Let $K$ be a field and $Vec_K$ – category of all vector spaces over $K$. I want to prove that every vector space with continuum basis is an inverse limit of countable inverse system of finite-dimensional vector spaces.
$$\dots \leftarrow V_i \leftarrow V_j \leftarrow \dots$$
I need a hint about how I can construct such inverse system since all I came up with gave me a limit with countable basis. Thanks in advance!
Best Answer
This is not true without additional hypotheses on $K$. To compute the inverse limit of a system of finite-dimensional vector spaces, you can just observe that it is dual to the direct limit of the dual system (this is immediate from the universal properties; see for instance this answer). In particular, the inverse limit of any system of finite-dimensional vector spaces is a dual vector space, so is isomorphic to $K^S$ for some set $S$. But the dimension of $K^S$ is always $|K^S|$ if $S$ is infinite (see this answer on MO). In particular, if $|K|>2^{\aleph_0}$, then any inverse limit of finite-dimensional vector spaces which is infinite-dimensional must have dimension greater than $2^{\aleph_0}$.
On the other hand, if $|K|\leq 2^{\aleph_0}$, then it is true (and it is true not just for some sequential inverse limit of finite-dimensional vector spaces but every sequential inverse limit of finite-dimensional vector spaces which is infinite-dimensional). Indeed, for sequential systems, the colimit of the dual system must be countable-dimensional, and so if it is infinite-dimensional the inverse limit will be isomorphic to $K^{\mathbb{N}}$ which has dimension $|K|^{\aleph_0}=2^{\aleph_0}$ since $|K|\leq 2^{\aleph_0}$. For a very concrete example, let $V_n=K^n$ where the maps $V_{n+1}\to V_n$ are the projections that drop the last coordinate, so the inverse limit is $K^{\mathbb{N}}$ in a very natural way.