Aleph

[Warman]

how many aleph sets are there? I mean is there an aleph 1837748392 or a number of alephs that tends towards infinity without reaching it or even an infinite number of alephs?

 

[ActuallySpaceMan42]

Infinite

 

[Cat275]

Well it's hard to tell.

Aleph omega is a limit of aleph alpha (a limit of infinite sequence of alephs) and is a supremum of aleph-n's which prove there exist infinitely many alephs but well after that we have aleph omega + 1 which is unlike omega + 1 it is actually cardinality greater than aleph omega until aleph omega + omega to aleph omega*aleph omega to aleph omega omega to aleph omega omega omega to aleph omega omega omega omega all of which are on a kappa=aleph-kappa since it's the fixed point of all sequence of alephs after aleph omega and below.

So i'd say more than an aleph omega amount of it atleast.

 

[Guardian_Doge]

Well it's hard to tell.

Aleph omega is a limit of aleph alpha (a limit of infinite sequence of alephs) and is a supremum of aleph-n's which prove there exist infinitely many alephs but well after that we have aleph omega + 1 which is unlike omega + 1 it is actually cardinality greater than aleph omega until aleph omega + omega to aleph omega*aleph omega to aleph omega omega to aleph omega omega omega to aleph omega omega omega omega all of which are on a kappa=aleph-kappa since it's the fixed point of all sequence of alephs after aleph omega and below.

So i'd say more than an aleph omega amount of it atleast.

I'll eat you, use the symbols.

 

[Cat275]

I'll eat you, use the symbols.

I refuse to be served and no im lazy.

 

[Warman]

Well it's hard to tell.

Aleph omega is a limit of aleph alpha (a limit of infinite sequence of alephs) and is a supremum of aleph-n's which prove there exist infinitely many alephs but well after that we have aleph omega + 1 which is unlike omega + 1 it is actually cardinality greater than aleph omega until aleph omega + omega to aleph omega*aleph omega to aleph omega omega to aleph omega omega omega to aleph omega omega omega omega all of which are on a kappa=aleph-kappa since it's the fixed point of all sequence of alephs after aleph omega and below.

So i'd say more than an aleph omega amount of it atleast.

wow thank you very much and so from which aleph would we have an inaccessible cardinal?

 

[Cat275]

Inaccessible cardinals normally implies alephs but not vice versa so I doubt you can find a inaccessible in a aleph unless if you have and use some unique axioms that is not proven to be inconsistent.

 

[sen_]

how many aleph sets are there? I mean is there an aleph 1837748392 or a number of alephs that tends towards infinity without reaching it or even an infinite number of alephs?

well, aleph numbers are defined by transfinite recursion where we define ℵ₀ as the smallest infinite cardinal and the cardinality of all countably infinite sets, we can define ℵα for each ordinal α, It follows from the Axiom Of Choice and the Well-Ordering Theorem that every infinite set is equinumerous with an aleph number. And consequently that for any cardinal κ ∃! α : κ = ℵα (which reads: for any cardinal kappa there exists exactly one ordinal alpha such that kappa is equivalent to aleph-alpha). Hence we can estabilish that the function f: ORD → ℵ is a bijection from the ordinals to the infinite cardinals. So the answer to your question is that there are "class-many" Aleph Numbers.

 

[Cat275]

Every set has a class many numbers (not talking about properclasses) so the answer you gave is very vague. (Will admit my answer is also vague to an extent)

But im assuming you mean as many as Ord since Ord=the 1st worldly (in cardinality) unless if we use an axiom that states let Ord be omega or if we assume even bigger and stronger sets or quite frankly deny aoi.

Op you can also check here btw for a more straightforward answer.
Assuming "there is no 1st worldly" or not assuming it's existence then all aleph numbers should be equal to it in cardinality.

 

[Iiliminal]

shinra transcends math

 

[sen_]

Every set has a class many numbers (not talking about properclasses) so the answer you gave is very vague. (Will admit my answer is also vague to an extent)

you're right about that, the answer is quite vague, but by "class-many" i mean as many as ORD.

 

[Cat275]

you're right about that, the answer is quite vague, but by "class-many" i mean as many as ORD.

Right. So you essentialy mean it's as big as the "least worldly" in cardinality. (no existence assumed)

 

[sen_]

Right. So you essentialy mean it's as big as the "least worldly" in cardinality. (no existence assumed)

if you're talking about how many Aleph numbers there are then no, they have no cardinality because they form a Proper Class, the best you could do to define a "Cardinality" is using the definition that |X| = |Y| Iff f: X → Y is a bijection, we know that there is a bijection between ORD and the Aleph numbers, thus we can "define" the "Cardinality" of {ℵα : α ∈ ORD} to be ORD itself (Cardinality is in quotes because both are not sets in the first place), ORD is not a set because of the Burali-Forti Paradox, and in ZFC there exists a well-ordering of every set, hence |{ℵα : α ∈ ORD}| = κ ⇒ κ is not a cardinal.

 

[Cat275]

You can use agc and establish the 1st worldly would be equal to ord or simply do V(a) where V(a) is the least worldly so my point stands.

I think the problem here is that I wasn't clear. I didn't mean cardinality in the usual sense since proper classes don't have your usual cardinality is that a better and clear elaboration or do you wan't me to elaborate even further?

 

[Warman]

well, aleph numbers are defined by transfinite recursion where we define ℵ₀ as the smallest infinite cardinal and the cardinality of all countably infinite sets, we can define ℵα for each ordinal α, It follows from the Axiom Of Choice and the Well-Ordering Theorem that every infinite set is equinumerous with an aleph number. And consequently that for any cardinal κ ∃! α : κ = ℵα (which reads: for any cardinal kappa there exists exactly one ordinal alpha such that kappa is equivalent to aleph-alpha). Hence we can estabilish that the function f: ORD → ℵ is a bijection from the ordinals to the infinite cardinals. So the answer to your question is that there are "class-many" Aleph Numbers.

and therefore to pass from one aleph to another it would be necessary to take an aleph square? (example: 2^aleph 3=aleph 4)

 

[Cat275]

Huh that's like almost completely unrelated to the answer.

 

[Warman]

Huh that's like almost completely unrelated to the answer.

i know but i thought asking here was better than opening another thread

 

[Cat275]

Right. I just got confused by the therefore.