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sometimes when i feel useless and stupid to the extreme i go to khan academy and do the absolute beginners maths and im like "wow im so good at this"

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Returning all my proofs as:

There would be no beauty in this world if this were false.

There is beauty in this world.

QED

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The thing about operator algebras is that all the literature is written like its algebra. By which I mean they'll write stuff like "you can easily see that [something I definitely can't see easily]". Analysis books have never done me dirty like this.

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Oh btw I didn't update you guys about my new thesis project: I'll be working on quantum measurements and the metaphors surrounding them

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Huge fan of functional analysis and anything functional analysis related. Differential geometry is fine but I prefer when differential geometry is used in combination with something (e.g. Lie groups or GR). Taking noncommutative geometry next semester which apparently takes a bit of both worlds? Anyways.

Algtop sounds cool but it just doesnt fit with the courses I am currently taking, so there's a chance I won't ever do any serious algtop even though it sounds really fun!

now I just need to know peoples favorite math classes, am I getting the right impression that mathblr adores algebraic topology?

also very curious about peoples opinions on functional analysis and on differential geometry -w-

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Have seen, imo depends on what kind of functional analysis the prof is involved with

Are all people who do functional analysis allergic to using im(T) and ker(T) to denote the image and the kernel of an operator T or is it just my lecturer?

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~~ Demonstrandum ~~

The last exam of the gauntlet: Algebraic Geometry 1. Although this is definitely a master-level course, it does not necessarily require (graduate) commutative algebra, which is good since I dropped that course early since I didn't enjoy it (although some lemmas were used in proofs). This also means that the course did not cover schemes and stuff, but rather stayed with abstract varieties: A topological space equipped with a structure sheaf. Anyways, too much yapping.

Q1: Let X, Y be varieties, with Y separated, and S a dense set in X. Show that equality of two morphisms on S is equivalent to equality of two morphisms on X. Use this to show that f(X) is dense iff f*: g mapsto g ° f is injective on the structure sheaf O_Y(Y). Honestly, BANGER question but I am a functional analysis-head, so am very biased here. Very cool imo I love dense sets, even if this question did not a priori have much to do with geometry 9/10.

Q2: Show that some sheaf of sets is indeed a sheaf of sets, and prove or disprove that it is a sheaf of abelian groups. Not a very exciting question, especially cause the sheaf seemed somewhat arbitrary, but I quite like sheafs after taking this course so 7.5/10.

Q3: Given some irreducible variety X given by an explicit, but too cumbersome to write here, closed set in P^1 x P^1 (over alg closed field k), determine that X is smooth. Then do some one-form trickery on X so that you can in the end glue things together and find the genus of X. To be honest, I get why they ask such questions, since being able to calculate explicit examples is important. Unfortunately, they chose polynomials from hell and everything went wrong when trying to prove smoothness, not just for me but for more people I spoke to. In the end I need to respect the question but it was a bit too much highschool algebra for my liking 7/10.

Q4: Some Riemann-Roch, Serre duality and divisor stuff on an irreducible smooth projective variety. I am 99% confident I got all the points here, but to be honest, I am still not sure what RR and Serre duality actually mean. Making the question is fine but my understanding sucks. Oh well, maybe Riemann surfaces next semester will help 7/10.

Overall, I think it was an 8 exam if the polynomials were nice to me. I also think it is unfair to judge an exam based on my inability to manipulate polynomials, so an 8 it is. Quite a fun exam otherwise and I quite liked the course even though it was conceptually the hardest one I took this semester.

About to take three math exams in a three day period. I studied well enough I think but still going to be a challenge. As such, I will rate every question on a scale of 1-10 after the exam, but not based on their difficulty but rather on their interestingness

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~~ Erat ~~

What's more fun than ordinary differential equations? That's right, partial differential equations! Although this was really the course I enjoyed the least over the last semester, but whatever. There were two parts to the course: One on general undergraduate-level theory (not a prereq) and one on elliptic theory and Sobolev spaces. Closed book, 3 smaller questions for each part.

Q1: Had us consider a PDE connected to a Sturm-Liouville type ODE via separation of variables, and had us solve said PDE (rare) via a L2 coefficients expansion. Not too exciting and not why I did the course, any physicist could do this (albeit less rigorously) 6/10.

Q2: For the so-called Linear Swift-Hohenberg equation, derive a solution formula for solutions using the Fourier transform. Use this to find some constants so that the solution with u_0 is some cosine exploded as time goes to infinity. Honestly I really like using the Fourier transform to find qualitative results of solutions, but unfortunately this one had us calculate the Fourier transform of the cosine, which you need distribution theory for to do rigorously (dirac deltas), which was not at all covered in the course... Otherwise cool question 7/10 could have been an 8.

Q3: State Hille-Yosida theorem and show the solution from last exercise is a semigroup. Nothing special, but free points and makes sense they ask this 7/10.

Q4: For a given elliptic PDE, show that for C^2 solutions a certain integral property holds, and define a weak solution accordingly. Then show there always exists a weak solution in H^1 (using Lax-Milgram). A classic question, basically the point of the course for me. Quite a lot of calculating and passing through limits, but rewarding in the end 8.5/10.

Q5: Show a certain function has a weak derivative and show some inequality for smooth functions, which you can extend to W^1,p functions. For this I just straight up goofed. Was supposed to be an easy question but CLEARLY positives and negatives don't cancel out, so I screwed everything up. Decent question otherwise though 7/10.

Q6: Show some bound on the H2 norm of a solution to `elliptic operator u = f` compared to the L2 norm of u and f. Honestly just barely had time for this question so I quickly scrabbled somethings that some terms would cancel. Also had a subequestion to state Fredholm's alternative which I did cause that is easier, just remembering. Maybe a cool question? Idk really 6.5/10.

Overall, a fine exam but nothing which really blew my mind. To be honest this was the course I was least hyped about anyways, so in the end that only makes sense. Was representative though so also a solid 7.

About to take three math exams in a three day period. I studied well enough I think but still going to be a challenge. As such, I will rate every question on a scale of 1-10 after the exam, but not based on their difficulty but rather on their interestingness

[positivelyprime]

~~ Quod ~~

I like big groups and I cannot Lie. First exam: Lie groups. In the one (1) past exam we got three large questions to answer, now we got five smaller questions in this exam (possibly because they switched to close-book this year.

Q1: Show that SO(p,q) is a Lie group (corresponding to a metric with (p,q)-signature) and determine its Lie algebra. Also show that a certain set is a homogeneous space for SO(2,1). A classic question for warming up, nothing special but testing the skills you learned. Somehow forgot definition of homogeneous space 7.5/10.

Q2: Show that for connected Lie group G, phi equals psi iff their derivatives at identity are equal. Not sure why we got this question? It was almost too easy? Nice 10 points though but not that interesting 6.5/10.

Q3: Three-parter which had us investigate the Lie algebra and exponential mapping of R\{0} (semidirect product) R, and investigate left and right Haar measures on this group. Was very similar to a homework exercise I made, so I think I got this one fully correct. Also relevant to the course, but did not necessarily culminate in a grand finale, still 7.5/10.

Q4: A 4-part question, culminating in proving the Schur orthogonality relations. A fair question, we spent an hour(?? time hard) on this in class and it was cool to revisit the proof, which I luckily remembered enough to get most of the parts correct 8/10.

Q5: Here's where things get funky: Investigate representations SL(2) and its Lie algebra sl(2) using four subquestions. The final result was that SL(2) had no unitary representations! Now unfortunately, we spent basically all of the representation theory on compact groups and unfortunately did not really get an outlook on non-compact groups. I guess this question was supposed to make up for that. The idea was cool, but at the end the question got a bit vague and tbh I am not really convinced yet why this should be true now, but okay. Would have been cool if better executed (also skill issue on my part) but for now 7/10.

Final review: Solid 7 exam, but quite tough. I think I ended up getting around an 8? Hopefully. But really everyone I talked to (~8 out of ~15-20 students) thought it was a hard exam so I really wonder what kind of grades there will be.

About to take three math exams in a three day period. I studied well enough I think but still going to be a challenge. As such, I will rate every question on a scale of 1-10 after the exam, but not based on their difficulty but rather on their interestingness

[positivelyprime]

About to take three math exams in a three day period. I studied well enough I think but still going to be a challenge. As such, I will rate every question on a scale of 1-10 after the exam, but not based on their difficulty but rather on their interestingness

[positivelyprime]

I never really understood this business of classes and sets. I get that certain collections cannot be called a set (for instance, the Russell's Paradox). Ok then what exactly are classes? Do they behave like sets? Why should I care if a category is small !!

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Can highly recommend this game, was obsessed with it like a year ago. Also spoilers under read more:

Once you unlock the levels where they combine rules shit gets real. A single puzzle could easily take an hour for me (although I am stupid)

this game is fucking unbelievable

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Cardinal Numbers

Nyaa!~

I have too many drafts :c

Hi everyone! Today, we are going to learn about infinite cardinal numbers! I hope this'll be fun! ^^

I was also planning to explain something about the axiom of choice, but I decided that it's better if that becomes its own blog-post.

So, first: what is a cardinal number?

✦ Motivation ✦

Well, cardinal numbers are numbers that represent the size of a set of things. For example, {owo, uwu} is a set of cardinality 2: it has two elements, owo and uwu.

However, we can't simply count the elements of a set to figure out its cardinality. What if we come across a set like {0, 1, 2, 3, ...}? I mean... you can try counting the elements, but it's gonna take you a veryy long time.

So! What do we do?

Well,,, first, observe that cardinals can't exist in a vacuum. If we say that {owo, uwu} has cardinality 2, what does that actually mean? ‘2’ is just a symbol we use to describe the cardinality of {owo, uwu}, but has no meaning on its own. Where it gets interesting is when we compare the cardinality of {owo, uwu} to {10, 3}: they're the same! Now, that is interesting.

So, we use ‘2’ to describe the cardinality of sets like {owo, uwu} and {10, 3}, and we use this symbol so we can more easily recognize when two sets have the same cardinality, without having to search for statements like “this set has the same size as this other set” exhaustively having to apply transitivity of (cardinal) equality.

But what makes {owo, uwu} and {10, 3} so similar that they can be considered ‘equinumerous’?

Well,,, observe that the cardinality of a set doesn't change when we replace some of the elements with other elements: the cardinality of a set is completely independent from what the members of that set actually are. If we have a set like {catgirl lucy, my dearest friend, Semi}, and I replace ‘my dearest friend’ with ‘my worst enemy’ (so we get {catgirl lucy, my worst enemy, Semi}), the size of the set doesn't change!

I hope these two observations are enough to motivate the definition of a cardinal number:

～ Cardinal Equality ～

[Definition] Two sets, A and B, have the same cardinality, denoted |A| = |B|, iff there exists a bijection f: A → B.

There are three kinds of functions that will be important to us in this blog-post:

An injection is a function f: A → B from some set A to some set B so that, for all x,y ∈ A, if f(x) = f(y), then x = y.

A surjection is a function f: A → B from some set A to some set B so that, for all y ∈ B, there is some x ∈ A so that f(x) = y.

A bijection is a function f: A → B from some set A to some set B that is both an injection and a surjection.

A bijection thus pairs every element of A with a unique element of B, and vice versa.

So now, we need to show that this definition of cardinal numbers actually makes sense. I.e., that it is an equivalence relation:

An equivalence relation is a relation ~ such that:

~ is reflexive: x ~ x for all x

~ is symmetric: if x ~ y then y ~ x

~ is transitive: if x ~ y and y ~ z, then x ~ z

...

What?

OK, fine... I'll give you the answer: reflexivity of cardinal equality is witnessed by the identity function on a set, symmetry is from the functions inverse and transitivity comes from composition. I do want you to start thinking about how to solve these problems in the future!

Given an equivalence relation ~ on some class C, we can define a function F with domain C so that, for all x,y ∈ C, F(x) = F(y) iff x ~ y. If ~ fails one of the conditions of an equivalence relation, such a function does not exist.

This time, I will leave you to ponder why~

So! We can choose some function |·| that maps sets X to objects |X| such that |X| = |Y| iff X and Y have the same cardinality, the choice of mapping doesn't really matter.

The cardinality of ℕ, the set of natural numbers, is denoted ℵ₀ (‘aleph-nought’), this is also the cardinality of ℤ, the set of integers, and ℚ, the set of rational numbers. ℵ₀ is an infinite cardinal (not 0, 1, 2, 3, etc), but it isn't the only infinite cardinal!

|P(ℕ)| = 𝔠, the continuum, is another infinite cardinal. It is the cardinality of the set of subsets of ℕ, i.e. the set of sets of natural numbers.

Why are these cardinals different? Well, given some function f: ℕ → P(ℕ), try to find some subset of ℕ that is not in the range of f~ Thus, show that f cannot be surjective!~

Here is a hint for when you get stuck: if there is some x for which A and B disagree on whether they contain it (x is in A but not in B, or x is in B but not in A), then A and B are different.

Now, that we have gone over the basics of cardinal equality, let's look at something more exciting~ cardinal comparison:

⊰ Cardinal Comparison ⊱

[Definition] For two sets A and B, |A| ≤ |B| iff there exists an injection f: A → B.

≤ is a partial order on cardinal numbers:

≤ is reflexive: x ≤ x for every cardinal x

≤ is transitive: if x ≤ y and y ≤ z, then x ≤ z

≤ is antisymmetric: if x ≤ y and y ≤ x, then x = y

Reflexivity and transitivity are easy exercises for the reader~ (the proof is literally just the same as refl and trans for cardinal equality :p). Antisymmetry, on the other hand, is a lot more difficult!

That cardinals are antisymmetric is known as the Shröder-Bernstein theorem. The proof may be hard to follow, so good luck! :D

Let A and B be sets and suppose there are injections f: A → B and g: Β → A. We want to define a bijection h: A → B.

We can see that f and g form chains a₁ ↦{f} b₁ ↦{g} a₂ ↦{f} b₂ ↦{g} ... alternating between elements of A and elements of B. Since g is an injection, for every a ∈ A, we either have that it comes from some b (i.e. g(b) = a), or from nothing. The same holds for f. So every element in A and B belongs to a unique chain ... ↦ * ↦ * ↦ ... defined by f and g. Some of these chains have a sudden start (i.e. some a ∈ A or some b ∈ B that doesn't have an inverse under g or f), and some of these chains are infinite.

For a ∈ A, we can define h(a) based on what kind of chain a belongs to. If the chain a belongs to starts in A, we can set h(a) = f(a). If the chain a belongs to starts with some element in B, we can set h(a) = g⁻¹(a) (i.e. we set h(a) to an element b in B such that g(b) = a). If a belongs to a chain that goes infinitely far left, we do whatever. I'm just going to choose h(a) = f(a).

I'll leave it to you to verify that h is indeed a bijection.

We can also define strict comparison:

[Definition] For cardinals x and y, x < y iff x ≤ y and x ≠ y.

Since ℵ₀ ≠ 𝔠, as shown in the previous chapter, and n ↦ {n} forms an injection from ℕ to P(ℕ), we have ℵ₀ < 𝔠. In fact, for every cardinal x, there is some cardinal y so that x < y.

Here is some basic terminology for partial orders:

[Definition] Let ≤ be a partial order on some class P and let A ⊂ P be a subclass of P. A maximum element of A is some x ∈ A so that, for all y ∈ A, x ≤ y → x = y. A minimal element of A is some x ∈ A so that, for all y ∈ A, y ≤ x → x = y. The greatest element of A, if it exists, is some x ∈ A so that, for all y ∈ A, y ≤ x. The least element of A, if it exists, is some x ∈ A so that, for all y ∈ A, x ≤ y. An upper bound of A is some x ∈ P so that, for all y ∈ A, y ≤ x. A lower bound of A is some x ∈ P so that, for all y ∈ A, x ≤ y. The infimum of A, denoted inf(A), if it exists, is the greatest lower bound of A. The supremum of A, denoted sup(A), if it exists, is the least upper bound of A.

Given x,y ∈ P, the meet of x and y, often denoted x ∧ y, is the infimum of {x,y}. The join of x and y, often denoted x ∨ y, is the supremum of {x,y}. If every pair of elements in a poset (partial ordered set) has a meet and a join, then that poset is called a lattice. x and y are comparable iff x ≤ y or y ≤ x, x ⊥ y is used to denote that x and y are incomparable. Examples of lattices are: the powerset lattice (P(X),⊂) of any set X, where x ∧ y = x ∩ y and x ∨ y = x ∪ y, linear orders like ℚ, where x ∧ y = min(x,y) and x ∨ y = max(x,y), etc.

Without assuming the axiom of choice, two cardinals needn't have a meet or join. Now you know random useless stuff about posets!!

Oh, and: ℵ₀ is a minimum infinite cardinal. Proof is left as an exercise~

❈ Cardinal Arithmetic ❈

Cardinal numbers are numbers, so it'd make sense if we could do arithmetic on them. And we can!

[Definition] For sets A and B, |A| + |B| is the cardinality of the set A ⊔ B, i.e. the disjoint union of A and B. Members of A ⊔ B are (0,a) and (1,b) for a ∈ A and b ∈ B.

[Definition] For sets A and B, |A| · |B| is the cardinality of the Cartesian product A × B of A and B. Members of A × B are ordered pairs (a,b) for a ∈ A and b ∈ B.

Arithmetic on finite cardinals works as you'd expect: 6 + 3 = 9 and 12 · 2 = 24. Addition and multiplication on infinite cardinals, however, is actually quite boring, as x+y and xy are simply max(x,y). Well, that is, if you assume AC. Without AC, cardinal arithmetic can actually get very interesting (and weird af).

I am not that familiar with cardinal arithmetic without the axiom of choice, so you'll have to do more research on that on your own if you're interested!

Cardinal addition and multiplication are commutative, x+y = y+x, xy = yx, associative, (x+y)+z = x+(y+z), (xy)z = x(yz), and multiplication distributes over addition, x(y+z) = xy+xz.

Personally, I think cardinal exponentiation is more interesting than addition or multiplication:

[Definition] For sets A and B, |A|^|B| is the cardinality of the set of functions from B to A.

For cardinals x and y, if x ≥ 2, then we can prove that x^y > y. 𝔠, the cardinality of the continuum, is equal to 2^ℵ₀ and to ℵ₀^ℵ₀. Also, it's the cardinality of the set of real numbers ℝ. Here's a vid I found that explains why.

I think I might be getting too eepyy to write. I'll write more tomorrow.

[zzz]

Good morning!

Cardinal exponentiation has all the properties you'd think it has: x^y · x^z = x^(y+z) and (x^y)^z = x^(yz). Also, 0⁰ = 1.

There are also infinite sums and infinite products:

[Definition] Σ_(i ∈ I) A_i is the set of tuples (i,a) for i ∈ I and a ∈ A_i. Σ_(i ∈ I) |A_i| = |Σ_(i ∈ I) A_i| is the cardinality of this set.

[Definition] Π_(i ∈ I) A_i is the set of functions f with I as domain and f(i) ∈ A_i for all i. Π_(i ∈ I) |A_i| = |Π_(i ∈ I) A_i| is the cardinality of this set.

Σ_(i ∈ {1,...,k}) x_i = x₁ + ... + xₖ and Π_(i ∈ {1,...,k}) x_i = x₁ · ... · xₖ, so this is a valid extension of sums and products. The sum of x many y's, i.e. Σ_(i ∈ I) x for |I| = y, is the same as the product of x and y. The product of x many y's, Π_(i ∈ y) x, is equal to x^y. If X is a family of sets, I also sometimes write ΣX and ΠX for Σ_(A ∈ X) A and Π_(A ∈ X) A.

Here is a fun fact that idk where to place in this blog post: a Dedekind infinite cardinal is a cardinal x = |A| for which there exists an injection f: A → A that is not surjective. In other words, x+1 > x. Without the axiom of choice, infinite Dedekind finite cardinal numbers (aka mediate/Dedekind cardinals) can exist, and these cardinals are incomparable to ℵ₀.

Here is a fun question: is the product of any number of non-zero cardinals always non-zero?

Well?~

◈ Axiom of Choice ◈

The axiom of choice (AC) states that the product of any number of non-zero cardinals is always non-zero. I.e. for any family X of non-empty sets, ΠX is non-empty (a member of ΠX is called a choice function for X).

Eh... I didn't really plan what to write in this chapter besides what the axiom of choice is. Oh, well!

...

Wait-

Wait, WHAT THE F---

So.. apparently, you cannot take infinite products of cardinals without assuming the axiom of choice. ℵ₀ is the cardinality of ℕ, but ℕ is not the only set with cardinality ℵ₀. If A has cardinality ℵ₀, then in how many ways does it have cardinality ℵ₀? Well, if a bijection f: A → ℕ exists, then 𝔠 many of those bijections exist. If we take ℵ₀ different sets A₀, A₁, A₂, ... all with cardinality ℵ₀, we have, for each k, multiple bijections f: Aₖ → ℕ. We would expect |Π_(k ∈ ℕ) Aₖ| = ℵ₀^ℵ₀ = 𝔠, but how do we choose a bijection f: Aₖ → ℕ for each natural number k? If we only had a finite amount of Aₖ, this wouldn't be a problem. However, we have an infinite amount of Aₖ... so we need to make an infinite amount of choices. So, we can define the set X = {{f | f: Aₖ → ℕ is a bijection} | k ∈ ℕ}, and a choice function for X gives us a bijection for each Aₖ! But... we need AC to prove such a choice function exists.

...

HOW AM I SUPPOSED TO LEARN AD WHEN THIS F----D UP SH-- HAPPENS?

..sorry for screaming at you.

Luckily, infinite cardinal addition still works as normal without AC... probably.

Maybe.

Nope, it doesn't.

【 Aleph Cardinals 】

As you can see, we have run out of symbols to put on the sides of the chapter titles.

Anyways! So, a well order is an order (A,≤) where:

≤ is a partial order on A.

≤ is total/linear: for all x and y in A, x and y are comparable by ≤.

≤ is well-founded: all non-empty S ⊂ A have a minimum element.

Well, ig those random facts about posets weren't completely useless.

I might make a blog post about ordinals later, which'll go more in-depth on well-orders and linear orders.

[Definition] An aleph cardinal is an infinite cardinal number |A| for which there exists a well-order (A,≤).

An example of an aleph cardinal is ℵ₀: ℵ₀ = |ℕ| and the usual order on ℕ is a well-order. The n-th aleph cardinal is written as ℵₙ, starting at 0th of course :3

The statement ‘ℵ₁ = 𝔠’ is known as the continuum hypothesis (CH), which is both unprovable and undisprovable. I might make a blog post about forcing in the future! ^^

Although cardinals (without AC) don't need to be linearly ordered, aleph cardinals are linearly ordered and even well-ordered. A proof of this is left as an exercise~ Because, well, of course it is :P

The well-ordering principle states that every set A has a well-order (A,≤). Equivalently, every infinite cardinal is an aleph cardinal.

It turns out: AC and the well-ordering principle are equivalent! Although AC, when written in its original form, seems more obviously true (to me at least), I think the well-ordering principle is a lot more useful.

I might go more in depth on why this is in my post about ordinal numbers. (why AC and the well ordering principle are equivalent)

❖ Cofinality ❖

There are two definitions of cofinality: one for cardinals, and one for ordinals. Since this blog-post is about cardinals, I'll give the one for cardinals:

[Definition] The cofinality of a cardinal x = |A| is the minimum cardinality of a partition X of A into sets of cardinality <x.

I often use cf(x) to denote the cofinality of x and cof(x) to denote the class of cardinals with cofinality x, though people also often use cof(x) to denote the cofinality of x. A partition of a set A is a set X such that all members of X are subsets of A, none of the members of X intersect and the union of all members of X is A.

We have cf(0) = 0, cf(1) = 1, cf(2) = 2 and for every finite n > 2, cf(n) = 2. 3 = |{a,b,c}| can be partitioned into {{a},{a,b}}, the partition {{a,b,c}} doesn't work as this has a set {a,b,c} of cardinality 3, which is not < 3.

[Definition] A cardinal κ is regular iff cf(κ) = κ.

Equivalently, every partition of κ either has cardinality κ or an element of cardinality κ.

We have cf(cf(x)) = cf(x) for every cardinal number x, so cf(x) is always regular.

0, 1 and 2 are examples of finite regular cardinals, though often, 2 is not regarded as regular. ℵ₀ also is a regular cardinal.

A cardinal that is not regular is called a singular cardinal. An example of a singular cardinal is ℵ_ω, which is the sum of ℵₙ for finite n. cf(ℵ_ω) = ℵ₀ as {ℵ₀,ℵ₁,ℵ₂,...} is a partition of ℵ_ω = Σ_(n ∈ ω) ℵₙ into ℵ₀ many cardinals each of which is <ℵ_ω.

Here is the ordinal definition of cofinality, which is (a lot) more often used if you have the axiom of choice:

[Definition] The cofinality of an order (A,≤) is the minimum cardinality of a cofinal subset S ⊂ A. S is cofinal if ∀x ∈ A ∃y ∈ S, x ≤ y.

Assuming the axiom of choice, the cofinality of a cardinal κ is often defined as the cofinality of the minimum ordinal α such that |α| = κ. An ordinal is the order-type of a well-order, and one ordinal is less than another if there exists an order preserving injection from one to the other. cf(κ) with the above definition agrees with the one I gave earlier for infinite κ, but if κ > 0 is finite, then cf(κ) is 1 instead of 2. I might talk more about this in my blog-post about ordinal numbers. For now, we'll just use the partition definition of cofinality.

If x is infinite Dedekind finite, then cf(x) is 2, which I think is kinda funni. We can also have cf(𝔠) = 2 if we omit choice :3 Though with choice, cf(𝔠) is uncountable (i.e. >ℵ₀).

I don't know how to end this blog, so here is some random terminology:

A cardinal x is finite iff κ is 0, 1, 2, 3, etc.

A cardinal x is countable iff it's ≤ℵ₀. It is uncountable if it's not ≤ℵ₀.

An alpeh cardinal is an infinite well-orderable cardinal.

A strong limit cardinal is a cardinal x for which, for all y < x, we have 2^y < x.

ℶ_0 = ℵ_0, ℶ_(n+1) = 2^(ℶ_n) and ℶ_λ = sup{ℶ_α | α < λ} for limit ordinal λ. A cardinal of the form ℶ_n is called a beth cardinal and ℶ₁ = 𝔠.

Assuming AC, x⁺ is the next cardinal after x, called the successor cardinal. AC is needed to ensure there exists a cardinal directly after κ.

The continuum hypothesis (CH) states that ℵ₁ = 𝔠. The generalized continuum hypothesis (GCH) states that 2^x = x⁺ for all infinite x. Both are unprovable and undisprovable from the usual axioms of ZFC.

Assuming AC, a weak limit cardinal (or simply limit cardinal) is a cardinal x for which, for all y < x, y⁺ < x. Assuming GCH, weak and strong limit cardinals are the same.

A strongly inaccessible cardinal (or simply, inaccessible cardinal) is a regular strong limit cardinal. Assuming AC, a weakly inaccessible cardinal is a regular weak limit cardinal. The existence of an inaccessible cardinal implies the consistency of ZFC, and thus, by Gödel's incompleteness theorems, ZFC cannot prove the existence of an inaccessible cardinal.

That's all I had to say about cardinals (for now), bye!~

[positivelyprime]

@isomorbism

[positivelyprime]

I'm a very open-minded guy if you use the discrete topology