Can an arbitrary vector space over $\mathbb{F}$ be written as $\mathbb{F}^X$ for some set $X$?

Question

• Each tuple $(x_1, \ldots,x_n) \in \mathbb{F}^n$ can be thought of as a function $f: \{1,\ldots,n\} \rightarrow \mathbb{F}$ where $f(i) = x_i$ for $i \in \{1,\ldots,n\}$. Similarly any $f: \{1,\ldots,n\} \rightarrow \mathbb{F}$ corresponds to a unique tuple in $(x_1,\ldots,x_n) \in \mathbb{F}^n$.

$$\mathbb{F}^n := \mathbb{F}^{\{1,\ldots,n\}} := \{f\ | f: \{1,\ldots,n\} \rightarrow \mathbb{F}\} $$

• In general, we define $\mathbb{F}^X := \{f\ | f: X \rightarrow \mathbb{F} \}$. Consider vectors $f,g \in \mathbb{F}^X$. The vector sum $h = f+g$ satisfies $h(x) = f(x) + g(x)$ for all $x \in X$. For any scalar $k \in \mathbb{F}$, the scalar multiplication with a vector $f$ is the vector $h = k\cdot f$ which satisfies $h(x) = kf(x)$ for all $x \in X$. We can verify that $\mathbb{F}^X$ along with the above defined vector addition and scalar multiplication operations indeed form a vector space.

In the last example $X$ could be any set. For instance if $X = \mathbb{N}$ then the vector space would consist of tuples of infinite length. If $X = \mathbb{R}$ then the vector space would consist of set of all real valued functions. My question is given an arbitary vector space over $\mathbb{F}$ can we write it as $\mathbb{F}^X$ where $X$ is some set. This seems true for finite dimensional vector spaces as any $n$ dimensional vector space is isomorphic to $\mathbb{F}^n$. What about an arbitary infinite dimensional vector space?

Answer 1

This is not true. Take the vector space over $F_2$ (or any other finite field) with a countable base (i.e. it is made of finite linear combinations of the type $\sum_{i=0}^{\infty}\alpha_ie_i$, where $\alpha_i\in F_2$ and $e_i$ are base vectors). This whole vector space is countable.

On the other hand, $F_2^X$ is finite if $X$ is finite, and uncountable otherwise.

Answer 2

As already told you, this is not true. What is true is the following:

If $V$ is a vector space over a field $F$, then there exists a set $X$ such that $V$ is isomorphic to $F^{(X)} := \{f \in F^X : f^{-1}(F \setminus \{0\}) \textrm{ is finite}\}$, which is a subspace of $F^X$.

Indeed, by the axiom of choice, $V$ has a basis, say $B$. Thus, the function $\phi : F^{(B)} \to V$, that sends $f \in F^{(B)}$ to the linear combination $$\sum_{b \in f^{-1}(F \,\setminus \{0\})} f(b)b,$$ is a bijective linear map: Each vector in $V$ can be written in a unique way as a linear combination of basis vectors, in other words, for each $v \in V$ there is a unique $f \in F^{(B)}$ such that $v = \phi(f)$. Linearity is easy to prove.