\(\mathbb{F}^{n}\) not being a field kinda sucks, so we made an object called a “vector space” which essentially does everything a field does except without necessitating a multiplicative inverse
We defined something called \(\mathbb{F}^{S}\), which is the set of functions from a set \(S\) to \(\mathbb{F}\). Turns out, \(\mathbb{F}^{S}\) is a Vector Space Over \(\mathbb{F}\) and we can secretly treat \(\mathbb{F}^{n}\) and \(\mathbb{F}^{\infty}\) as special cases of \(\mathbb{F}^{s}\).
All vector spaces \(\mathbb{F}^{n}\) and \(\mathbb{F}^{\infty}\) are just special cases \(\mathbb{F}^{S}\): you can think about those as a mapping from coordinates \((1,2,3, \dots )\) to their actual values in the “vector”
The way Axler presented the idea of “over” is a tad weird; is it really only scalar multiplication which hinders vector spaces without \(\mathbb{F}\)? In other words, do the sets that form vector spaces, apart from the \(\lambda\) used for scalar multiplication, need anything to do with the \(\mathbb{F}\) they are “over”? The name of the field and what its over do not have to be the same—“vector space \(\mathbb{C}^2\) over \(\{0,1\}\)” is a perfectly valid statement
If lists have finite length \(n\), then what are the elements of \(\mathbb{F}^{\infty}\) called? “we could think about \(\mathbb{F}^{\infty}\), but we aren’t gonna.”
Why is \(1v=v\) an axiom, whereas we say that some \(0\) exists? because we know 1 already, and you can follow the behavor of scalar multiplication
what’s that thing called again in proofs where you just steal the property of a constituent element?: inherits
