How are $F^n$ and $F^\infty$ special cases of $F^S$? [duplicate]

Question

I'm teaching myself Linear Algebra using Axler's Linear Algebra Done Right book and I'm confused about some definitions mentioned in the book.

The book defines $F$ as $\mathbb{R}$ or $\mathbb{C}$ where $F$ stands for a field.

$F^n$ is the set of all lists of length $n$ of elements of $F$: $$F^n = \{(x_1, x_2, ..., x_n): x_j \in F \text{ for } j = 1, 2, ..., n\}.$$

$F^\infty$ as the set of all sequences of elements of $F$: $$F^\infty= \{(x_1, x_2, ...): x_j \in F \text{ for } j \in 1, 2, ...\}.$$

The book then introduces $F^S$ as below:

• If $S$ is a set, then $F^S$ denotes the set of functions from $S$ to $F$.
• For $f, g \in F^S$, the sum $f + g \in F^S$ is the function defined by $(f + g)(x) = f(x) + g(x)$ for all $x \in S$.
• For$\lambda \in F$ and $f \in F^S$, the product$\lambda f \in F^S$ is the function defined by $(\lambda f)(x) = \lambda f(x)$ for all $x \in S$.

The book says that $F^n$ and $F^\infty$ are special cases of the vector space $F^S$ but I don't really get why that's the case. How can we express lists as functions from $S \to F$? Won't they then be functions from $S$ to $F^n$ (where $n$ is the length of the list)?

Answer 1

An element $(x_1,\cdots,x_n) \in F^n$ may be thought of as a function $$ f : \{1,2,\cdots,n\} \to F \text{ such that } f(i) = x_i $$ This is analogous to how we define functions in $F^S$ for more general sets $S$ (indeed, the above characterization is essentially the identification $F^n := F^{\{1,2,\cdots,n\}}$), and all of the properties you expect under one representation you will get out of the other. (You may more appropriately think of this as an isomorphism than an equality, up to your definition of $F^n$.)

A sequence $(x_i)_{i=1}^\infty \in F^\infty$ may likewise be thought of as a function of domain $\mathbb{N}$, handled analogously.