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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!

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  • $\begingroup$ Just to make it clear, a "continuum basis" for $V$ means that a minimal set of algebraic generators for $V$ has cardinality $2^{\aleph_0}$? $\endgroup$ – Fosco Dec 25 '19 at 19:54
  • $\begingroup$ @Fosco yes, I meant exactly it $\endgroup$ – Gleb Chili Dec 25 '19 at 19:56
  • $\begingroup$ 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? $\endgroup$ – Fosco Dec 25 '19 at 19:59
  • $\begingroup$ @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 $\endgroup$ – Gleb Chili Dec 25 '19 at 20:24
  • $\begingroup$ 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 $\endgroup$ – Fosco Dec 25 '19 at 20:27
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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.

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