Is the category of all axioms a small category or a large category?
I don't know what a category is but aren't there only like ten axioms in ZFC? Sounds small to me.
>>9356081>he thinks that ZFC is the smallest theory>not Peano Arithmeticlmao that's like 2 axioms, 0 being a natural number and that 0 has a successor.
Sorta related:What do working theoreticians think of HoTT?I can't take it seriously because HoTT-Coq sounds like gay porn.
>>9356081zfc has literally infinite axioms dude
>>9356345You may want to rethink that statement
>>9356083>this retarded system>PeanoHow do you get arithmetic from these axioms? How do you know it exists? In your system: 1=4, 1=/=4, 1=/=1, 1=0, etc.
>>9356365>1=4Well no, because 4 isn't the first (1) successor of 0. If σ is the successor function, then you get the number four (4) at the fourth recursion: σ(σ(σ(σ(0)))) = 4.
Now I'm interested, how would that category work?I know there's a category of all mathematical statements, where a morphism from P to Q amounts to a proof of Q from P (under some ambient axiomatic system).You could consider the category of all axioms (say, in ZFC) as a full subcategory, but it doesn't sound particulary interesting.
>>9356373You didn't prove sigma is surjective
>>9356399Axiom 1:(i) ℕ is a set(ii) 0 ∈ ℕ(iii) σ : ℕ --> ℕAxiom 2: ~(∃n ∈ ℕ | σ(n) = 0)(I.e. zero is not a successor)Axiom 3: ∀x,y ∈ ℕ : σ(x) = σ(y) ==> x=y(Sigma injective)Axiom 4: Suppose S ⊆ ℕ such that(i) 0 ∈ S, and(ii) n ∈ S ==> σ(n) ∈ S, for an arbitrary n ∈ ℕ.Then:S ⊆ ℕ ⋀ 0 ∈ S ⋀ [∀n(n ∈ S ==> σ(n) ∈ S)] ==> S = ℕAxiom 4 makes σ surjective over its codomain ℕ by induction
>>9356432>N is a setWrong
>>9356432>ℕ is a setNot really. First, because Peano is independent of set theory and second because if "N is a set" was an axiom then there is the implicit axiom of "sets exist" which means that implicitly you are assuming all the axioms of set theory, and that defeats the purpose of the exercise.Retarded undergrad students may internet Peano as the set of axioms that say "N is a set" but in reality the whole of axioms of Peano tell us that N may be treated like a set. For example, the first axiom tells us that [math] 1 \in \mathbb{N} [/math] is a valid statement.This does not mean that N is a set in the sense of formal set theory, all it says is that within this theory you can invoke the sentence [math] 1 \in \mathbb{N} [/math] when writing a proof. You may even interpret this axiom informally as "1 exists".
>>9356435Norman pls go
>>9356373How do you fuck up this much with only two axioms?
>>9355961This is even worse than the phenotype meme
"all axioms" will depend on the language you want to talk about.Given a set theory U, you can take any set X and consider the group G = (X, id_X) with one element (the identity). Of course, those groups are all isomorphic. Nevertheless, it means you have a group for each set, and thus the class of all such object doesn't form a set. I suppose in a similar way, you could attempt to argue that for each set X, you can take the predicate P_X the uniquely characterizes X, and take e.g. "there exists and A such that P(A)" as an axiom. Then all such axioms alone would form a class and the corresponding category wouldn't by U-small.If I needed to make sense of the question, really, I'd take it you speak of the syntactic category for a theoryhttps://ncatlab.org/nlab/show/syntactic+category
>>9355961The set of all strings of finite length is countable. The set of all axioms is a subset of the set of all strings of finite length.>inb4 >set
>>9356611>The set of all strings of finite length is countable.define "string"
>>9356611What if the strings use characters from an infinite alphabet?
>>9355961If they are finite sentences in a finite or countably infinite alphabet it is small
>>9356622Functions from {1..n} to an alphabet of characters.>>9356624All axioms that can be stated by humans are stated in finite languages.
>>9356787*languages with a finite alphabet
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>>9356803Sorry, I did not mean to post this here.
