Countably dimensional vector space as a limit Consider for example the countable product of $1$-dimensional vector spaces
$$V := \prod_{n\ge0}k\cdot e_n,$$
$k$ a field. This space is of uncountable dimension: if $\{v_1,v_2,\ldots\}$ were a countable basis, let $v_{n,j}$ denote the coordinates of $v_n$, that is
$$v_n = \sum_{j\ge1}v_{n,j}e_j.$$
By swapping elements and using finite linear combinations à la Gauss, we can assume without loss of generality that $v_{i,j} = 0$ for all $j<i$. Now consider
$$v := \sum_{n\ge0}\left(\sum_{j=0}^nv_{n,j}\right)e_n\in V.$$
It is clear that $v$ cannot be written as a finite linear combination of element of the chosen set of elements.
My question is:

Can one obtain a countably dimensional vector space as an arbitrary (category theoretical) limit of finite dimensional spaces? For example as the kernel of some linear map?
If not, how can one prove it?

 A: Let me begin with two Lemmas.

Lemma 1: Let $(X_\alpha)$ be an inverse system of compact $T_0$ spaces such that the maps of the system are all closed maps.  Then the inverse limit $X=\varprojlim X_\alpha$ is compact.

Proof: See A.H. Stone, Inverse limits of compact spaces, General Topology and its Applications (1979), Theorem 5.

Lemma 2: Let $(V_i)_{i\in I}$ be a family of vector spaces and suppose $x_1,\dots,x_n\in \prod_{i\in I} V_i$ are linearly independent.  Then there exists a finite subset $S\subseteq I$ such that the images of $x_1,\dots,x_n$ in $\prod_{i\in S}V_i$ are still linearly independent.

Proof: We use induction on $n$; the case $n=0$ is trivial.  Given a subset $S\subseteq I$,  we say that a statement about elements of $\prod_{i\in I} V_i$ is true "over $S$" if it is true when you project everything to $\prod_{i\in S} V_i$.
Suppose the Lemma is true for $n$ and let $x_1,\dots,x_{n+1}\in \prod_{i\in I} V_i$ be linearly independent.  By the induction hypothesis, we can find a finite subset $S\subseteq I$ such that $x_1,\dots,x_n$ are linearly independent over $S$.  Suppose that no finite set $T$ exists such that $x_1,\dots x_{n+1}$ are linearly independent.  Given any finite set $T\supseteq S$, a linear relation between the $x_k$ over $T$ must involve $x_{n+1}$, so there exist scalars $a_1^T,\dots,a_n^T$ such that $$x_{n+1}=\sum_{k=1}^n a_k^T x_n$$ over $T$.  Moreover, these scalars are actually unique, since $x_1,\dots,x_n$ are linearly independent over $S$ (and hence also over $T$).  This uniqueness implies that if $T\subseteq T'$, $a_k^T=a_k^{T'}$ for all $k$.  This then implies that $a_k^T$ is independent of the choice of $T$, since any two $T$'s can be compared to their union.  Thus there exist scalars $a_k$ such that $$x_{n+1}=\sum_{k=1}^n a_k x_n$$ over any finite set $T$ containing $S$.  This implies that actually $$x_{n+1}=\sum_{k=1}^n a_k x_n$$ in $\prod_{i\in I}V_i$, which contradicts the assumption that $x_1,\dots,x_{n+1}$ was linearly independent.

Now I will prove that a countably infinite dimensional vector space cannot be a limit of finite dimensional vector spaces.  Here's the executive summary: if an infinite dimensional vector space is a limit of finite dimensional vector spaces, it is an inverse limit of finite dimensional vector spaces.  Given such an inverse limit, using Lemma 2 you can find a sequence of spaces in it whose dimension goes to infinity, and then the inverse limit of that sequence has uncountable dimension.  But by a compactness argument using Lemma 1, the inverse limit of the whole system surjects onto the inverse limit of the sequence.
Now for the details.  Suppose $F:I\to \mathtt{Vect}_k$ is a diagram of finite-dimensional vector-spaces whose limit $L$ is infinite dimensional.  Note that $L$ is naturally a subspace of the product $V=\prod_{i\in I} F(i)$ over all objects of $I$.  For each finite set of objects $S\subset I$, let $V_S=\prod_{i\in S} F(i)$.  Give each $V_S$ a topology by saying that a closed set is a finite union of affine subspaces of $V_S$.  Since $V_S$ is finite dimensional, this topology is Noetherian, and in particular is compact.  The projection maps between the spaces $V_S$ for different values of $S$ are also continuous and closed.  By Lemma 1, if we topologize $V$ as the inverse limit of the $V_S$, $V$ is compact.  Moreover, $L$ is a closed subset of $V$, since for each morphism in $I$ the condition that an element of $V$ respect that morphism defines a basic closed set in the topology of $V$.  Thus $L$ is compact as well.
Since $L$ is infinite-dimensional, we can choose an infinite sequence $(x_n)_{n\in\mathbb{N}}$ of linearly independent elements of $L$.  By Lemma 2, we can choose finite subsets $S_n\subset I$ for each $n$ such that $x_1,\dots,x_n$ are linearly independent when projected to $V_{S_n}$.  We may also assume that $S_n\subseteq S_{n+1}$ for each $n$.
Now let $L_n$ be the image of $L$ in $V_{S_n}$.  By our choice of $S_n$, $\dim L_n\geq n$ for all $n$.  The vector spaces $L_n$ naturally form an inverse system $$\dots\to L_2\to L_1\to L_0.$$  Let $L_\omega$ be the inverse limit of this system.  There is a natural linear map $f:L\to L_\omega$, which I claim is surjective.  Indeed, given an element $y\in L_\omega$, for each $n$ the set of $x\in L$ which agree with $y$ on $L_n$ is a nonempty closed subset of $L$.  These closed subsets have the finite intersection property, so their intersection is nonempty by compactness of $L$.  But their intersection is just $f^{-1}(\{y\})$.  Thus $f^{-1}(\{y\})$ is nonempty for any $y\in L_\omega$, so $f$ is surjective.
Thus to show that $L$ has uncountable dimension, it suffices to show that $L_\omega$ has uncountable dimension.  But $L_\omega$ is easy to understand, since the maps $L_{n+1}\to L_n$ are all surjective.  If $d_n=\dim L_n$, we can choose bases for $L_n$ for each $n$ such that the map $L_{n+1}\to L_n$ maps the first $d_n$ basis vectors of $L_{n+1}$ to the basis vectors of $L_n$ and the remaining basis vectors of $L_{n+1}$ to $0$.  It is then clear that the inverse limit of this system can be identified with $k^\mathbb{N}$ (the coordinates corresponding to the coefficients with respect to our bases on the $L_n$).  Since $k^\mathbb{N}$ has uncountable dimension (see https://mathoverflow.net/a/168624/75, for instance), we're done.
A: We will use the following easily proven fact:

Fact: The dual of a vector space cannot have countable dimension.

Let $V:=\varprojlim_\alpha V_\alpha$ be a limit of finite dimensional vector spaces. Then we can replace the diagram given by the $V_\alpha$ and the various morphisms between them with an isomorphic one where the the spaces are given by the duals $W_\alpha^*$ of some finite dimensional vector spaces $W_\alpha$. Then we have
\begin{align}
V:=&\ \varprojlim_\alpha V_\alpha\\
=&\ \varprojlim_\alpha W_\alpha^*\\
=&\ \varprojlim_\alpha \hom\left(W_\alpha,k\right)\\
=&\ \hom\left(\varinjlim_\alpha W_\alpha,k\right)\\
=&\ \left(\varinjlim_\alpha W_\alpha\right)^*
\end{align}
which cannot be countably dimensional because of the fact stated above.
