The vector space $F^{X}$ of all functions $X \to F$ with pointwise operations, and $F^{n}$ as the case $X = n = \{0, 1, \dots, n-1\}$

Let $F$ be a field (Field) and let $X$ be any set. Write

$F^{X} \;:=\; \{\, f \;:\; f \text{ is a function } X \to F \,\},$

and for $f \in F^{X}$ write $f(x)$ for its value at $x \in X$. Two elements of $F^{X}$ are equal exactly when they agree at every point of $X$. Define

$(f + g)(x) \;:=\; f(x) + g(x), \qquad (\lambda f)(x) \;:=\; \lambda\, f(x), \qquad 0_{F^{X}}(x) \;:=\; 0_F,$

for $f, g \in F^{X}$, $\lambda \in F$ and $x \in X$, the operations on the right being those of $F$. These are the pointwise operations.

These rules really are the required data. For $f, g \in F^{X}$ the assignment $x \mapsto f(x) + g(x)$ is a function $X \to F$, so $+$ is a binary operation $F^{X} \times F^{X} \to F^{X}$ (Binary operation on a set; associativity, commutativity, and a subset closed under the operation); for $\lambda \in F$ and $f \in F^{X}$ the assignment $x \mapsto \lambda f(x)$ is a function $X \to F$, so scalar multiplication is a map $F \times F^{X} \to F^{X}$; and $0_{F^{X}}$, the constant function at $0_F$, is an element of $F^{X}$.

$F^{X}$ is a vector space over $F$ (Vector space over a field). Each axiom is an equation between elements of $F^{X}$, hence holds exactly when it holds at every $x \in X$ after evaluation, and there it is the corresponding field axiom applied to the values $f(x), g(x), \lambda, \mu$:

• associativity and commutativity of $+$, and $f + 0_{F^{X}} = f$, come from the same laws for $+$ in $F$; the additive inverse of $f$ is $x \mapsto -f(x)$, which lies in $F^{X}$ and satisfies $f + (-f) = 0_{F^{X}}$ pointwise. This is axiom (V1);
• $\bigl(\lambda(f+g)\bigr)(x) = \lambda(f(x)+g(x)) = \lambda f(x) + \lambda g(x)$ is (V2), by distributivity in $F$;
• $\bigl((\lambda+\mu)f\bigr)(x) = (\lambda+\mu)f(x) = \lambda f(x) + \mu f(x)$ is (V3), by distributivity in $F$;
• $\bigl((\lambda\mu)f\bigr)(x) = (\lambda\mu)f(x) = \lambda(\mu f(x))$ is (V4), by associativity of multiplication in $F$;
• $(1_F f)(x) = 1_F f(x) = f(x)$ is (V5), by the multiplicative identity law in $F$.

The case $X = n$

A natural number is a von Neumann natural (The natural numbers $\mathbb{N}$ (von Neumann)), that is a set, and $n = \{\, m \in \mathbb{N} : m < n \,\} = \{0, 1, \dots, n-1\}$ (On $\mathbb{N}$ the order is membership: $m < n \iff m \in n$). Taking $X = n$ therefore gives

$F^{n} \;=\; \{\, x \;:\; x \text{ is a function } n \to F \,\},$

whose elements are written $x = (x_0, x_1, \dots, x_{n-1})$ with $x_i := x(i)$ for $i < n$. The coordinates are indexed from $0$, because $0 \in n$ whenever $n \ne 0$ and $n \notin n$ always. The operations read

$(x + y)_i = x_i + y_i, \qquad (\lambda x)_i = \lambda x_i \qquad (i < n),$

and the zero of $F^{n}$ is the tuple all of whose coordinates are $0_F$.

The two boundary cases. $\mathbb{N}$ contains $0$, so $n = 0$ is a genuine case. Since $0 = \varnothing$, the set $F^{0} = F^{\varnothing}$ has exactly one element, the empty function; that element is $0_{F^{0}}$, so $F^{0}$ is the zero space $\{0_{F^{0}}\}$, not the empty set. For $n = 1$ we have $1 = \{0\}$, and the map $F^{1} \to F$ sending $x$ to its single coordinate $x_0$ is a bijection satisfying $(x+y)_0 = x_0 + y_0$ and $(\lambda x)_0 = \lambda x_0$; we use it to read $F^{1}$ as $F$ where convenient. (No general notion of isomorphism of vector spaces is available on this page, and none is claimed here: what is asserted is exactly the displayed bijection and the two displayed equations.)

• The same set $F^{X}$ also carries a ring structure, and the two must not be conflated. For a ring $R$ and a set $X$, The ring $R^{X}$ of all functions from a set $X$ into a ring, with pointwise operations equips the set $R^{X}$ of all functions $X \to R$ with pointwise addition and pointwise multiplication. Taking $R = F$, the underlying set is literally the same set $F^{X}$ as here, and the addition is literally the same operation, $(f+g)(x) = f(x) + g(x)$ in both. What differs is the second operation:

  	second operation	type
  ring of functions	$(fg)(x) = f(x)\,g(x)$	$F^{X} \times F^{X} \to F^{X}$
  vector space (here)	$(\lambda f)(x) = \lambda\, f(x)$	$F \times F^{X} \to F^{X}$

  Neither is a special case of the other, since they do not even have the same domain: one multiplies two functions, the other multiplies a function by a scalar. They agree in the following sense, and this is the whole of the relation between them: for $\lambda \in F$ let $c_\lambda \in F^{X}$ be the constant function at $\lambda$; then $\lambda f$ and the ring product $c_\lambda f$ have the same value $\lambda f(x)$ at every $x$, so they are equal. Both structures are present on $F^{X}$ at once, and nothing on this page uses the ring product.

• Why $F^{n}$ is defined this way rather than as "$n$-tuples". An $n$-tuple is already a function on an index set, and taking that index set to be the natural number $n$ itself makes the coordinates, the finite sums of The product $g_0 g_1 \cdots g_{n-1}$ of a finite list in a monoid, by recursion, with the empty product ($n = 0$) equal to the identity and the induction arguments below all run over the same object. The price is that every index starts at $0$, and that $F^{0}$ is a one-element space; both are recorded above so that no statement on this page or its companion is quietly restricted to $n \ge 1$.

• $X$ is arbitrary. Nothing above assumes $X$ finite, countable, or nonempty. The case $X = \mathbb{N}$ gives the space of all families of scalars indexed by $\mathbb{N}$, and the case $X = m \times n$ gives the matrices of The vector space $M_{m \times n}(F) := F^{\,m \times n}$ of $m$ by $n$ matrices over a field, with entrywise operations.