Countably dimensional vector space as a limit

Question

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?

Answer 1

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.

Answer 2

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.

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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.