Do we have to prove the existence of every single set we use in mathematics?

Do we have to prove the existence of every single set we use in mathematics?

Other urls found in this thread:

en.wikipedia.org/wiki/Definable_real_number#Definability_in_models_of_ZFC
en.wikipedia.org/wiki/Empty_set#Axiomatic_set_theory
twitter.com/SFWRedditVideos

The correct, rigorous, formal answer is yes.

However, the real world practice is no. Very few people are actually aware of undefinable sets and those that are actively choose to ignore them because they open a Pandoras box of questions and theorems that make the mathematical community uncomfortable.

anything between { and } is a set, what is this ambiguity you're talking about?

There are so many things wrong with this I don't even know where to begin.

yes please
mathematics isn't black magic where you call a superior power every time you say "let S be..."

But within your field, you prove that a set is not empty before you go assuming things about it right?

depends on your philosophy.
if you can show that a set is not empty (somehow), but you are incapable of showing at least one element of that set, I could be less inclined to accept your subsequent arguments. Most modern mathematicians don't care for that though, they just want the properties of the set.

It can be empty, but you should realize that before going further studying the properties of the empty sets :)

Not really. It isn't uncommon to start by assuming a thing exists, proceeding to prove things about it, and then leaving the job of finding an example of the thing up to someone else.

the empty set sure has a lot of properties.

I like the distinction the intuitionists use for handling that. They say "non-empty" when one uses a positive proof by contradiction (because in the raw language of set theory, non-empty is a double negation) and "inhabited" for when you've actually shown that a set contains something.

It's clopen!

I like the zero ring better.

No, because given any axiom system (i.e. foundation for a proof system) there will always be sets whose existence we cannot prove.

For example, set-theorists often talk about the set [math]0^\sharp[/math], but provably cannot prove from ZFC that it exists.

I like this and if I ever use set language I'm gonna inhabited.
> I like the zero ring better.
Why?

>But within your field, you prove that a set is not empty before you go assuming things about it right?

As a combinatorialist, not often. That's because I work with sets of combinatorial objects which are almost always nonempty and it is almost always trivial to give an example of such an object.

If in a proof you want to say something about a non-trivial set then you should at least show that it is non-empty, usually by constructing one of its most trivial elements.

I've never heard of undefinable sets. What do you mean by them, just stuff which can't be constructed using zfc?

shoo norbert shoo

Yes.

Except for the set of natural numbers which you get by the axiom of infinity of ZF, and the empty set which you also get by an axiom of ZF.

If by "existence" you mean "nonemptiness" or even "nontriviality" then technically no because sometimes people go for contradictions by defining a set and showing it's empty or degenerate (in whatever context that makes sense).

Not just ZFC, the problem affects any axiomatic system really. Essentially logic has a bunch of restrictions built in in order for it to play nice and for a number of logic theorems to work. These restrictions make it so that the set of all logical statements you can write is countably infinite.

As such, if you're arguing that there exist an uncountably infinite number of sets (such as the real numbers if you regard them as sets) then you clearly end up with a problem where only a countable subset can be defined. What ends up happening as a result is that you can never actually talk about an undefinable number, they are in a sense "unthinkable". You can only talk about these numbers as members of a larger set.

en.wikipedia.org/wiki/Definable_real_number#Definability_in_models_of_ZFC

This line of thinking puts the existence of many sets such as the reals on tenuous footing. I also believe it may be possible to stretch the argument to disprove the axiom of choice on some uncountable sets.

If u a bitch

It's much better practice to include known objects that obey whatever property you're looking at, or at least construct non-retarded examples.

There's no axiom in ZFC that guarantees the existence of the empty set.

You can prove its existence by starting with some defined set (say the set of naturals, which as you said is assumed to exist), then using the axiom of comprehension on a contradiction as the predicate (say, x≠x).

it might be very difficult to construct such objects that have the property you're working with

for example, a theorem of goldfeld states that if there exists any elliptic curve of rank at least 4, then one gets a useful lower bound for class numbers

this was proven before any curve was known to have such a property, but it certainly led people to look harder for such curves knowing the payoff

Some texts give an axiom of the empty set. Though it's arguably redundant.
en.wikipedia.org/wiki/Empty_set#Axiomatic_set_theory

I'm not the guy you were responding to but personally I'm not entirely convinced that this axiom is sufficient but the reasons are too pedantic to get into.

Me again. Realized that was poorly worded. By "this axiom" I meant that the axiom of restricted comprehension + another set.

Existence is a matter of ontology and not a mathematical question.

>not a mathematical question.

>2016
>Not knowing than most top tier and important mathematical proofs have to do with existence and uniqueness

We could fucking rename mathematics as a whole as 'Existence theory' and it would technically be correct.

wew lad

How's junior year of undergrad treating you?

>ontology
Wrong type of existence, buddy.

I'm not far in formal math so I don't have a terribly exciting reason. I think it's amusing to have the additive and multiplicative identities equal, not that it means much.

Yes.

It's not redundant. That's pretty much the same as the axiom "there exists a set". Some people take this as an axiom of first order logic, but I disagree with the convention. Empty models have a right to exists, damn it.

What the wiki article and that user are talking about is that if you already have another axiom that gives you any other set (such as the naturals) then you can combine that with the set comprehension axiom in order to produce the empty set as an empty subset of the set you already have.

Personally I find such claims suspect but they're entirely acceptable within the context of normal people math.