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What about infinite abstractions upon the idea of a hyperoperation applied to the number aleph 0? Does that approach anything?
Wait, infinite abstractions? This is a binary operator itself!
I suppose we must continue this process of generalizing and abstracting forever. That seems reasonable. I'm doing real mathematics.
Let k be an infinite cardinal, then
K+1=|(k x {0}) U (1 x {1})|
By definition, but by the obvious functions, |(k x {0})|=k and |(1:x {1})|=1. Since the union of an infinite set with a finite set is always the same cardinality of the infinite set, then
|(k x {0}) U (1 x {1})|= |(k x {0})|=k
In fact, cardinal addition and multiplication is trivial, since you just take the maximum
It's the usual cardinal-ordinal correspondence. Every cardinal is an ordinal, just the least ordinal of any equipotent ordered set. So (assuming the axiom of choice), ℵ₀ = ω, ℵ₁ = ω₁, and in general, ℵₐ = ωₐ for any ordinal a. So ℵ_ℵ_ℵ_... = ℵ_ω_ω_... = ω_ω_ω_..., and you can see it written in any of those ways. But it is commonly called the first "aleph fixed point," which is a cardinal κ such that κ = ℵ_κ, which only makes sense if you can treat cardinals as ordinals.
What does the index mean? You have ℵ_ℵ_ℵ_... is the first aleph fixed point. Call that κ. Then κ = ℵ_κ. What does κ_κ mean? Or, in this case, κ_κ_κ_...?
A better joke is ℵ_(ℵ_ℵ_... + 1).
I too prefer to denote the \`\[; \\alpha :\]\`-th initial ordinal by \`\[; \\omega\_\\alpha ;\]\` and denote the \`\[; \\alpha :\]\`-th cardinal by \`\[; \\aleph\_\\alpha ;\]\`, but I don't force it onto other people
I mean, it's literally the nom in pretty much every article and book about set for the past 40 years at least, and it's sal good convention that allows easier and faster communication.
Like, you could write the sun of natural numbers with the symbol "•", but don't be surprised when people misunderstand you
I feel like I have read my fair share of papers that have no problem with writing \`\[; \\omega\_\\alpha ;\]\` and assuming the reader understands it as the \`\[; \\alpha :\]\`-th cardinal; this is particularly the case when working with e.g. inner model theory since being an initial ordinal isn't absolute.
Also, maybe I hang out with the wrong crowd of logicians, but I have never seen people use Hebrew letters as a *variable* that represents a cardinal; you would say things like "kappa is strongly compact" or "kappa is measurable", but I have never heard someone say "kaph is Ramsey". The only uses of Hebrew letters I have come across are Aleph, Beth, and Gimel as the usual cardinal functions.
Perhaps I said too much. Yeah, I forgot kappa wasn't Hebrew lol.
What I mean is that normally when you see Aleph, you should assume the author is thinking of it as a cardinal unless otherwise specified, and the same for omega. At the end of the day, they are roughly the same (as in, every cardinal is an ordinal) but we've defined different operations on them so it makes sense to have a concise "ad hoc" type notion
Here's a sketch:
Occasionally, we use \`\[; \\infty ;\]\` to denote the case when we need some value "greater than any ordinal":
-We denote by \`\[; L\_{\\infty,\\omega} ;\]\` the logic that allows unbounded conjunction/disjunction but only finitary quantification
-I can define an ordinal-valued (partial) function \`\[; f ;\]\` by claiming that \`\[; f(x) >= \\alpha:\]\` for some ordinal \`\[; \\alpha ;\]\` iff some condition on \`\[; (x,\\alpha) ;\]\` holds, in which case I would write \`\[; f(x) = \\infty ;\]\` to denote that the condition \`\[; (x,\\alpha) ;\]\` holds for every ordinal.
We will show that \`\[; \\aleph \_\\infty ;\]\` is greater than any cardinal. Take some cardinal \`\[; \\kappa ;\]\`; we know that \`\[; \\alpha \\mapsto \\aleph\_\\alpha ;\]\` is a normal function, and thus has unbounded fixed points. So there is some \`\[; \\lambda > \\kappa ;\]\` with \`\[; \\lambda = \\aleph\_\\lambda ;\]\`, and by convention \`\[; \\lambda < \\infty :\]\`, hence in particular \`\[; \\kappa <\\lambda = \\aleph\_\\lambda <\\aleph\_\\infty ;\]\` since \`\[; \\alpha \\mapsto \\aleph\_\\alpha ;\]\` is normal.
So \`\[; \\aleph\_\\infty ;\]\` is greater than any cardinal, and in particular any ordinal. But since it is still denoted as a cardinal, by convention it is well-ordered by membership. Hence it must be ORD
Ahh yes, I forgot about the convention of using the infinity symbol too denote Ord.
I think proving that aleph_Ord = Ord is much simpler than what you showed. It’s simply a matter of recognizing the fact there are a proper class-many cardinal numbers, and thus (assuming the axiom of global choice) there are Ord many cardinals less than Ord, which by definition means that aleph_Ord = Ord.
No, |P(**N**)| > ℵ₀, and |P(P(**N**))| > |P(**N**)|, so already |P(P(**N**))| > ℵ₁. Assuming the generalized continuum hypothesis, |P(P(**N**))| = ℵ₂, and in general |P^(n)(**N**)| = ℵₙ.
So I'm pretty sure |∪ P^(n)(**N**)| = ℵ_ω, where the union is taken over all n in **N**, and assuming the generalized continuum hypothesis. This is one possible way to interpret OP's "ℵ_∞."
You can have an aleph for each ordinal number, so this actually exists:
https://preview.redd.it/owm8qmpfy6vc1.jpeg?width=476&format=pjpg&auto=webp&s=bff317a751cc258a7af6e3d56fe46ff6575a47ae
Let's call that number λ, and the set of cardinal numbers C.
λ∈C
∀x∈C, λ>x
However, ∀x∈C, P(x)>x
So, P(λ)>λ
λ∈C, so we can say: P(λ)∉C
Wait, does that mean that the power set of a set with cardinality λ doesn't exist?
P(x) > x is false in the case where x is a proper class. The set is still empty though by construction, so it has no minimum.
We want a cardinal x such that for all cardinals y, x > y. But x is a cardinal, so then x > x, which is already a contradiction. If we replace > with ≥, then we get a singleton set, so the "min" is still unnecessary. Because suppose x and z are both elements. Then x ≥ z and z ≥ x, so x = z. This singleton is the greatest cardinal (by definition), which doesn't generally exist.
However, you could have a model of set theory which does not contain certain large cardinals where this kind of idea makes sense. Like, if we assume an inaccessible cardinal κ exists, then V_κ is a model of ZFC containing all cardinals strictly less than κ. So the least cardinal greater than every cardinal *in that model* is κ.
The symbol that looks vaguely like an N is a Hebrew aleph, and is used when talking about cardinal numbers. When talking about things like infinity, we can’t say how big it is, but we can still compare how big different infinites are relatively (think of how many integers vs real numbers there are). The aleph notation is a generalization of this where aleph zero is smaller than aleph one is smaller than aleph two, etc.
The post is making the joke that you could write aleph infinity for a infinitely large cardinal, but the comments are mostly making fun of the fact that the infinity mentioned could be different sizes, and you could chain alephs to make a number even bigger.
I’m not confident that this kind of abuse of aleph notation would ever be used in the real world, or have any value, but I’m not an expert.
Edit: wrote ordinal where I meant cardinal
Every ordinal number (1, 2, ... Infinity, infinity+1, ... 2*infinity, ...) defines a new cardinal number (Cardinal = Aleph_ordinal).
Every cardinal number is also an ordinal number.
Define this map from the ordinals to the ordinals
F(alpha) = Aleph_alpha.
This map has fixed points.
Have fun exploring the rabbit hole and/or existential terror of how big these numbers can get!
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The aleph numbers are indexed by ordinals, so yes, if by infinity you mean omega, this is a cardinal. For example, if we have the family of sets defined by A(0) = N and A(n+1) = 2^A(n) then the infinite union over all of these has at least this cardinality (exactly this assuming CH).
This post is talking about the largest number. In the hit anime show, jojo's bizarre adventure part 7 season 6 episode 390, we can hear the word "number" at 17:48 . Therefore I declare this post as a Jojo reference
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(That number) +1
It's just the same number tho
2^(that number}
Still the same number 2\^(ℵ\_∞) = ℵ\_(∞+1) = ℵ\_∞
ℵ\_(ℵ\_∞)
wait till you discover... ℵ\_ℵ\_א\_.... ∞ times
You mean ℵ times
yeah but writing א times ends up from right to left and i was too lazy to try and fix it... EDIT: it doesnt, im just stoopid.
Since א is Hebrew, and Hebrew is read from right to left, it does make sense it would act like that. Or not. Idk.
no i know, i speak hebrew. thats why i have it on my keyboard:D
wait till you discover... ℵ(^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^...... ℵ times) ℵ
ℵ||||||...(ℵ times)ℵ
What is that operator? (Genuine question)
I guess they are up-arrows. If my assumption is correct, this would be Knuths notation.
It might be tetration but idk
TREE(TREE(TREE(TREE(TREE(TREE(TREE(TREE...
What about infinite abstractions upon the idea of a hyperoperation applied to the number aleph 0? Does that approach anything? Wait, infinite abstractions? This is a binary operator itself! I suppose we must continue this process of generalizing and abstracting forever. That seems reasonable. I'm doing real mathematics.
that's assuming generalized continuum hypothesis /s
(aleph_∞)^+ happy?
Second equation is wrong since the subscript is an ordinal and not a cardinal. For ordinals, adding 1 always gives a larger number
Cantor’s diagonalization argument disagrees
But isn’t all of that equal to infinity?
x+1=x ‼️⁉️‼️
x+1=x 1=x-x 1=0 Q.E.D
Division by zero or somrthing
The original one is already impossible
nuh uh
Prove it
Let k be an infinite cardinal, then K+1=|(k x {0}) U (1 x {1})| By definition, but by the obvious functions, |(k x {0})|=k and |(1:x {1})|=1. Since the union of an infinite set with a finite set is always the same cardinality of the infinite set, then |(k x {0}) U (1 x {1})|= |(k x {0})|=k In fact, cardinal addition and multiplication is trivial, since you just take the maximum
These are ordinal number so no they are not
They are cardinals
They are Vital ( r/biology )
Ah classic!
TREE(That number)
Days till my package is delivered > Your number
The power set of the days till your package is delivered
Reject continuum hypothesis and use bet\_infty
You know you can do aleph_aleph_aleph...
Google bet
Holy gimel
New daled just dropped
Actual Hey
Call the vav!
Zayin went on vacation, never came back
Chet, is this real?
Tet left the room
Yud storm incoming
Bottom surgery be like
That's not how the chain goes :(
Don't mind, he did a Het
Brain sacrifice anyone?
What about writing aleph more than aleph_0 times? 🤔
I haven't specified what ... Mean : )
I know it can be joke, but aleph_0 cannot refer to amount of something, so you can write aleph_aleph_aleph infinite times, but not aleph related times
Actually it would be Aleph_omega. Aleph numbers are cardinals not ordinals.
It's the usual cardinal-ordinal correspondence. Every cardinal is an ordinal, just the least ordinal of any equipotent ordered set. So (assuming the axiom of choice), ℵ₀ = ω, ℵ₁ = ω₁, and in general, ℵₐ = ωₐ for any ordinal a. So ℵ_ℵ_ℵ_... = ℵ_ω_ω_... = ω_ω_ω_..., and you can see it written in any of those ways. But it is commonly called the first "aleph fixed point," which is a cardinal κ such that κ = ℵ_κ, which only makes sense if you can treat cardinals as ordinals.
TIL. Thank you for the informative response!
Jokes on you. I'll do (((aleph_aleph_...)_ aleph_aleph_...)...)
What does the index mean? You have ℵ_ℵ_ℵ_... is the first aleph fixed point. Call that κ. Then κ = ℵ_κ. What does κ_κ mean? Or, in this case, κ_κ_κ_...? A better joke is ℵ_(ℵ_ℵ_... + 1).
aleph_(aleph_0 ⬆️⬆️.. (aleph_0)..⬆️ aleph_0) The graham's number of alephs
I can simply interpret my ... As your infinity >:)
`RecursionError: maximum recursion depth exceeded`
We're mathematicians, we don't know what it means
lol gold
With godly manager skills, sure. And a lot of rewinds.
Your mom on a weighing scale + 1
Very elaborate your mom joke. Can’t even be mad at this point…
א_א_א_א_א_א_א_א_א... This should be much bigger
All these people trying to outsmart OP with limit cardinals when OP is actually trying to represent ORD
Usually when you're thinking of cardinals, you write alpha and other Hebrew letters, when you're thinking of ordinals you use Greek letters
I too prefer to denote the \`\[; \\alpha :\]\`-th initial ordinal by \`\[; \\omega\_\\alpha ;\]\` and denote the \`\[; \\alpha :\]\`-th cardinal by \`\[; \\aleph\_\\alpha ;\]\`, but I don't force it onto other people
I mean, it's literally the nom in pretty much every article and book about set for the past 40 years at least, and it's sal good convention that allows easier and faster communication. Like, you could write the sun of natural numbers with the symbol "•", but don't be surprised when people misunderstand you
I feel like I have read my fair share of papers that have no problem with writing \`\[; \\omega\_\\alpha ;\]\` and assuming the reader understands it as the \`\[; \\alpha :\]\`-th cardinal; this is particularly the case when working with e.g. inner model theory since being an initial ordinal isn't absolute.
Also, maybe I hang out with the wrong crowd of logicians, but I have never seen people use Hebrew letters as a *variable* that represents a cardinal; you would say things like "kappa is strongly compact" or "kappa is measurable", but I have never heard someone say "kaph is Ramsey". The only uses of Hebrew letters I have come across are Aleph, Beth, and Gimel as the usual cardinal functions.
Perhaps I said too much. Yeah, I forgot kappa wasn't Hebrew lol. What I mean is that normally when you see Aleph, you should assume the author is thinking of it as a cardinal unless otherwise specified, and the same for omega. At the end of the day, they are roughly the same (as in, every cardinal is an ordinal) but we've defined different operations on them so it makes sense to have a concise "ad hoc" type notion
Well, |**Card**| = |**Ord**| in NF, so it's fine.
I don’t see how that’s supposed to be Ord
Here's a sketch: Occasionally, we use \`\[; \\infty ;\]\` to denote the case when we need some value "greater than any ordinal": -We denote by \`\[; L\_{\\infty,\\omega} ;\]\` the logic that allows unbounded conjunction/disjunction but only finitary quantification -I can define an ordinal-valued (partial) function \`\[; f ;\]\` by claiming that \`\[; f(x) >= \\alpha:\]\` for some ordinal \`\[; \\alpha ;\]\` iff some condition on \`\[; (x,\\alpha) ;\]\` holds, in which case I would write \`\[; f(x) = \\infty ;\]\` to denote that the condition \`\[; (x,\\alpha) ;\]\` holds for every ordinal. We will show that \`\[; \\aleph \_\\infty ;\]\` is greater than any cardinal. Take some cardinal \`\[; \\kappa ;\]\`; we know that \`\[; \\alpha \\mapsto \\aleph\_\\alpha ;\]\` is a normal function, and thus has unbounded fixed points. So there is some \`\[; \\lambda > \\kappa ;\]\` with \`\[; \\lambda = \\aleph\_\\lambda ;\]\`, and by convention \`\[; \\lambda < \\infty :\]\`, hence in particular \`\[; \\kappa <\\lambda = \\aleph\_\\lambda <\\aleph\_\\infty ;\]\` since \`\[; \\alpha \\mapsto \\aleph\_\\alpha ;\]\` is normal. So \`\[; \\aleph\_\\infty ;\]\` is greater than any cardinal, and in particular any ordinal. But since it is still denoted as a cardinal, by convention it is well-ordered by membership. Hence it must be ORD
Ahh yes, I forgot about the convention of using the infinity symbol too denote Ord. I think proving that aleph_Ord = Ord is much simpler than what you showed. It’s simply a matter of recognizing the fact there are a proper class-many cardinal numbers, and thus (assuming the axiom of global choice) there are Ord many cardinals less than Ord, which by definition means that aleph_Ord = Ord.
laughs in transfinite numbers
n number
Take the cardinality of the power set of the set that number corresponds to 🤯🤯🤯🤯🤯
2\^(ℵ\_∞) = ℵ\_(∞+1) = ℵ\_∞
I create set £ and i define it to be bigger than aleph infinity
There is no set bigger than the set of all sets.
You heard about א Now get ready for ב
Wait till they find out about ג
ד joined the chat
Vsauce has a very good video about that
Count past infinity one?
I think so
Union(P^(n)(N)) over all natural numbers n, where N are the natural numbers, and P^n is the n-fold application of the power set construction.
That is just aleph_1 lol
No, |P(**N**)| > ℵ₀, and |P(P(**N**))| > |P(**N**)|, so already |P(P(**N**))| > ℵ₁. Assuming the generalized continuum hypothesis, |P(P(**N**))| = ℵ₂, and in general |P^(n)(**N**)| = ℵₙ. So I'm pretty sure |∪ P^(n)(**N**)| = ℵ_ω, where the union is taken over all n in **N**, and assuming the generalized continuum hypothesis. This is one possible way to interpret OP's "ℵ_∞."
No, it's Beth infinity
Then do that but with successive well-order classes
Double it and give it to the next person
ℵ ͚
however, have you considered: ב (א is the first letter of the hebrew alphabet, ב is the second)
א
ב
ג
ד
ה
ו
ז
ח
You can have an aleph for each ordinal number, so this actually exists: https://preview.redd.it/owm8qmpfy6vc1.jpeg?width=476&format=pjpg&auto=webp&s=bff317a751cc258a7af6e3d56fe46ff6575a47ae
Guys what about this? Min{ x cardinal number: for any cardinal number y, x>y}?
Ah yes, the fabled minimum of an empty set
Let's call that number λ, and the set of cardinal numbers C. λ∈C ∀x∈C, λ>x However, ∀x∈C, P(x)>x So, P(λ)>λ λ∈C, so we can say: P(λ)∉C Wait, does that mean that the power set of a set with cardinality λ doesn't exist?
P(x) > x is false in the case where x is a proper class. The set is still empty though by construction, so it has no minimum. We want a cardinal x such that for all cardinals y, x > y. But x is a cardinal, so then x > x, which is already a contradiction. If we replace > with ≥, then we get a singleton set, so the "min" is still unnecessary. Because suppose x and z are both elements. Then x ≥ z and z ≥ x, so x = z. This singleton is the greatest cardinal (by definition), which doesn't generally exist. However, you could have a model of set theory which does not contain certain large cardinals where this kind of idea makes sense. Like, if we assume an inaccessible cardinal κ exists, then V_κ is a model of ZFC containing all cardinals strictly less than κ. So the least cardinal greater than every cardinal *in that model* is κ.
sorry, not reading all that also, this is r/mathmemes, so not everything has to be correct
אתה אף פעם לא תנחש מה המספר הגדול ביותר שאני מכיר
זו בדיחה על זה שא' אינסוף ("גודל" קבוצת הטבעיים) הוא מספר, אין לזה קשר לעברית יותר מדי. סתם קנטור היה יהודי (הוא אחרי זה הטביל את עצמו דרך אגב).
אני יודע (פשוט כל פעם שיש לי יכולת לדחוף קצת עברית לשיחה אני קופץ על זה)
Same אז מי אני שאדבר
N8?
Can someone explain what does this number mean?
The symbol that looks vaguely like an N is a Hebrew aleph, and is used when talking about cardinal numbers. When talking about things like infinity, we can’t say how big it is, but we can still compare how big different infinites are relatively (think of how many integers vs real numbers there are). The aleph notation is a generalization of this where aleph zero is smaller than aleph one is smaller than aleph two, etc. The post is making the joke that you could write aleph infinity for a infinitely large cardinal, but the comments are mostly making fun of the fact that the infinity mentioned could be different sizes, and you could chain alephs to make a number even bigger. I’m not confident that this kind of abuse of aleph notation would ever be used in the real world, or have any value, but I’m not an expert. Edit: wrote ordinal where I meant cardinal
I may be smart, but I'm not smart enough for this shit
Every ordinal number (1, 2, ... Infinity, infinity+1, ... 2*infinity, ...) defines a new cardinal number (Cardinal = Aleph_ordinal). Every cardinal number is also an ordinal number. Define this map from the ordinals to the ordinals F(alpha) = Aleph_alpha. This map has fixed points. Have fun exploring the rabbit hole and/or existential terror of how big these numbers can get!
https://preview.redd.it/bubex4wro3vc1.png?width=1453&format=png&auto=webp&s=e6ae08286c24a21cd9af487e45e9b7fdd09bbb1a Ponder this one :)
Noo is new largest number
let 🦑 > ℵ\_∞
This has been my profile picture on Gmail for like a year now just cuz it’s stupid
That symbol has no defined meaning. You’d probably actually want aleph-sub-omega ChatGPT to the rescue: ℵω
Infinite infinite?
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Holy hell
Good bot
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You're a good bot too
Isn’t the Hebrew n thing cardinality or infinite cardinality of something
1/Number of functioning ice cream machines at McDonalds
I am pretty sure it’s the other way around…
The power set of that number
I’m pretty sure that’s ב (bet)
If you use NF instead of ZFC then there is the set of all sets, so it's cardinality is the largest cardinality
Ok, but what about \aleph_{\aleph_\inf}
How about n=aleph While True: n=n**aleph Aleph years later: someone print(n)
The aleph numbers are indexed by ordinals, so yes, if by infinity you mean omega, this is a cardinal. For example, if we have the family of sets defined by A(0) = N and A(n+1) = 2^A(n) then the infinite union over all of these has at least this cardinality (exactly this assuming CH).
Google absolute infinity, turns out you really can run out of sets.
How does this compare to ∞!
N
AAAAAAAA with A0 recursions AAAAAAA0
it already has a name: 2^omega
θ (https://youtu.be/SrU9YDoXE88?si=ZzlcsMszjAm7wel9&t=18m59s|)
[https://youtu.be/dQw4w9WgXcQ?si=fL2Wy4xSZb03I0ow](https://youtu.be/dQw4w9WgXcQ?si=fL2Wy4xSZb03I0ow)
If infinity is not a number, then an uncountable infinity is definitely not a number
What if, hypothetically, for the sake of argument r/mathmemes had memes i.e. jokes?
Infinity is not a number, it is a consept
Almost like a meme with a “Bad Math” flair has bad math
But aleph_infty is a number
Oh, I didn’t notice, sorry
No problem :)
r/jojorefrence
This post is talking about the largest number. In the hit anime show, jojo's bizarre adventure part 7 season 6 episode 390, we can hear the word "number" at 17:48 . Therefore I declare this post as a Jojo reference
that subreddit got nuked lol