TITLE: 'Free Vector Space' and 'Vector Space'
QUESTION [20 upvotes]: In this very nice book, author has defined vector space as set of functions $f : S \rightarrow F$ where $S$ is a finite set and $F$ is a field. It turns out that this definition is closely resemble definition of Free Vector Space.
There are one or two threads on other online community regarding this. While I try to undersand it, could anyone point out some differences between vector spaces defined this way and defined in standard ways e.g. Hoffman and Kunze.
Can one give a small concrete example of constructing a vector space using this definition and its implication. Any tutorial or link will be appreciated.

REPLY [5 votes]: This is not the definition of an arbitrary vector space, for (at least) two reasons.
1) The requirement that $S$ be finite means your vector space -- let me call it $F^{(S)}$ -- is a finite-dimensional vector space.  So infinite-dimensional vector spaces are missing.  For this, one can extend the construction to an arbitrary set $S$, but we need to slightly redefine $F^{(S)}$ to be the set of all functions $f: S \rightarrow F$ such that $f^{-1}(F \setminus \{0\})$ is finite (finitely nonzero functions).  Note that this is precisely the definition in the planetmath article on Free Vector Spaces that you link to.
2) This definition captures all vector spaces up to isomorphism only.  It is much closer to the concept of "vector space with distinguished basis".  Indeed, if $V$ is an $F$-vector space, then taking $S$ to be a basis of $V$, we get a map from $V$ to $F^{(S)}$ by mapping each basis element $e \in S$ to the characteristic (or "delta") function $1_e: S \rightarrow F$: namely, for $e' \in S$, $1_e(e') = 1$ if $e = e'$ and $0$ otherwise.  This is consistent with the universal mapping property underlying the definition of "free vector space", i.e., every vector space can be viewed as (or more accurately, canonically endowed with the structure of) the free vector space on any one of its bases in this way.  
In terms of examples: for finite sets $S$, from a concrete perspective this is close to just considering the vector space $F^n$ of $n$-tuples of elements of $F$.  Indeed, arguably $F^n$ is an example of this with $S = \{1,\ldots,n\}$.  In general, the only difference I can see is that the finite set $S$ does not come with a distinguished ordering, so whereas $F^n$ has a canonical ordered basis $e_1,\ldots,e_n$, $F^{(S)}$ has a canonical unordered basis $\{1_s \ | \ s \in S\}$.