Lecture 1: Set Theory

Hello, Veeky Forums. This will be the first lecture in a series I will be giving here on Brave New Algebra. We will be starting from rudimentary set theory. This thread is meant to be a discussion place, so if anybody has any questions, feel free to ask. Shitposting is accepted but discouraged.

A set is naively a collection of things. We will start by imposing the axioms of Zermelo-Fraenkel set theory, coupled with the axiom of choice (sue me). But first, some definitions.

Two sets are equal, written A=B, if, given any element x, x is in A if and only if x is in B.

The product of two sets A and B is the set AxB:={(a,b) : a in A, b in B}. We will impose axioms to ensure that these two operations are allowed and produce sets.

A set A is a subset of another set B if, given any element x, x in A implies that x is in B. Exercise: prove that A=B if and only if A and B are subsets of one another. Exercise: show that the empty set {} is a subset of every set.

The union of two sets A and B is the set AUB:={x : x in A or x in B}.

The intersection of two sets A and B is the set A∩B:={x: x in A and x in B}.

A function f:A->B is a subset of AxB such that if (a,b) and (a,b') are both in f, then b=b'. Intuitively, a function assigns each element in its domain (A) to a unique element in its codomain (B). A function is injective if (a,b) and (a',b) in f imply that a=a'. A function is surjective if for every element b in B, there is an element (a,b) in f. Notation-wise, if (a,b) is in f, we write f(a)=b. Exercise: can there be a function that is both injective and surjective?

Now, the axioms.

1. Given sets A and B, there is a set AUB so that x is in AUB if and only if x is in A or x is in B.

2. Given any two sets A and B, there is a set C so that A and B are elements in C.

3. Given any set A, there is a set P(A):={S : S is a subset of A}.

4. Two sets are equal if they contain the same elements.

5. Every set A contains an element B such that A and B are disjoint. (That is, they have an empty intersection: they share no elements.)

6. (informally:) Given any formula p in the language of set theory defined over elements of a set X, there is a subset of X written {x : x in X and p(x)}.

7. There exists a set S such that the empty set is an element in S and if s is in S, then sU{s} is also a member of S. This axiom says that there exists an infinite set.

8. The product of a family of nonempty sets contains at least one element. (This is the axiom of choice.)

Let's discuss ordinals.

Definition. An ordinal is a set S such that every element in S is also a subset of S, and so that set membership defines a strict well-ordering on S.

Exercise. Show that the empty set is an ordinal, and show that no element is less than the empty set (it is the minimal ordinal).

With {}=0 the smallest ordinal, we can inductively define the successor of an ordinal S(k) to be the powerset of k. Ordinals form a proper class, and they will probably turn out to be useful somewhere down the line. I figured we should have them under our belt.

Time to talk about functions.

Let f:A->B be a function. For any element b in B, the fiber f^-1(b) is the set of elements in a such that f(a)=b.

Exercise. Show that a function is surjective if and only if the fiber over every element in the codomain is nonempty. Show that a function is injective if and only if the fiber over every element in the codomain is isomorphic to either the point (the set {*}) or the empty set. Characterize a bijective function (one which is both surjective and injective) in terms of fibers.

Given a function f:A->B, the image of f is the subset of B on elements with nonempty fiber. Show that a function surjects onto its image, and that an injection has image isomorphic to the domain.

Generalizing fibers, the preimage of a subset B' of B is the set of elements a in A such that f(a) is in B'. The preimage of the image of a function is the domain. The preimage of a single element is the fiber over that element. The preimage of the empty set is the empty set.

Given sets A and B, the set of functions from A to B is often of interest. We denote this set B^A (imagine the functions "dropping" from A to B). There is an evaluation function ev:Ax(B^A)->B which is defined by ev(a,f)=f(a). It takes an element in A and a function from A to B and yields the evaluation of that function at that element. Later on, we will characterize function sets using universal properties.

Given sets A and B, there are two projection functions p_A:AxB->A and p_B:AxB->B, defined by p_A(a,b)=a and p_B(a,b)=b. Projections are always surjective.

Exercise. Show that the fiber of the projection onto A (as above) is always canonically isomorphic to B, and visa versa.

Again, we will later show that a product and its projections satisfy a universal property. This will allow us to morally move the definition of a product to other settings using the language of category theory.

I think this pretty much concludes the lecture. I was thinking about talking about cardinals, but they will be easier to discuss later anyways. Size issues are really just a nuisance in category theory. Thanks guys. Any questions?

If not, then the next lecture will be tomorrow. We're talking about the basics of category theory! Come prepared.

>prove that A=B if and only if A and B are subsets of one another

A inside of B and B inside of A is equivalent to x in A implies x in B AND x in B implies x in A which by definition is x in A if and only if to x in B.

>Exercise: show that the empty set {} is a subset of every set.

We have the proposition x element of {} implies x element of A (arbitrary set).

x element of {} is a false statement and as we know, an implication that starts with a falsehood is always true. Therefore indeed x element of {} implies x element of A.

>Exercise: can there be a function that is both injective and surjective?

Consider the set {a} and {b} and the function {(a,b)}, by the definition it is injective and surjective.

>Every set A contains an element B such that A and B are disjoint.
Explain

>Exercise. Show that the empty set is an ordinal, and show that no element is less than the empty set (it is the minimal ordinal).

x element of {} implies x subset of {} is trivially true as there is no element in {} so we have an implication that starts with a false statement. ez pz.

Assume there exists some set A so that it is an element on the empty set and is a subset of the empty set.

But the empty set has no elements and therefore this is a contradiction and thus A does not exist.

ez pz.

disjoint means that they don't share any elements

otherwise, every element of A has an element of A. so there's an arbitrarily infinite containment chain.

for example, say x in A, y in A, y in x. then y in x in A. but since y in A, y has an element z in A. then z in y in x in A

and so on

I know that but if we take an element(say x) from a set(say s) won't that be present in the intersection of x and s. Give an example maybe?

>Exercise. Show that a function is surjective if and only if the fiber over every element in the codomain is nonempty.

Assume there exists son b in B such that the fiber of b is the empty set. That means that there is no a in A so that f(a) = b and therefore it contradicts the definition of surjective.

The other way around, if f is surjective then that means every b has an element a so that f(a) = b and that completes the 'only if' part of the proof.

>isomorphic
wew lad, where did you previously define that?

>canonically isomorphic
Holy shit nigga chill. You throwing this shit at me with no definition? God damn.

>I think this pretty much concludes the lecture

Congrats, no one learned anything from your lecture because you asked questions with terms you have not defined.

Say the set A is {1,2}. Say the element B is 1. Then A intersection B is {1} . similarly for 2. So how are A and B disjoint?

x won't have an element in the intersection.
the intersection of {x} and s is {x} but the intersection of x and s is something else.

for example take 3 := {0,1,2}
then 3n2 = {0,1}, 3n1 = {1} and 3n0={}

You are confusing subsets with elements: note that there is a difference between the set X and {X}.

the intersection of 1 and {1,2] is empty
1 is {0}, not {1}

Sorry: two sets are isomorphic if there is a bijection between them. There is not a real definition for canonical isomorphism in terms of the fibers, but you will see what I mean.

The fiber of p_A over a is the set {(a,b): b in B}. How would you construct a bijection between any such fiber and B?

I think I got it now. Thanks

>This axiom says there exists an infinite set.
Opinion dismissed

Wilberger pls

you haven't defined what sets are, or what elements are.

there's no way you can discuss elements of sets when we have no idea what you're talking about.

Instead of trying to use fancy words to describe what you're doing. Start with definitions and theorems. Fuck the context

>we can inductively define the successor of an ordinal S(k) to be the powerset of k
Aren't those the cardinals?

According to ZFC, elements are sets themselves. A set is any object in a model of ZFC. That's enough for a well-defined mathematical theory. The axioms are defined for the language with just one relation symbol (for inclusion), and any collection of objects in this language that satisfy the axioms above form a universe for ZFC. Any such object is a "set."

I'm avoiding deep technicalities because none of it will be important once we get into categories. Lawvere's ETCS is a better system for set theory I think; I just wanted to introduce the language and intuitions of set theory so that they will carry over when we discuss homotopy fibers and such.

Also what about limit ordinals, i.e. those which cannot be represented by successor functions?

Shit, you are right. I started writing a section on caridnals before switching to ordinals, and forgot to change it.

The successor of an ordinal k is the set kU{k}.

Suppose we have two sets A and B and define the operation set difference as A - B = {x ∈ A : x ∉ B}.

Now, given the sets X and {X}, what is X - {X}?

A weak limit ordinal is one which is not the successor of another ordinal and not 0.

Ordinals won't show up too much as we progress in subsequent lectures. I am not too concerned with getting into them. I can organize a more focused lecture on them if people are interested.

The axiom of foundation guarantees that X-{X} will be empty for all X (because X is the only element in {X} and cannot be an element of itself).

you mean X - {X} will be X for all X

How do I see that something is NOT a set?

I read that the class of all vector spaces or the class of all groups are too large to be sets. But I don't see why.

Right, I was describing {X}-X. Sorry. The same argument applies in both cases.

>A function f:A->B is a subset of AxB such that if (a,b) and (a,b') are both in f, then b=b'.

If I give you a set of ordered pairs f such that if (a,b) and (a,b') are both in f, then b=b', is f still a function? Or does a function have to contain knowledge of its domain and codomain? In the latter case, a function should be a triple (f, A, B).

>A function f:A->B is a subset of AxB such that if (a,b) and (a,b') are both in f, then b=b'.
Actually that's only a partial function because you didn't mention that for each a in A there has to be a b in B such that (a,b) is in the function.

A class won't be a set if it violates any of our axioms. The axiom 5 above, called the axiom of foundation, guarantees that no set can be an element of itself, and this can be used to show that the "set" of all sets, which must definitionally contain itself, is not actually a set as it violates 5.

Oh, right right. Thanks.

(Sorry for all of the errors guys; if you haven't noticed, I am trying to rush through foundations to get to the meat.)

>2. Given any two sets A and B, there is a set C so that A and B are elements in C.

You mean such that A and B are the *only* elements in C.

if you can't prove it's a set, it's probably not a set
we know of one object that isn't a set: the class of all cardinals. why? well, what's its cardinal?
so if you want to prove something isn't a set, make an injection from class of all cardinals
for example the class of vector spaces. for any set X we can make X a Q-vector space of basis X. for instance for X = {a,b,c} it's a vector space of dimension 3. So just take the injection that takes X to the vector space generated by X for each cardinal X (or hell, each set X) and there you go, it's too big

>8. The product of a family of nonempty sets contains at least one element.

You have not defined what a family is.

a "family" an "element" a "set" an "object", etc etc are all different names for a set

Ah, very nice and elegant. Thank you.

We can keep the looser version then just use comprehension, right?

> The product of two sets A and B is the set AxB:={(a,b) : a in A, b in B}.

This is not well defined since you have not defined what (a,b) means.

> The union of two sets A and B is the set AUB:={x : x in A or x in B}. The intersection of two sets A and B is the set A∩B:={x: x in A and x in B}.

These are not well defined as they stand. You have to use axiom 6 to define them properly.

Also, you have not defined what a product of a "family" is.

Not really. You will run into problems when trying to define what (a,b) means if you keep the looser version.

Okay, define (a,b) = { {a}, {a,b} }

if you use comprehension you don't

Come on guys.

A nonempty family of nonempty sets is a nonempty set S such that for all s in S, s is nonempty. The product [math] \Pi S = T [/math] is uniquely determined by the existence of projections p_s:T->S for every s, so that given any set T' and any set of functions q_s:T'->s, there is a unique function h:T'->T so that for all s p_s(h)=q_s. Semantically, this product is the set of tuples with entries in each of the s, and projections forget the other entries. No, I'm not going to build up tuples. This is an overview of set theory.

If, in assuming that that something is an element of a set, you get a contradiction, then that something is not a set. For example, if the collection of all sets is a set, then it is an element of itself and that leads to contradiction (because of the axiom of foundation), so it can't be a set.

>The product ΠS=T is uniquely determined by the existence of projections p_s:T->S

You have not defined what projections are and what T is exactly.

Sounds like you guys have an ample grasp on foundational set-theory. Excellent.

We're going over category theory tomorrow. Pretty much all of the special definitions in set theory are subsumed by limits and colimits in the category of sets as laid out in ETCS. These nuances are not the focus of this series, and when we get to the cooler stuff a lot of it won't matter at all.

A lot of higher categorical constructions work modulo variations in strictness of the coherence involved. We'll discuss it in the lecture on coherence laws. You'll see.

T is not a bound variable. The projections are just functions, uniquely characterized by the universal property I just gave.

That's true but without zfc, or stating propositional logic, you leave the audience wondering what's going on.

Everything else is fine though

Can you show me how to make it work with comprehension?

That characterization is not enough to define these functions. Can you show how to get these "projections" from the axioms?

when can we expect this lecture? how about in 15-20 min?

You're missing the empty set axiom. Without this axiom, you cannot construct any other sets.

>Size issues are really just a nuisance in category theory.

I don't see the issue. Just define you categories relative to a universe and everything should play out well.

I'm working on a paper right now. I didn't expect this much conversation on this thread; my plan was to give the next lecture tomorrow. If not, I can probably do it when I get home around midnight. I want to do it during a higher-traffic time, though.

The empty set can be handled by the axiom of comprehension, just select the empty formula.

It's enough to define everything up to unique isomorphism, which is all we will need as we go further. I don't know how to build up to it from set-theoretic foundations alone, and I don't care because I know that Set is complete and cocomplete.

That's what I plan on doing. Grothendieck universes are my preferred reconciliation for size issues.

>No, I'm not going to build up tuples. This is an overview of set theory.

If you're going to do an overview, do it properly. Otherwise, why even bother?

> The empty set can be handled by the axiom of comprehension, just select the empty formula.

No it can't because comprehension needs an existing set in order to apply it.

>just select the empty formula.
Also, what exactly do you mean by an "empty formula"?

that version of infinity says a set exists
x such that x != x

>I don't know how to build up to it from set-theoretic foundations alone

That's fine. As long as you realize there are holes in your foundations (which you should fill up at some point).

>that version of infinity says a set exists

Your axiom of infinity already assumes the existence of the empty set, so you cannot use it to define the empty set without the definitions becoming circular.

Okay, there is an empty set. Thanks.

Because it doesn't matter up to equivalence when we start looking at various categories of sets. Even more sacrilegious, we'll move on to infinity categories and the necessary equivalences will be even looser. As I have already said, I want to touch on set theory because fibers and images will be important when talking about homotopy theory.

I meant the unique formula that no elements in the domain of discourse satisfy. I guess in logic it's just "false."

He's actually right, because the axiom of infinity references the empty set by saying that there is a set containing the empty set. Unless we can take that axiom as also guaranteeing the existence of {} implicitly? I wasn't careful when I wrote it all out. (I really don't care.)

Categorically, the semantics don't actually matter modulo unique isomorphism, but you are right of course.

>It's enough to define everything up to unique isomorphism, which is all we will need as we go further.

Unless you define what a "unique isomorphism" before-hand, this makes no sense.

> That's what I plan on doing. Grothendieck universes are my preferred reconciliation for size issues.

I think you're better off just ignoring foundations entirely. You've left too many things badly specified and undefined.

Yeah. It just felt wrong to. People are going to ask for the nitty gritties of anything I work with. Had to start somewhere.

Regarding unique isomorphisms, we'll define them tomorrow. They are isomorphisms that are unique in carrying over projection maps in this case. Universal properties will be handled rigorously in the language of categories.

>the unique formula that no elements in the domain of discourse satisfy

This is very vague. How do you know there is a unique formula and what exactly is the "domain of discourse". Do you want to start talking about models of set theory?

This is perhaps the trickiest axiom to understand. It's just there to ensure we don't have infinite containment [math] A \ni A_1 \ni A_2 ... [/math]

We don't want sets to have this kind of behavior.
Note this axiom prevents a set from containing itself, for instance.

I don't want to get into a talk about model theory. Let's stick with the axiom of the empty set and call it good.

While stating correct things, the other posters miss the point. The reason the that axiom is so important is that it says the [math]\in[/math]-relation is well-founded. This well-foundedness is vital for sets to form a cumulative hierarchy inductively built from nothing.

>Your axiom of infinity already assumes the existence of the empty set
Not really. The formal axiom starts with:
>There exists a set A such that there exists an a in A, and for every x, x is not an element of a

Which shows the empty set is an element without creating any circularity.


By the way, OP, I really like your idea of making lectures. Do you mind if I make one too for basic set theory? I know that your purpose is to rush to categories, and I think it'll be nice to teach the fundamentals for people who are new to the field.

Reminder that FOl is so pleb that people need to work with non-empty carrier. this is why plebs like to claim that ''there exists something in my set theory (aka there is a set)'' which cannot be demonstrated and they feel proud that all their constructions are not reduced to the empty set.

there are logics with empty carrier, but plebs despise them.

But what if you have an isomorphic homomorphic bijective surjection onto a mapped projection in R3?

let X = {1}
{X} - X = {{1}} - {1} = {{1}} = {X}

It's the same, what comes right after - is a set of elements you substract. So by doing {X} - X, you're not subtracting X from {X}, but the elements of X from {X}.

What if X = {X}?

When you have a set, you can create de group from this set with the group of permutations of its elements, so if there was a "set of all groups", you could make a group based on that set, which would have to contain said group, ie itself, impossible with foundation.

You can't necessarily say what is and what is not a set though, when using ZFC you actually have to use a meta theory from which you "borrow" some sets that go with ZFC, a model. So ZFC does not necessarily recognize as sets some things that could be considered as such.

Please, do lecture! I'd love that and you'd surely do better than I.

Contradicts foundation

Whoops, missed the earlier discussion abt the axiom of foundation. My bad.

You're doing the lord's work, OP.

Given these axioms, how does one take a union of infinitely many sets?

Suppose you want to take a union of all sets in S (a family of sets). Then x is in US if and only if there is at least one element of S which contains x.

You can't.

s/formal/correct

So, are you using comprehension or something? Given those axioms, how does one construct the set that you describe?

Actually, the well-foundedness if vital if extensionality is to work properly.

You are correct I think. Anybody know if ZFC requires the axiom of union to be for infinite unions? I thought finite would work because we have the axiom of pairing, but that simply shuffles the issue around.

>You can't necessarily say what is and what is not a set though, when using ZFC you actually have to use a meta theory from which you "borrow" some sets that go with ZFC, a model.

The part about "borrowing" is wrong and doesn't make much sense.

Yes, you require the union axiom to allow "infinite unions".

Yes, it seems the OP is considering functions to be triples (f, A, B). So I guess OP uses f:A->B as a shorthand for (f, A, B).

>>If I give you a set of ordered pairs f such that if (a,b) and (a,b') are both in f, then b=b', is f still a function?
no, your writing lacks ''for all a and b, b' ''

Great! I'll see when I have the time, I'll probably have something ready in a few weeks.

In set theory as fundations, sets are not defined, just as in category theory as fundations, categories are not defined.

in category theory as fundations, you can define a set, just like you can define a category in set theory as fundations

Why do I need to care about axiomatization of set theory? I'm a math grad student and I always used naive set theory, treating sets as a black box on which all operations simply work, and I never ran into any problems.

fundations ask the question ''why do math?''. of course a pleb students like you is not meant to reflect on his work...

>the product of a family of nonempty sets

What is a "family"? A set?

>Ordinals form a proper class

What is a "class"?

>Show that the fiber of the projection onto A (as above) is always canonically isomorphic to B, and visa versa.

What is (canonically) isomorphic?

>saving thumbnails

Im curious, isn't the second axiom kind of weird?
Why pair? I would propose a simpler and equivalent axiom
2a. For any set A, there is a set B such that A is the only element of B.
With the axiom of summation this implies 2. and is simpler. Or am i missing something?

>I'm a math grad student
How is this possible without at least learning about AC?