am So in order to actually see the set of all sets you have to stop being a literalist for a bit.

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.