# Vector space with continuum basis as an inverse limit of finite dimensional vector spaces

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!

• Just to make it clear, a "continuum basis" for $V$ means that a minimal set of algebraic generators for $V$ has cardinality $2^{\aleph_0}$? – Fosco Dec 25 '19 at 19:54
• @Fosco yes, I meant exactly it – Gleb Chili Dec 25 '19 at 19:56
• You can obtain such a $V$ from a filtered co limit of cardinality $2^{\aleph_0}$, of finite dimensional spaces. Do you want a limit instead? Was it a typo? – Fosco Dec 25 '19 at 19:59
• @Fosco I want to show that every continuum basis space is a projective limit of countable inverse system of finite-dimensional spaces in the following sense: en.wikipedia.org/wiki/Inverse_limit – Gleb Chili Dec 25 '19 at 20:24
• Yes, my question was meant to avoid the possible misunderstanding of confusing projective limit (i.e. limit) VS inductive limit (i.e. colimit). Did you find the statement quoted somewhere? I see an evident reason why every space is a COlimit of finite spaces, I have never seen a similar result for limits – Fosco Dec 25 '19 at 20:27

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.