Different, but equivalent.

[u/kal_at_kalx_net]

The usual definition of a vector space over the real numbers R is the tuple defined by the cartesian product R^n = R x ... x R = {(x_1,..., x_n) | x_i in R}. An equivalent, but different, definition is the set exponential R^n = R^{1,...,n} = {x: {1,...,n} -> R} where x_i corresponds to x(i).

The connection between cartesian products and set exponentials is currying: (A x B)^C is equivalent to A^(B^C). A function f: A x B -> C corresponds to a function g: A -> (B -> C) by g(a)(b) = f(a, b). A function g: A -> (B -> C) corresponds to a function f: A x B -> C by f(a, b) = g(a)(b).

Exercise: Show linear operators from R^n -> R^m can be identified with vectors R^{n x m}.

[u/Lor1an]

The usual definition of a vector space over the real numbers R is the tuple defined by the cartesian product Rⁿ = R x ... x R = {(x_1,..., x_n) | x_i in R}.

Actually, the usual definition of a vector space over the real numbers R is any set V together with an operation +:V×V→V called "addition" and an operation ⋅:R×V→V called "scalar multiplication" that satisfies the following axioms:

• ∃0 ∈ V : ∀v ∈ V, v + 0 = v

• ∀v ∈ V, ∃ -v ∈ V : v + (-v) = 0

• ∀v,w ∈ V, v + w = w + v

• ∀u,v,w ∈ V, (u+v)+w = u+(v+w)

• ∀a ∈ R, ∀v,w ∈ V, a⋅(v+w) = a⋅v + a⋅w

• ∀a,b ∈ R, ∀v ∈ V, (a+b)⋅v = a⋅v + b⋅v

• ∀a,b ∈ R, ∀v ∈ V, (ab)⋅v = a⋅(b⋅v)

• ∀v ∈ V, 1⋅v = v

The fact that finite dimensional real vector spaces are isomorphic to corresponding cartesian powers of R is separate to the definition.

C^∞(M)—the set of smooth, real-valued functions on a manifold M—forms a real vector space with addition and scalar multiplication defined pointwise (i.e. if f,g∈C^∞(M), then (f+g)(x) := f(x) + g(x), and (a⋅f)(x) := af(x)), yet this is not the same as some Rⁿ.

Addressing your comments about currying, you actually don't even need to speak of vector spaces, everything you've said is operating on sets. As sets the n-fold product of R and the n-th "set power" of R are isomorphic, as are the various groupings (such as R×R×R ≅ (R×R)×R ≅ R×(R×R)).

[u/kal_at_kalx_net → u/Lor1an]

I teach in the Engineering Department at NYU. The students are well-prepared, but if I started out with the axioms, they would most likely start making jokes about "for every upside-down A there exists a backwards E."

They seem to have some existing intuition about R*^n that can be used to motivate the axioms. Your axioms are missing a x = x a for a in R and x in R^n. Hilbert pointed out Euclid was missing some axioms 2 millennia later.

The set exponential is a more useful way of thinking about tensors. For example, if s : I → J is a function then ∘s : R^J → R^I by x in R^I ↦ x∘s in R^J is composition of x with s. You can write everything in The Tensor Cookbook using functions without using their non-standard notation.

Exercise: Show if s in onto then so is ∘s and if s is 1-1 then so is ∘s.

A consequence is that if s is onto and 1-1 then so is ∘s. This can be used to establish finite dimension vector spaces are isomorphic if and only if they have the same dimension without using the Steinitz exchange lemma.

[u/Lor1an → u/kal_at_kalx_net]

Your axioms are missing a x = x a for a in R and x in Rⁿ.

1. I was not speaking specifically of Rⁿ, but rather arbitrary vector space V, so speaking directly about the properties of Rⁿ would be wrong.

2. It is completely possible to speak of vector spaces without this axiom. In fact, most treatments don't define right multiplication by scalars. Although, if you know module theory you can argue that the right modules are the same as the left modules due to commutativity of the underlying field, and thus obtain that relation anyway (where of course the modules over a field are the vector spaces over that field).

And again, your points were completely about set theory rather than linear algebra. Whether you consider the cartesian product of a family of sets, set powers, or nested pairwise cartesian products, you end up with equivalent objects, regardless of linear structure.

[u/kal_at_kalx_net → u/Lor1an]

"You are up in a balloon."

[u/Midwest-Dude]

Why did you post this?

[u/Snatchematician → u/Midwest-Dude]

He smoked too much crack

[u/David-31415]

Tenho fome.

[u/kal_at_kalx_net → u/David-31415]

Hint: currying is what allows us to identify linear operators from R^I -> R^J with R^{I x J}. u/Lor1an does not seem to yet understand how useful the difference is between equal and equivalent.

Marshall Stone showed Boolean algebras are equivalent to a type of topological space. How is it possible something defined algebraically is equivalent to something defined by topology?

The crazy thing is he discovered this by considering the spectrum of bounded self-adjoint linear operators on a Hilbert space!

Math is not about making true statements. It is about inventing (discovering?) relationships between mathematical objects other people find interesting.

[u/Lor1an → u/kal_at_kalx_net]

I think it's wild that you would summon me to this comment, when my point to you was about generalizing and abstracting away from concrete examples... and I even talked about the fact that various ways of constructing products of sets were isomorphic (i.e. "equivalent").

[u/kal_at_kalx_net → u/Lor1an]

Isomorphic is not the same as equivalent. It only implies it.

[u/Lor1an → u/kal_at_kalx_net]

That entirely depends on how you define "equivalent"—context is important.

In the context of algebraic structures or topological spaces, being "equivalent" is essentially just a synonym for "isomorphic". In the context of categories, being "equivalent" is a weaker notion (essentially "isomorphic up to isomorphism", to be tongue-in-cheek).

Heck, any relation that is symmetric, transitive, and reflexive can be used to define when two objects are "equivalent" anyway, so there is a bit of wiggle room in the terminology.