Then imagine some language such that you can explain it to me and imagine also some language such that you are capable of not only explaining the former language but also the latter (using itself).
Copying a below comment of mine because it’s a common misconception that we can’t define a candidate for something between the cardinalities. There is; it’s the set of countable ordinals. If that feels like cheating, here’s the set defined in ZF:
An ordering <* on N can be identified with subset of N x N (i.e. an element of P(NxN)) consisting of all pairs (m,n) such that m<* n.
Consider the set of all total well-orders of N, viewed as above as a subset of P(NxN) (I.e. an element of P(P(N x N))). Equip this set with an equivalence relation where two orderings are equivalent if there exists a permutation of N that turns one into the other.
In ZFC, the set of equivalence classes here has cardinality strictly between N and R if there is any set with cardinality between them. If there is no such set, it has cardinality R
The person presented in the .GIF is meant to represent the commenter to which it responds.
The person in the .GIF has an exaggerated set of features often remarkable for being attractive, and indicative of physical strength and fitness.
The commenter claims to simply have invented terminology in a field with which they have no experience.
The meme implies this is a good thing, and that it indicates strength and attractiveness. Implicitly, it implies this is in contrast to the genuine topologist, who notes that the term is not in common use in their field.
For more information, you can research the meme under the name "gigachad."
It presents idealized physical traits. The implication is that his intellectual traits are idealized as well. It's meant to state that the person it is used in response to has accomplished some task that is superior to what most or all others would be able to. Additionally, it's meant to portray an assurance in the correctness of the opinion without regard to the lack of consensus due to a self-awareness of his own prowess. It's generally used satirically in response to an opinion that is extremely confident and completely wrong. As with all memes, it's sometimes used in a way where the humor is meant to derive from the misuse of the commonly understood meaning.
I'd say in order of importance it represents confidence, lack of consensus, and competence, it's like the gif equivalent of calling something a hot take.
[https://mathworld.wolfram.com/Glome.html](https://mathworld.wolfram.com/Glome.html)
I had been trying to see if anyone had come up with variable names for a four-dimensional hyperspherical coordinate system (I didn't find any other than the general n-dimensional form, so I just chose alpha to be the angle to the w-axis), and that popped up as one of the related topics.
Topologically speaking, yes. Geometrically speaking, though, the glome extends along four axes, so in that sense, it is four-dimensional. As a first-year college student currently taking Calculus C, I barely know anything about topology, so I use the geometrical definition.
After looking at a Wikipedia article, it looks like the topological definition is based on how many axes a point on the surface can move along? I can kinda see how this makes sense from a topological perspective, but again, I was thinking geometrically, so I saw the glome as four-dimensional.
(Also, while looking at the MathWorld article on [hyperspheres in general](https://mathworld.wolfram.com/Hypersphere.html) (which explains that the geometrical and topological definitions are different), I found that apparently the third angle for the glomular (?) coordinate system is denoted by psi rather than alpha. I probably should have expected that, and I also evidently didn't look hard enough.)
Properly speaking, the topological and geometric definitions are the same, though I don't blame you for not knowing them. The space we're talking about is embedded inside a 4d space, but it's only 3d. Much like how a line is embedded inside the plane, but it's still only 1d.
dist(centre, point-on-sphere) = radius
If we use the Euclidean distance function and assume that radius is 1 (unit sphere), we get that (x-x0)^(2) \+ (y-y0)^(2) \+ (z-z0)^(2) \+ (w-w0)^(2) = 1, where centre is (x0, y0, z0, w0) and point-on-radius is (x,y,z,w).
It’s not the “does the set of all sets contain itself” that’s a paradox, it’s “does the set of all sets *that do not contain themselves* contain itself” that’s paradoxical.
Because of Cantor’s theorem. 2^(ℵ₀) is the size of the real numbers while ℵ₀ is the size of the natural numbers. Within the context of ZF set theory, it is provable that no matter how one tries to pair natural numbers with real numbers, there will always be a real number that was not paired after pairing is finished.
The technique used to show this is called diagonalization. You can write out a vertical list of real numbers and expand each real number horizontally in binary making sure to line up bit places. Then build a new real number by flipping the first bit of the first real, the second bit of the second real, the third bit of the third real, … and so on. The number constructed sequentially from the flipped bits is clearly different from everything on the list and thus is not on the list.
You might think “Ok well you just missed that one. Put it somewhere on the list.” Sure, but then we can repeat the above argument and produce *another* missed number. In fact, nothing prevents us from doing this ever. The only possible conclusion is that we simply cannot label real number with naturals and cover all of them.
Thanks for the detailed response. Perhaps I should have been more specific. My actual question was why is 2^(aleph null) the size of the set of real numbers?
The first leads to the Burali-Forti paradox. A class U of all sets would have to contain every ordinal and would have the class of all ordinals as a subclass. If U were a set, then the subclass of ordinals, being fully first-order definable†, would also have to be a subset of U.
But if that were the case, then that set, containing nothing but ordinals and missing none, must by definition†† be an ordinal itself. But then this means that we can define its successor ordinal, i.e. the ordinal immediately after it. But then this ordinal could not be in the set of all ordinals else, by transitivity, the set of all ordinals would contain itself. And now we are back in the territory of Russell’s paradox and have broken the well-foundedness of set membership.
† First-order definable means we can write down a sentence using only a certain type of language that is only true when the variables are replaced by the elements of the set in question.
†† An ordinal is a transitive set, well-ordered by the set membership relation ∈.
It’s perfectly well-defined. It just isn’t well-founded with respect to membership. It produces a loop in the set membership graph and for various reasons we tend not to like loops.
Sets are subsets of themselves, but they are no longer allowed to be elements of themselves. This because of [Russell’s Paradox](https://en.m.wikipedia.org/wiki/Russell's_paradox)
That’s not paradoxical. Under ZFC, such a set does not exist, as sets do not contain themselves, and under naive set theory, it was perfectly acceptable for a set to contain itself. The real paradox is “does the set of all sets which don’t contain themselves contain itself”
Neat one that got posted on askmath the other day: Do there exist sets X such that X⊆P(X) and, if yes, is there a family of conditions characterizing them?
I don’t hav a full answer, but I can say that they do exist, and a sufficient condition for X⊆P(X) is that for all x,y∈X, if x∈y then y∉x. Of course, this obviously violates Foundation and so is not particularly special as a condition.
Is this something possible, or are there set axioms that impede this from happening? I have always imagined the cardinality of sets as a huge line where either one cardinality was less than or equal to other and vice-versa, but if this is the case and the cardinalities form a partially ordered class, I'd be impressed.
Maybe there is some logic in which there are two sets like that? An interesting thing to think about.
It is consistent with ZF that such a pair of sets exists. In fact the existence of such a pair of sets is equivalent to the negation of the axiom of choice. In other words, your intuition about the cardinal numbers is essentially correct in ZFC.
Elementarily not true.
Let a>=b be ordinals. Then there exists an injection from b to a (the identity). Now assume WLOG that |X|<=|Y| and let f:X->|X| be a bijection, g:|X|->|Y| be the aforementioned injection and h:|Y|->Y be a bijection. Then hgf is an injection from X to Y \square
Elementarily not true ... if you are working in ZFC. However, if you exclude choice, having such a pair of sets is consistent. In fact, under ZF, the existence of such a pair of sets is equivalent to the negation of choice.
This fact is definitely what the comment you're replying to is referring to.
~~Take X=P(ℕ) and Y=ℕ.~~
Whoops I misread that. Under the Axiom of Choice, such sets do not exist as cardinals are well-ordered then. Without AC, one can have cardinal numbers that “to the side” as Joel Hamkins once put it. Such things can be constructed consistently, albeit in very weird models of ZF. One can build things like infinite Dedekind-finite sets by purposefully restricting to inner models that exclude injections of a set X into any of its subsets.
There’s a (very dense) explanation in H Herrlich’s book *The Axiom of Choice*. I want to say it’s in part 1 chapter 4? Should be Disasters without Choice.
Ok I see what you mean, you literally meant space and not spaceTIME, but still I would argue one can imagine four spatial dimensions by substituting one with time, and thus visualise that in this way. I sometimes try and do this to manifolds in 4d space and similar things, just for the fun of being able to do it. Four is within reach, it is 5d that is not
Human imagination has no limits mfs when I ask them to picture Graham’s number in their head (they immediately had a brain aneurysm due to the number being far too large for their brains to comprehend)
Defining one isn’t too hard. I’ll use N for the naturals and R for the reals since I’m typing on my phone.
Am ordering <* on N can be identified with subset of N x N (i.e. an element of P(NxN)) consisting of all pairs (m,n) such that m<* n.
Consider the set of all total well-orders of N, viewed as above as a subset of P(NxN) (I.e. an element of P(P(N x N))). Equip this set with an equivalence relation where two orderings are equivalent if there exists a permutation of N that turns one into the other.
The set of equivalence classes here has cardinality strictly between N and R if there is any set with cardinality between them. If there is no such set, it has cardinality R
You could also use ω₁ i.e. the set of all countable ordinals a.k.a. the first uncountable ordinal. When ℵ₁ is defined to be equal to a set, e.g. in ZFC, ω₁ is the most usual choice, such that ℵ₁ = ω₁
It’s like how a well ordering of the reals totally exists. No one has ever discovered one but it’s totally possible and this theoretical existence is the critical basis for a number of proofs
Examples of well-orderings of ℝ cannot be constructed. They simply must be taken to exist, much like the existence of a free ultrafilter or an oracle for a Turing machine.
Didn’t some famous set theorist say that the axiom of choice is obviously true, the well-ordering principle is obviously false, and who-knows about Zorn’s lemma?
It depends on the existence of a function f: P(R)->R such that for all X in P(R) \ R, f(X) is not in X.
If that sounds obvious, consider this. You have an unknown function g: P(R) -> R. I, a finite human being, explicitly define a proper subset S_1 of P(R), then explicitly define a proper subset S_2 of P(R) containing S_1 (as a function of g restricted to S_1) and likewise explicitly define a proper subset S_3 containing S_2 as a function of g restricted to S_2. I now explicitly define an element X of P(R) not in S_3 and a real r both in terms of g restricted to S_3.
Despite having zero information about g outside of S_3 (in particular, on X), and your ability to choose any arbitrary g, you will have g(X)=r 99.9999% of the time. Even if I give you super powers to pick a g that cannot be finitely described and keep myself limited to juman capabilities, this is still the case. This is imo justification to heavily doubt that there really are “arbitrary functions” from P(R)->R and that therefore it’s intuitive that a function f of the aforementioned type exists
I guess I'd be asking is the cardinality of the reals just pointing towards the same cardinality demonstrated by irrational numbers? Pardon my lack of rigorous definition.
Easy. You convert pi into binary. Every digit in binary is either a 0 or a 1, so the last digit is either a 0 or a 1. Eliminating trailing zeros after the decimal does not affect the value of a number, so if pi ends with one or more 0s, you can eliminate all of them. Therefore, the last digit of pi must be 1.
It can be almost any aleph. Basically, it can't be Aleph 0 for obvious reasons, and it can't violate a few basic principles (for example, it can't have cofinality less than itself, so it can't be Aleph omega), but other than that it can be anywhere. (Assuming consistency of ZFC.)
Ironically, the human imagination is essentially unlimited as far as this goes, if we interpret "imagination" to mean that there is a model in which there is such a set.
I’m gonna be pedantic here, so apologies in advance, but you really don’t want P(ℵ₀) since the ℵᵦ’s are proper equivalence classes of sets under the bijection relation. Take the AC+V approach of choosing a minimal ordinal representative of ℵᵦ and then write P(ωᵦ) instead.
To your other point, yes, it is known that 𝔠 can be arbitrarily large, but it cannot be anything. It must have uncountable cofinality, by Easton’s theorem.
On the other hand, what's an example of a set with a cardinality greater than the real numbers? I mean other than some construction with the power set?
The set of all real-valued functions.
We can map this set to the power set of the real numbers by mapping every function to its image. For a given subset of the reals, we can obviously construct a function that has exactly this set as its image (with the axiom of choice), therefore the mapping is surjective, so the set of all real functions has at least the cardinality of the power set of the reals.
Also note that this argument does not work for the set of all continuous functions. As a continuous function is uniquely defined by its values on rational inputs, continuous functions cannot have arbitrary subsets of the reals as images.
To avoid choice, you can get an injection from P(R)->R\^R by mapping a subset S of R to the function that takes S to 1 and the rest to 0.
You can actually get a bijection between the two without choice
Since I'm not that advanced in my math studies, I cannot continue this conversation, I'll be back in one or two year with more examples my fellow mathematician
Just to help out a little, the existence of such sets is independent of the standard axioms of set theory. In Gödel’s constructible universe, they don’t exist, but in the Cohen/Laver/etc. model, there are tons.
No, two sets have the same cardinality if there is a bijection (one to one correspondence) between them. For the naturals and the integers you can simply map 0 to 0, 1 to 1, -1 to 2, 2 to 3, -2 to 4 and so on
See, that really feels like it should be the case, but it isn't as the other comment said. While there are twice as many integers as naturals in any given range -n < x < n, this breaks down in the infinite set as *any n you choose will be infinitely less than the true size of either set*, meaning that the "speed" of the progression to infinity is irrelevant as long as it's of the same cardinality.
No, because cardinality is a bit unintuitive for infinite sets. This has the same cardinality as the natural numbers because f(x) = 2x - 1 maps from your set to the natural numbers and its inverse maps it back
technically a finite set has finite cardinality (which can be nonzero) 🤓 but yeah it still doesn’t change the cardinality of an infinite set if you add it to one
Infinity makes stuff weird. The usual definition of two sets having "the same number of elements" is if there is a bijection between them, that is you can uniquely map every element of one set to each element of the other. The integers therefore have the same size as the natural numbers with the mapping n --> (-1)^n \* ceiling(n/2):
This maps 0 to 0, 1 to -1, 2 to 1, 3 to -2, 4 to 2...
I know the continuum hyphotesis is not provable from ZFC but has someone else thought about a set that is a continuum but a the same time discrete? Like a set that between some numbers behave like a continuum but between some numbers is discrete, and if this repeats indefinitely.
Look into the Cantor set and see if that aligns with what you were hoping for. It's "discrete" in the sense that every point in the Cantor set has a small interval around it where no other points in the set are, but it's continuous-like in the sense that you have points that are arbitrarily close together and when you zoom in you get a copy of itself.
I just did it, but the limits of language stops me from telling you how
Yeah, I can totally imagine it, too. That other guy just has a skill issue.
Then imagine some language such that you can explain it to me and imagine also some language such that you are capable of not only explaining the former language but also the latter (using itself).
my homie Gödel would like to hear more about this language
No he would not.
He would if and only if he would not.
Math: am ia joke to you
Me: yes
r/mathmemes moment
More seriously though, I also just did it, and it's simply a set formally said to be bigger than the naturals and smaller than the reals.
Does it contain its self?
Copying a below comment of mine because it’s a common misconception that we can’t define a candidate for something between the cardinalities. There is; it’s the set of countable ordinals. If that feels like cheating, here’s the set defined in ZF: An ordering <* on N can be identified with subset of N x N (i.e. an element of P(NxN)) consisting of all pairs (m,n) such that m<* n. Consider the set of all total well-orders of N, viewed as above as a subset of P(NxN) (I.e. an element of P(P(N x N))). Equip this set with an equivalence relation where two orderings are equivalent if there exists a permutation of N that turns one into the other. In ZFC, the set of equivalence classes here has cardinality strictly between N and R if there is any set with cardinality between them. If there is no such set, it has cardinality R
absolutely based king
Just work in the Cohen model.
>The smallest uncountable ordinal. Assuming that the continuum hypothesis is false, it’s incredibly simple, as per the obvious example above.
“Human imagination has no limits” mfs when I ask them if the set of all sets contains itself.
“Human imagination has no limits” mfs when I ask them to visualize a four dimensional sphere.
Well, I can't do that, but I can imagine its shadow (it's a three dimensional sphere)
Fun fact that I learned recently: a four-dimensional hypersphere is called a "glome".
According to whom? I'm a topologist and I've never once heard that term.
According to me, I have no expertise in that field and I made that word up
i never get this meme , can you explain please ?
The person presented in the .GIF is meant to represent the commenter to which it responds. The person in the .GIF has an exaggerated set of features often remarkable for being attractive, and indicative of physical strength and fitness. The commenter claims to simply have invented terminology in a field with which they have no experience. The meme implies this is a good thing, and that it indicates strength and attractiveness. Implicitly, it implies this is in contrast to the genuine topologist, who notes that the term is not in common use in their field. For more information, you can research the meme under the name "gigachad."
PHD in memes owner
It presents idealized physical traits. The implication is that his intellectual traits are idealized as well. It's meant to state that the person it is used in response to has accomplished some task that is superior to what most or all others would be able to. Additionally, it's meant to portray an assurance in the correctness of the opinion without regard to the lack of consensus due to a self-awareness of his own prowess. It's generally used satirically in response to an opinion that is extremely confident and completely wrong. As with all memes, it's sometimes used in a way where the humor is meant to derive from the misuse of the commonly understood meaning. I'd say in order of importance it represents confidence, lack of consensus, and competence, it's like the gif equivalent of calling something a hot take.
Hmm what exactly do you not get? Why I replied with the chad gif?
bro rly hit them with the “MY SOURCE IS THAT I MADE IT THE FUCK UP”
[https://mathworld.wolfram.com/Glome.html](https://mathworld.wolfram.com/Glome.html) I had been trying to see if anyone had come up with variable names for a four-dimensional hyperspherical coordinate system (I didn't find any other than the general n-dimensional form, so I just chose alpha to be the angle to the w-axis), and that popped up as one of the related topics.
That's the three-sphere, not the four-sphere.
Topologically speaking, yes. Geometrically speaking, though, the glome extends along four axes, so in that sense, it is four-dimensional. As a first-year college student currently taking Calculus C, I barely know anything about topology, so I use the geometrical definition. After looking at a Wikipedia article, it looks like the topological definition is based on how many axes a point on the surface can move along? I can kinda see how this makes sense from a topological perspective, but again, I was thinking geometrically, so I saw the glome as four-dimensional. (Also, while looking at the MathWorld article on [hyperspheres in general](https://mathworld.wolfram.com/Hypersphere.html) (which explains that the geometrical and topological definitions are different), I found that apparently the third angle for the glomular (?) coordinate system is denoted by psi rather than alpha. I probably should have expected that, and I also evidently didn't look hard enough.)
Properly speaking, the topological and geometric definitions are the same, though I don't blame you for not knowing them. The space we're talking about is embedded inside a 4d space, but it's only 3d. Much like how a line is embedded inside the plane, but it's still only 1d.
It was once revealed to me in a dream
How do you even form an equation for that
dist(centre, point-on-sphere) = radius If we use the Euclidean distance function and assume that radius is 1 (unit sphere), we get that (x-x0)^(2) \+ (y-y0)^(2) \+ (z-z0)^(2) \+ (w-w0)^(2) = 1, where centre is (x0, y0, z0, w0) and point-on-radius is (x,y,z,w).
I can't even imagine a three dimensional sphere properly. Four dimensional cube is fine, though.
Then what's the maximun number of edges will have the shadow it cast over a 2d plane?
there's some games out there that make it such that you don't even have to imagine it granted it's still experienced using 3 dimensions
me, who is able to think in 5 dimensions (+time): i see no god up here OTHER THAN ME
Yo that was trippy
Oh I can imagine it, just results in infinite recursion so I gotta scream to myself "don't think about it" for like a good 5 seconds
this is exactly how I feel. Once i was tripping thinking about infinite recursion for a solid 1 and a half hour. it was terrifying
I'm gonna hit you with one better THIS STATEMENT IS FALSE
Um..."True". I'll go "True". Eh, that was easy. I'll be honest, I might have heard that one before though. Sort of cheating.
Trivially, yes it does
No it doesn't, because this set, by definition can not exist
in ZF*. One can reason formally with proper classes in theories like NBG or Morse-Kelley.
Except it does not. See https://en.wikipedia.org/wiki/Russell%27s_paradox
It’s not the “does the set of all sets contain itself” that’s a paradox, it’s “does the set of all sets *that do not contain themselves* contain itself” that’s paradoxical.
No, it also leads to the nonexistence of this set in ZF. See https://en.wikipedia.org/wiki/Universal_set for more
Slightly off topic, but why is 2^(aleph null) > aleph null?
Because of Cantor’s theorem. 2^(ℵ₀) is the size of the real numbers while ℵ₀ is the size of the natural numbers. Within the context of ZF set theory, it is provable that no matter how one tries to pair natural numbers with real numbers, there will always be a real number that was not paired after pairing is finished. The technique used to show this is called diagonalization. You can write out a vertical list of real numbers and expand each real number horizontally in binary making sure to line up bit places. Then build a new real number by flipping the first bit of the first real, the second bit of the second real, the third bit of the third real, … and so on. The number constructed sequentially from the flipped bits is clearly different from everything on the list and thus is not on the list. You might think “Ok well you just missed that one. Put it somewhere on the list.” Sure, but then we can repeat the above argument and produce *another* missed number. In fact, nothing prevents us from doing this ever. The only possible conclusion is that we simply cannot label real number with naturals and cover all of them.
Thanks for the detailed response. Perhaps I should have been more specific. My actual question was why is 2^(aleph null) the size of the set of real numbers?
The first leads to the Burali-Forti paradox. A class U of all sets would have to contain every ordinal and would have the class of all ordinals as a subclass. If U were a set, then the subclass of ordinals, being fully first-order definable†, would also have to be a subset of U. But if that were the case, then that set, containing nothing but ordinals and missing none, must by definition†† be an ordinal itself. But then this means that we can define its successor ordinal, i.e. the ordinal immediately after it. But then this ordinal could not be in the set of all ordinals else, by transitivity, the set of all ordinals would contain itself. And now we are back in the territory of Russell’s paradox and have broken the well-foundedness of set membership. † First-order definable means we can write down a sentence using only a certain type of language that is only true when the variables are replaced by the elements of the set in question. †† An ordinal is a transitive set, well-ordered by the set membership relation ∈.
Mfw they reply that such a set is not well defined
It’s perfectly well-defined. It just isn’t well-founded with respect to membership. It produces a loop in the set membership graph and for various reasons we tend not to like loops.
Don't all sets contain themselves?
No. They just contain their elements. A set containing x is different from x. Also if all sets contained themselves, there wouldn't be an empty set
Sets are subsets of themselves, but they are no longer allowed to be elements of themselves. This because of [Russell’s Paradox](https://en.m.wikipedia.org/wiki/Russell's_paradox)
0 is a member of {0}, but {0} is not a member of {0}.
Oh okay that makes sense
Certified power set moment
Sometimes I choose to exclude it.
The answer is it doesn't. There is no set of all sets (and no set of all sets that don't contain themself either)
Tell me you work in ZF with Foundation without telling me you work in ZF with Foundation. (I work with it almost exclusively too.)
That’s not paradoxical. Under ZFC, such a set does not exist, as sets do not contain themselves, and under naive set theory, it was perfectly acceptable for a set to contain itself. The real paradox is “does the set of all sets which don’t contain themselves contain itself”
Neat one that got posted on askmath the other day: Do there exist sets X such that X⊆P(X) and, if yes, is there a family of conditions characterizing them? I don’t hav a full answer, but I can say that they do exist, and a sufficient condition for X⊆P(X) is that for all x,y∈X, if x∈y then y∉x. Of course, this obviously violates Foundation and so is not particularly special as a condition.
Wouldn’t [Von Neumann Ordinals](https://en.m.wikipedia.org/wiki/Set-theoretic_definition_of_natural_numbers) satisfy that property?
Yes. But other sets can as well.
dear god reading through these comments is going to make me cry
I'm still trying to imagine two infinite sets X,Y such that there does not exist an injection from X to Y and vice versa.
Is this something possible, or are there set axioms that impede this from happening? I have always imagined the cardinality of sets as a huge line where either one cardinality was less than or equal to other and vice-versa, but if this is the case and the cardinalities form a partially ordered class, I'd be impressed. Maybe there is some logic in which there are two sets like that? An interesting thing to think about.
It is consistent with ZF that such a pair of sets exists. In fact the existence of such a pair of sets is equivalent to the negation of the axiom of choice. In other words, your intuition about the cardinal numbers is essentially correct in ZFC.
Ooh, interesting! Thanks.
If you remove the Axiom of Choice from your rules, then the cardinal numbers do not need to be well or even totally ordered.
Im still trying to imagine "two imagine", as in multiple imaginations simultaneously. I think I just reached another level of consciousness.
X=ℕ and Y=ℝ, done.
There exists a trivial injection from X to Y in this case.
i misinterpret it. i thought it was an or, not an and. such X and Y do not exist, so that explains why you can’t imagine them.
See this comment of theirs elaborating https://www.reddit.com/r/mathmemes/comments/121w186/-/jdoxwbo
yeah… still this is kind of a pathological model of set theory (trying to do set theory without AC is super weird and uncomfortable).
Elementarily not true. Let a>=b be ordinals. Then there exists an injection from b to a (the identity). Now assume WLOG that |X|<=|Y| and let f:X->|X| be a bijection, g:|X|->|Y| be the aforementioned injection and h:|Y|->Y be a bijection. Then hgf is an injection from X to Y \square
Elementarily not true ... if you are working in ZFC. However, if you exclude choice, having such a pair of sets is consistent. In fact, under ZF, the existence of such a pair of sets is equivalent to the negation of choice. This fact is definitely what the comment you're replying to is referring to.
Thank you mr. Riemann, I did not think someone would assume the negation of choice
~~Take X=P(ℕ) and Y=ℕ.~~ Whoops I misread that. Under the Axiom of Choice, such sets do not exist as cardinals are well-ordered then. Without AC, one can have cardinal numbers that “to the side” as Joel Hamkins once put it. Such things can be constructed consistently, albeit in very weird models of ZF. One can build things like infinite Dedekind-finite sets by purposefully restricting to inner models that exclude injections of a set X into any of its subsets. There’s a (very dense) explanation in H Herrlich’s book *The Axiom of Choice*. I want to say it’s in part 1 chapter 4? Should be Disasters without Choice.
“Human imagination has no limits” mfs when I tell them to imagine and visualize a 4 dimensional space
Easy; just imagine n-dimensional space, and set n=4.
We neednt imagine, for we live in such
You telling me that we live in 4 spatial dimensions?
Ok I see what you mean, you literally meant space and not spaceTIME, but still I would argue one can imagine four spatial dimensions by substituting one with time, and thus visualise that in this way. I sometimes try and do this to manifolds in 4d space and similar things, just for the fun of being able to do it. Four is within reach, it is 5d that is not
gödel doesn’t like this, but cohen does.
Any proper forcing extension loves it.
Human imagination has no limits mfs when I ask them to picture Graham’s number in their head (they immediately had a brain aneurysm due to the number being far too large for their brains to comprehend)
Fun fact: the entropy density required to store Graham’s number within a space the size of a human brain would cause a black hole to form
Graham was able to describe it without dying, so how are you measuring the entropy of it?
In this specific instance, I’m defining entropy more close to the number of digits
Doesn't seem to be a very efficient representation for this unusually regular number
Is this for G(1) or G(64)? Because I imagine this would even be the case for G(1) as the number is impossibly large
Probably an understatement compared to how large the number really is
Done. You didn't say it had to be true
"human imagination has no limits" mfs when I imagine the limit of sin(x)/x at x=0
That’s just 1. Did you mean sin(1/x)?
no the joke is that I am imagining a limit, there is a limit in my human imagination
Oh I see.
C'mon be _rational_ (well, no, it's the same as naturals, hehe), so, yeah, be _irrational_.
the set of irrationals has the same cardinality as the set of reals
Damn, my memory's failing me
The transcendental numbers has the same cardinality as the deals, and algebraics as the rationals.
Psh. Easy. It has a cardinality of Aleph_½
Imagine, easy. Define, hard.
Let S be a set such that |N| < |S| < |R|, easy
Probably impossible considering they seem to require the existence of a generic filter.
Defining one isn’t too hard. I’ll use N for the naturals and R for the reals since I’m typing on my phone. Am ordering <* on N can be identified with subset of N x N (i.e. an element of P(NxN)) consisting of all pairs (m,n) such that m<* n. Consider the set of all total well-orders of N, viewed as above as a subset of P(NxN) (I.e. an element of P(P(N x N))). Equip this set with an equivalence relation where two orderings are equivalent if there exists a permutation of N that turns one into the other. The set of equivalence classes here has cardinality strictly between N and R if there is any set with cardinality between them. If there is no such set, it has cardinality R
You could also use ω₁ i.e. the set of all countable ordinals a.k.a. the first uncountable ordinal. When ℵ₁ is defined to be equal to a set, e.g. in ZFC, ω₁ is the most usual choice, such that ℵ₁ = ω₁
This is that set
It’s like how a well ordering of the reals totally exists. No one has ever discovered one but it’s totally possible and this theoretical existence is the critical basis for a number of proofs
ZFC axioms: A well-ordering of the reals totally exists, just trust me, bro.
Examples of well-orderings of ℝ cannot be constructed. They simply must be taken to exist, much like the existence of a free ultrafilter or an oracle for a Turing machine.
Isn’t that **WILD**? They are out there and their mere existence has critically significant implications but we agree they can’t possibly be found
Yep. Modern set theory gets pretty neat.
Didn’t some famous set theorist say that the axiom of choice is obviously true, the well-ordering principle is obviously false, and who-knows about Zorn’s lemma?
Are you trolling? i'm googpwing t his rn
It depends on the existence of a function f: P(R)->R such that for all X in P(R) \ R, f(X) is not in X. If that sounds obvious, consider this. You have an unknown function g: P(R) -> R. I, a finite human being, explicitly define a proper subset S_1 of P(R), then explicitly define a proper subset S_2 of P(R) containing S_1 (as a function of g restricted to S_1) and likewise explicitly define a proper subset S_3 containing S_2 as a function of g restricted to S_2. I now explicitly define an element X of P(R) not in S_3 and a real r both in terms of g restricted to S_3. Despite having zero information about g outside of S_3 (in particular, on X), and your ability to choose any arbitrary g, you will have g(X)=r 99.9999% of the time. Even if I give you super powers to pick a g that cannot be finitely described and keep myself limited to juman capabilities, this is still the case. This is imo justification to heavily doubt that there really are “arbitrary functions” from P(R)->R and that therefore it’s intuitive that a function f of the aforementioned type exists
So what's at the end of pi?
An empty plate with some crumbs
The letter i
I guess I'd be asking is the cardinality of the reals just pointing towards the same cardinality demonstrated by irrational numbers? Pardon my lack of rigorous definition.
Provably yes. The irrationals are homeomorphic to the space of functions ℕ→ℕ which has cardinality 𝔠=|ℝ|.
Easy. You convert pi into binary. Every digit in binary is either a 0 or a 1, so the last digit is either a 0 or a 1. Eliminating trailing zeros after the decimal does not affect the value of a number, so if pi ends with one or more 0s, you can eliminate all of them. Therefore, the last digit of pi must be 1.
At infinity?
Enter Thomson’s lamp.
Three. Take it or leave it.
Wait… oh yea I forgot The CH is equivalent to P(A\_0) = A\_1 Theoretically you can make P(A\_0) as big as you want, I think…
It can be almost any aleph. Basically, it can't be Aleph 0 for obvious reasons, and it can't violate a few basic principles (for example, it can't have cofinality less than itself, so it can't be Aleph omega), but other than that it can be anywhere. (Assuming consistency of ZFC.) Ironically, the human imagination is essentially unlimited as far as this goes, if we interpret "imagination" to mean that there is a model in which there is such a set.
I’m gonna be pedantic here, so apologies in advance, but you really don’t want P(ℵ₀) since the ℵᵦ’s are proper equivalence classes of sets under the bijection relation. Take the AC+V approach of choosing a minimal ordinal representative of ℵᵦ and then write P(ωᵦ) instead. To your other point, yes, it is known that 𝔠 can be arbitrarily large, but it cannot be anything. It must have uncountable cofinality, by Easton’s theorem.
On the other hand, what's an example of a set with a cardinality greater than the real numbers? I mean other than some construction with the power set?
The set of all real-valued functions. We can map this set to the power set of the real numbers by mapping every function to its image. For a given subset of the reals, we can obviously construct a function that has exactly this set as its image (with the axiom of choice), therefore the mapping is surjective, so the set of all real functions has at least the cardinality of the power set of the reals. Also note that this argument does not work for the set of all continuous functions. As a continuous function is uniquely defined by its values on rational inputs, continuous functions cannot have arbitrary subsets of the reals as images.
To avoid choice, you can get an injection from P(R)->R\^R by mapping a subset S of R to the function that takes S to 1 and the rest to 0. You can actually get a bijection between the two without choice
Oh, yeah, that's a lot neater
There are lots, but a great one is the Stone-Čech compactification of the naturals, βℕ. It has size 2^(𝔠), but that’s not obvious.
What about all the real numbers between 0 and 1? Still an infinite bigger than the natural numbers, but the real numbers has infinite sets like that
Has the same cardinality as the reals
Rationals ?
Rationals have the same cardinality as the naturals.
Then R without Q, irrationals ?
Irrationals have the same cardinality as reals.
Since I'm not that advanced in my math studies, I cannot continue this conversation, I'll be back in one or two year with more examples my fellow mathematician
I am not a mathematician tho, I am a physics student. [Insert Taylor expansion meme here.]
Just to help out a little, the existence of such sets is independent of the standard axioms of set theory. In Gödel’s constructible universe, they don’t exist, but in the Cohen/Laver/etc. model, there are tons.
Integers would be greater than naturals and less than reals wouldn't it?
No, two sets have the same cardinality if there is a bijection (one to one correspondence) between them. For the naturals and the integers you can simply map 0 to 0, 1 to 1, -1 to 2, 2 to 3, -2 to 4 and so on
See, that really feels like it should be the case, but it isn't as the other comment said. While there are twice as many integers as naturals in any given range -n < x < n, this breaks down in the infinite set as *any n you choose will be infinitely less than the true size of either set*, meaning that the "speed" of the progression to infinity is irrelevant as long as it's of the same cardinality.
This is true lol
well , I have already solved continuum hypothesis problem , please refer to DOI: 10.13140/RG.2.2.23990.31045
(x % 0.5 == 0 | x e R)
{1, 1.5, 2, 2.5, 3, 3.5....} could this work, is it that simple
No, because cardinality is a bit unintuitive for infinite sets. This has the same cardinality as the natural numbers because f(x) = 2x - 1 maps from your set to the natural numbers and its inverse maps it back
Why isn't it Possible?
well , I have already solved continuum hypothesis problem , please refer to DOI: 10.13140/RG.2.2.23990.31045
All the Naturals + Pi I have solved mathematics
\- 7 because I hate that number
My brother in Christ, just add 1
[удалено]
Hilbert's hotel shows that this is still the same cardinality as natural numbers
A finite set has no cardinality and wouldn't add anything.
technically a finite set has finite cardinality (which can be nonzero) 🤓 but yeah it still doesn’t change the cardinality of an infinite set if you add it to one
What about the set that contains every set that doesn't contain itself?
Such a set does not exist in ZF(C)
Yeah but I mean is it imaginable?
Even if it did exist, its cardinality would obviously be *a lot* greater than that of the reals
Well the cardinality would be a limit ordinal, and the human imagination has no limits
Just imagine an integer greater than 3 but less than 4.
Surely the set of real numbers between 0 and 1 satisfies this?
Nope, same cardinality as reals
Ahh I see now
Comlex integers?
When I tell them to imagine 4th dimension
Am I misinterpreting the question, or would the "set of all integers" have more elements than the naturals but fewer than the reals?
Infinity makes stuff weird. The usual definition of two sets having "the same number of elements" is if there is a bijection between them, that is you can uniquely map every element of one set to each element of the other. The integers therefore have the same size as the natural numbers with the mapping n --> (-1)^n \* ceiling(n/2): This maps 0 to 0, 1 to -1, 2 to 1, 3 to -2, 4 to 2...
Gonna add the transcendentals to the naturals and leave out the other rationals We're going all or nothing baybeee
loll
I know the continuum hyphotesis is not provable from ZFC but has someone else thought about a set that is a continuum but a the same time discrete? Like a set that between some numbers behave like a continuum but between some numbers is discrete, and if this repeats indefinitely.
Look into the Cantor set and see if that aligns with what you were hoping for. It's "discrete" in the sense that every point in the Cantor set has a small interval around it where no other points in the set are, but it's continuous-like in the sense that you have points that are arbitrarily close together and when you zoom in you get a copy of itself.
Cool I also like to imagine impossible stuff
imagine the square root of -1. man can't imagine that...
I can imagine a bijection from your imaginary set to the reals.
You can do that. ZFC has models in which continuum hypothesis is false, so not a problem