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

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)?

• Does this answer your question? Intuition behind $F^n$ and $F^\infty$ being examples of function spaces, $F^S$. Many other duplicates can be found using Approach$0$. May 17 at 6:45
• @AnneBauval I checked it out but didn't really understand it, that's why I created a new question. Thanks for the note though. May 17 at 7:01
• Your question is a FAQ and this link was far from the only duplicate! If that one did not suit you, just choose your prefered one among the numerous good hits in the approach0 query. May 17 at 7:07
• @AnneBauval I didn't know such a website (Approach0) even existed, I'm still new to Math StackExchange. May 17 at 7:08
• I guessed you didn't, whence my link. Did you find in it a duplicate which suits you better? May 17 at 7:44

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.