In order theory, we have the concept of a lattice, which is defined as consisting of an underlying set $L$ together with two binary operations $\wedge$ and $\vee$. Now when $L$ is finite, the concept of a lattice works great. For example, if $X$ is a finite set, then the lattice generated by the singletons of $\mathcal{P}(X)$ is $\mathcal{P}(X)$. However, if $X$ is infinite, then this is not the case. Thus, a more useful concept in the infinite case is the concept of a complete lattice, which consists of an underlying set $L$ together with two unary operation $\bigwedge$ and $\bigvee$ defined on the powerset of $L$. Sure enough, we have that for $X$ an arbitrary set, not necessarily finite, the complete lattice generated by the singletons of $\mathcal{P}(X)$ is $\mathcal{P}(X)$.
I reckon vector spaces have a similar issues to lattices. When a vector space is finite-dimensional, everything is great. However for infinite-dimensional ones, we really need some sort of unary summation operator $\sum.$
Here's an example where this really matter. Lets adopt, for the moment, all the usual definitions (in particular, let basis and linear combination mean what they usually mean, etc.) and allow $V$ to denote the vector space of all functions
$$[0,1] \rightarrow \mathbb{R}.$$
Furthermore, consider the family $e : [0,1] \rightarrow [0,1] \rightarrow \mathbb{R}$ such that for all $i \in [0,1]$ all $x \in [0,1]$ we have $$x = i \rightarrow e_i(x) = 1,\quad x \neq i \rightarrow e_i(x) = 0.$$
It follows that $e$ cannot span $V$, since only those functions $[0,1] \rightarrow \mathbb{R}$ with finite support can be written as a linear combination of the elements of $e$. So $e$ is not a basis for $V$.
However, it is obvious how to write any function $f : [0,1] \rightarrow \mathbb{R}$ as an "infinitary" linear combination using $e$. Namely
$$f = \sum_{i \in [0,1]} f(i)e_i.$$
So I think that we need a notion of "complete vector space," meaning a vector space equipped with an infinitary summation operator $\sum$ subject to certain reasonable constraints. Furthermore, concepts like "linear combination" and "basis" really need to be defined differently in a complete vector space, because infinite sums are allowed.
However, its not clear how to formalize this idea.
Has anyone succeeded in formalizing the notion of a "complete vector spaces"?
Discussion. Let us first note that this is probably not a topological idea, for the following reason. Suppose $X$ is a set, possible infinite, and $K$ is a field not equipped with any particular topology. Then there ought to exist a "complete vector space" of all functions $X \rightarrow K$. However as far as I can see, there is no topology in sight.
Secondly, if $V$ is a complete vector space over a field $K$, then it is not clear what the domain of $\sum$ should be. Taking the domain as $\mathcal{P}(V)$ isn't going to work; after all, if $x \in V$ is nonzero, we don't want to demand that the sum of $\{kx \mid k \in K\}$ exist. That would be too much to ask! So the domain of $\sum$ really needs to be a proper subset of $\mathcal{P}(V).$ Actually, this isn't really the natural thing to do. It would be better if the domain of $\sum$ were a proper subset of the collection of all multisets in $V$.
I learned in the comments that actually, topological vector spaces do the trick. The key point is that a field $K$ can always be viewed as a topological field with the discrete topology, and thus it can be viewed as a topological vector space over itself. This isn't very exciting if we're only interested in $K^n$ for finite $n$, since $K^n$ just ends up having the discrete topology. However, if we're interested in infinite powers like $K^I$ for $I$ an infinite set, then things become much more interesting. In particular, $K^I$ no longer has the discrete topology (see also, product topology versus box topology). This means, among other things, that we can uniquely decompose any $f \in K^I$ as an infinite sum
$$f=\sum_{i \in I}f(i)e_i,$$
where $e$ is defined as in the original question.