$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$

# 4.4: Vectors, Lists and Functions: $$\mathbb{R}^{S}$$

$$\newcommand{\vecs}{\overset { \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

Suppose you are going shopping. You might jot down something like this on a piece of paper: We could represent this information mathematically as a set, $$S=\{\rm apple, orange, onion, milk, carrot\}\, .$$

There is no information of ordering here and no information about how many carrots you will buy. This set by itself is not a vector; how would we add such sets to one another?

If you were a more careful shopper your list might look like this: What you have really done here is assign a number to each element of the set $$S$$. In other words, the second list is a function
$$f:S\longrightarrow {\mathbb R}\, .$$
Given two lists like the second one above, we could easily add them -- if you plan to buy 5 apples and I am buying 3 apples, together we will buy 8 apples! In fact, the second list is really a 5-vector in disguise.

In general it is helpful to think of an $$n$$-vector as a function whose domain is the set $$\{1,\dots,n\}$$. This is equivalent to thinking of an $$n$$-vector as an ordered list of $$n$$ numbers. These two ideas give us two equivalent notions for the set of all $$n$$-vectors:
$${\mathbb{R}}^{n} :=\left\{ \begin{pmatrix}a^{1} \\ \vdots \\ a^{n}\end{pmatrix} \middle\vert \, a^{1},\dots a^{n} \in \mathbb{R} \right\} =\{ a:\{1,\dots,n\}\to \mathbb{R}\} := \mathbb{R}^{ \{1,\cdots,n\} }$$
The notation $$\mathbb{R}^{ \{1,\cdots,n\} }$$ is used to denote functions from $$\{1,\dots,n\}$$ to $$\mathbb{R}$$. Similarly, for any set $$S$$ the notation $$\mathbb{R}^{S}$$ denotes the set of functions from $$S$$ to $$\mathbb{R}$$:
$$\mathbb{R}^{S}:=\{ f:S\to \mathbb {R}\}\, .$$
When $$S$$ is an ordered set like $$\{1,\dots,n\}$$, it is natural to write the components in order. When the elements of $$S$$ do not have a natural ordering, doing so might cause confusion.

Example 48

Consider the set $$S=\{*, \star, \# \}$$ from chapter 1 review problem 9. A particular element of $$\mathbb{R}^{S}$$ is the function $$a$$ explicitly defined by
$$a^{\star}=3, a^{\#}=5, a^{*}=-2.$$
It is not natural to write
$$a=\begin{pmatrix}3 \\ 5 \\ -2\end{pmatrix} ~{\rm or} ~a=\begin{pmatrix}-2\\ 3 \\ 5\end{pmatrix}$$
because the elements of $$S$$ do not have an ordering, since as sets $$\{*, \star, \# \}=\{*,\star,\#\}$$.

In this important way, $$\mathbb{R}^{S}$$ seems different from $$\mathbb{R}^{3}$$. What is more evident are the similarities; since we can add two functions, we can add two elements of $$\mathbb{R}^{S}$$:

Example 49 Addition in $$\mathbb{R}^{S}$$

If $$a^{\star}=3, a^{\#}=5, a^{*}=-2$$ and $$b^{\star}=-2, b^{\#}=4, b^{*}=13$$
then $$a+b$$ is the function
$$(a+b)^{\star}=3-2=1, (a+b)^{\#}=5+4=9, (a+b)^{*}=-2+13=11\, .$$

Also, since we can multiply functions by numbers, there is a notion of scalar multiplication on $$\mathbb{R}^{S}$$:

Example 50 Scalar Multiplication in $$\mathbb{R}^{S}$$

If $$a^{\star}=3, a^{\#}=5, a^{*}=-2$$,
then $$3a$$ is the function
$$(3a)^{\star}=3\cdot3=9, (3a)^{\#}=3\cdot5=15, (3a)^{*}=3(-2)=-6\, .$$

We visualize $$\mathbb{R}^{2}$$ and $$\mathbb{R}^{3}$$ in terms of axes. We have a more abstract picture of $$\mathbb{R}^{4}$$, $$\mathbb{R}^{5}$$ and $$\mathbb{R}^{n}$$ for larger $$n$$ while $$\mathbb{R}^{S}$$ seems even more abstract. However, when thought of as a simple "shopping list'', you can see that vectors in $$\mathbb{R}^{S}$$ in fact, can describe everyday objects. In chapter 5 we introduce the general definition of a vector space that unifies all these different notions of a vector.