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

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

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);
• $(\lambda (f+g))(x)=\lambda (f(x)+g(x))=\lambda f(x)+\lambda g(x)$ is (V2), by distributivity in $F$;
• $((\lambda +\mu )f)(x)=(\lambda +\mu )f(x)=\lambda f(x)+\mu f(x)$ is (V3), by distributivity in $F$;
• $((\lambda \mu )f)(x)=(\lambda \mu )f(x)=\lambda (\mu f(x))$ 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: $m<n⟺m\in n$). 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 $n\mathrm{\ne }0$ and $n\mathrm{\notin }n$ always. The operations read

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

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 $0=\mathrm{\varnothing }$, the set $F^0=F^{\mathrm{\varnothing }}$ 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.