Have you read any book about Set Theory or Mathematical Logic? And saying that it's "simply wrong" is wrong :D

I don't know (have not touched/worked with) the Peano axioms. But from what I have found here https://mathworld.wolfram.com/PeanosAxioms.html is that the theory they imply (because they make a statement about every subset of the domain) is from second order and I have not worked yet in the domain of second order logic. But just give me a month and I will give you a well defined set of axioms in a second order language.

For now all I can do is give you a formal definition of Peano arithmetic from what I have understood from the page I linked.

Let N be a set. Let z be an element of N. Let s be a function from N to N. We say that <N, z, s> is Peano arithmetic if:

(∀ x ∊ N)[s(x) ≠ z]

(∀ x ∊ N)(∀ y ∊ N)[s(x) = s(y) ⇒ x = y]

(∀ M ∊ P(N))[z ∊ M & (∀ x ∊ M)[s(x) ∊ M] ⇒ M = N]

Now I'll give you the definitions which I give to my students for a Field and Linear space over a Field. Since they are abstract and I'll let you on your own come with a logically correct definition for a Hilbert space. I'll just say that it's not hard at all :)

Let F be a set. Let z be an element of F. Let e be an element of F. Let sum be a function from FxF to F. Let neg be a function from F to F. Let prod be a function from FxF to F. Let inv be a function from F\{z} to F. We say that <F, z, e, sum, neg, mul, inv> is a Field if:

(∀ a ∊ F)(∀ b ∊ F)(∀ c ∊ F)[sum(a, sum(b, c)) = sum(sum(a, b), c)];

(∀ a ∊ F)[sum(z, a) = a & sum(a, z) = a];

(∀ a ∊ F)[sum(a, neg(a))= z & sum(neg(a), a) = z];

(∀ a ∊ F)(∀ b ∊ F)[sum(a, b) = sum(b, a)];

(∀ a ∊ F)(∀ b ∊ F)(∀ c ∊ F)[prod(a, prod(b, c)) = prod(prod(a, b), c)];

(∀ a ∊ F)(∀ b ∊ F)[prod(a, b) = prod(b, a)];

(∀ a ∊ F)[prod(e, a) = a & prod(a, e) = a];

(∀ a ∊ F)(∀ b ∊ F)(∀ c ∊ F)[prod(sum(a, b), c) = sum(prod(a, c), prod(b, c))];

(∀ a ∊ F)[a ≠ z ⇒ prod(a, inv(a)) = e & prod(inv(a), a) = e].

Let K = <F, z, e, sum, neg, mul, inv> and K be a Field. Let V be a set. Let n be an element of V. Let vsum be a function from VxV to V. Let vneg be a function from V to V. Let sprod be a function from FxV to V. We say that <V, n, vsum, vneg, sprod> is a Linear space over the Field K if:

(∀ a ∊ V)(∀ b ∊ V)(∀ c ∊ V)[vsum(a, vsum(b, c)) = vsum(vsum(a, b), c)];

(∀ a ∊ V)[vsum(n, a) = a & vsum(a, n) = a];

(∀ a ∊ V)[vsum(a, vneg(a)) = n & vsum(vneg(a), a) = n];

(∀ a ∊ V)(∀ b ∊ V)[vsum(a, b) = vsum(b, a)];

(∀ a ∊ V)(sprod(e, a) = a];

(∀ a ∊ F)(∀ b ∊ F)(∀ v ∊ V)[sprod(a, sprod(b, v)) = sprod(prod(a, b), v)];

(∀ a ∊ F)(∀ u ∊ V)(∀ v ∊ V)[sprod(a, vsum(u, v)) = vsum(sprod(a, u), sprod(a, v))];

(∀ a ∊ F)(∀ b ∊ F)(∀ v ∊ V)[sprod(sum(a, b), v) = vsum(sprod(a, v), sprod(b, v))].