What is a set?

Thanks, I'll look into it!

so if I print one of those symbols and shove it up my ass it becomes a set? how absurd

Welcome to set theory!

>the "yep, it's complicated!" zero-value defeatist-answer

A set is an undefined notion if you're using Set Theory as your foundation. If you're using something else, then you might be able to define a set, but then you'll have other undefined notions.

Any formal system has to have undefined notions. It can't be turtles all the way down.

Extensionality
[math] \forall X\,\forall Y\,[\,X=Y\quad\Leftrightarrow\quad \forall z(z\in X\ \Leftrightarrow\ z\in Y)\,] [/math]

Pairing
[math] \forall x\,\forall y\,\exists Z\,\forall z\,[\,z\in Z\quad\Leftrightarrow\quad z=x ~or~ z=y\,] [/math]

Union
[math] \forall X\,\exists Y\,\forall y\,[\,y\in Y\quad\Leftrightarrow\quad\exists Z(Z\in X {~ and ~} y\in Z)\,] [/math]

Empty set
[math] \exists X\,\forall y\,[\,y\notin X\,] [/math] this set X is denoted by [math] \emptyset [/math]

Infinity
[math] \exists X\,[\,\emptyset\in X { ~and~ } \forall x(x\in X\Rightarrow x\cup\{ x\}\in X)\,] [/math]

Power set
[math] \forall X\,\exists Y\,\forall Z\,[\,Z\in Y\quad\Leftrightarrow\quad
\forall z(z\in Z\ \Rightarrow\ z\in X)\,] [/math]

Replacement
[math] \forall x\in X\,\exists!y\,P(x,y)\quad \Rightarrow \quad [\,\exists Y\,\forall y\,(y\in Y\ \Leftrightarrow\ \exists x\in X\,(P(x,y)))\,] [/math]

Regularity
[math] \forall X\,[\,X\ne\emptyset\quad\Rightarrow\quad\exists Y\in X\,(X\cap Y=\emptyset)\,] [/math]

Constructibility
[math] V=L [/math]

>not including the weaker forms of union and powerset that follow from subset, which you omit presumably because it follows from replacement
>listing the empty set axiom even though it follows from the underlying logic
>listing constructibility and not choice, (the first implies the second but also implies a lot of stuff you might not want, such as GCH or there being no uncountable measurable cardinals)

wew lad

This.

Except you don't have to worry about the domain of discourse in your logic being a notion of collection. Just take a formalist approach and consider the bunch of strings on your piece of paper and rewrite rules. The fact that the word "domain of discouse" lingers over those letters we call terms of the logic and that once you formalized the logic they make up a kind (of set) comes after the fact and doesn't hinder from defining set membership as in

>How do you define a thing without using set or collection?
It is a group of things.
A list filled with items
A series of items layed out in a ordered way

>It is a group of things.
define group

>A list filled with items
this is bad because lists are countable (real numbers aren't)

>A series of items layed out in a ordered way
the complex numbers aren't ordered

>Why wouldn't you want the clean GCH
>Why would you want absurdly huge sets like 0# of no use or interest

Because you want to keep it open and not deny them outright. Contemporary set theory research is in large cardinals, consistency strength, and that sort of thing. Also (I don't know too much about it) supposedly some interesting things happen when CH fails. Some mathematicians even claim that it ought to be false, and that the value of the continuum ought to be aleph_2.

so what happens when you include godel's incompleteness theorem? does it become impossible to make a truly rigorous form of set theory in this sense because sets are so encompassing tha the semantic framework can lead to unfalsifiable shit like "flowers are pretty" or is this an issue with 1st order logic in the general matter outside of set theory?

Being too general handicaps what you can say about a system. All mathematicians can live happily within V=L with most questions that you would care about having definite answers.

People who don't hold V=L should be treated like mathematicians who deny choice or infinite sets.

You still have to deal with a notion of a collection. Right at the beginning when you start defining your formal language (the situation only gets worse when you start defining a proof system and logic over said language) you have to give a formal grammar (collection of rules) and an alphabet (collection of characters). The difference here is that we don't have any set operations, theorems, or arguably even notation of set theory: and by collection we're concretely referring to a "bunch of things" (e.g. strings).

I do prefer the formalist approach but at the very bottom you are forced to introduce some amount of informalism.

It's a theorem, not an axiom. In general when we're working over a logic defined over a language given over a finite alphabet then the number of statements you can make is countable at most. So the number of theorems, proofs, definitions, etc.... is countable.

Now consider that you can define uncountable sets like the reals in set theory. This means that there will exist an uncountable number of reals that you can't define or make statements about (much less prove them). At best you can talk about these reals as elements in larger sets. This doesn't mean that these numbers don't exist exactly. Existence is given by the models of the theory and there exists a popular model where the real numbers exist (we just can't do very much with them, but so far that hasn'tnl been a problem for mathematics).

so they exist but not in a manner we can describe? sounds kinda philosophical.

Hence Wildberger deciding that it's not rigorous and we should try to construct rational versions of mathematics.

That said, it's worth noting that the set of algebraic numbers is a proper subset of the definable numbers (it's proper because for instance there are definable transcendental numbers that are not algebraic) but no one who deals with the algebraic numbers cares about this distinction so theres often a lot of bad cognitive noise in discussions on this topic.

Set (noun): a group of well defined and distinct objects

define 'well defined'

>distinct objects

sets aren't multisets

>a group
BZZZZZZZZZZZZZZT

Define 'define'

Hijacking this thread to conjecture the existence of a set V containing all sets.

what is the "smallest" foundation, in terms of axioms?

is there any lower bound on the amount of axioms to have a consistent theory?

(does this somehow violate or allude to the incompleteness theorems?)

this is solved in Homotopy type theory.

calling types sets to to fit your idea, you have a series of universal sets U1, U2, U3... with Ui a subset of U(this is solved in Homotopy type theory.

calling types sets to to fit your idea, you have a series of universal sets [math]U_1 \subset U_2 \subset U_3 \subset \cdots [/math]. when you define a set it lives in a universal set large enough to contain it. for example a set containing all sets in U1 lives in [math]U_2[/math], since it cant live in [math]U_1[/math], a set containing all sets that dont contain themselves in [math]U_2[/math] would live in [math]U_3[/math], and so on.

You then work with the union of all universal sets [math]U = \bigcup_i U_i[/math] when working with 'all' sets.i+1). when you define a set it lives in a universal set large enough to contain it. for example a set containing all sets in U1 lives in U2, since it cant live in U1, a set containing all sets that dont contain themselves in U2 would live in U3, and so on. You then work with the union of all universal sets Ui when working with 'all' sets.

But any set can be given a group structure :^)

By Cantor's diagonal argument [math]|P(V)| > |V|[/math] for any such set you might try to construct, contradicting the notion that V is a set of all sets. You may be interested in: NBG set theory (with proper classes), MK set theory (same), TG set theory (with inaccessible cardinals).

The trivial theory containing no axioms is clearly consistent. Most axiom schemas actually in use have countably infinitely many axioms, though nothing is stopping you from using a finite number of axioms. For example, the theory of groups has 3 axioms (associativity, identity, inverses).

>But any set can be given a group structure :^)
wrong, not the empty set

The group of order 0. It's like the field of order 1, but smaller :^)

>But any set can be given a group structure :^)
obviously true for non-empty finite sets but for weird large cardinals? whats the structure?

just found this
en.wikipedia.org/wiki/Group_structure_and_the_axiom_of_choice#A_ZF_set_with_no_group_structure
the statement "every set admits a group structure" requires the axiom of choice, strange

Yeah, basically you use choice to well-order the set and then you can just make an ordinal-like group structure from that. Without choice sets can act really weirdly and you're not guaranteed to be able to define any such operations on the elements of the set.

Lets say I met someone who said the following:

>I dont have any real math education, >but I am einstein/tesla league in rational/logic thinking.

What would you advise someone like that to do?

find a trusty sidekick named Watson and become a detective

engage in argunents on Veeky Forums with no knowledge on the topic

satisfying af desu

but doesn't the theory of groups rely on the theory of sets?

And when that someone also found a way to apply all those formula's in the same system? But without the use of calculus or math?

The theory of groups is most conveniently defined over a language with a binary function + and a distinguished element (or nullary function) 0, and consists of the axioms:
[math]\forall a,b,c (a+(b+c) = (a+b)+c)[/math]
[math]\forall a (a+0 = 0+a = a)[/math]
[math]\forall a \exists b (a+b = b+a = 0)[/math]
Of course the metalanguage is probably some sort of set theory, but that's no different than if you're working with a theory of sets. Model theory always requires some foundation to work with, whether you're dealing with theories of sets, theories of groups, or anything else.

>the complex numbers aren't ordered
The complex numbers cannot be an ordered field. That is not to say they cannot be ordered.

>define group
Anything modeled by a language which uses the [math]\in[/math] binary relation.

>this is bad because lists are countable (real numbers aren't)
Only if you insist the list have order type [math]\omega[/math]. Assuming choice any infinite set can be written out in a (possibly uncountably long) list corresponding to some well-ordering.

>the complex numbers aren't ordered
See above. Assuming choice, there is in fact a well-ordering on the complex numbers. This is different than being an "ordered field", which the complex numbers are not.

Does the set of all sets contain itself?