2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ You have proven that $|V|$ is uncountable, but so is $\Bbb R$, so this does not prove that $V$ is uncountably dimensional. If our definition of the product topology requires finite sums of basis vectors you are correct. If we allow infinite sums of basis vectors the space has countable dimension. $\endgroup$ – Ross Millikan Oct 4 '16 at 5:10
  • $\begingroup$ @RossMillikan I don't think I understand your remark... I am working in the category of vector spaces, so I'm not considering any topology. Also, I work with Hamel bases, so I only allow finite sums of basis vectors. $\endgroup$ – Daniel Robert-Nicoud Oct 4 '16 at 5:17
  • $\begingroup$ If you only allow finite sums you are correct, but the cardinality of the space is not sufficient. You can multiply each basis vector by uncountably many coefficients so the space is uncountable. That would also be true for a finite dimensional space. You are claiming that countably many vectors cannot span the space. $\endgroup$ – Ross Millikan Oct 4 '16 at 5:25
  • $\begingroup$ @RossMillikan Ah, I see. The sequences I consider are not linearly independent. But still, I'm pretty convinced that the space we obtain that way is uncountably dimensional. I'll think about a proof of that fact. $\endgroup$ – Daniel Robert-Nicoud Oct 4 '16 at 5:33
  • $\begingroup$ @RossMillikan Ok, now I have proven that the space $V$ has uncountable dimension. $\endgroup$ – Daniel Robert-Nicoud Oct 4 '16 at 12:13
3
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ You might be tempted to avoid the use of Lemma 1 by just putting the Zariski topology on each $V_S$, so that the inverse limit topology on $V$ is just the Zariski topology on (the $k$-points of) an infinite-dimensional affine space, and then you can make an algebraic argument that this is compact. However, it turns out that the Zariski topology on the $k$-points of an infinite-dimensional affine space is actually not compact in general! Note that Lemma 1 does not apply to the Zariski topology, since projections between affine spaces are not closed maps. $\endgroup$ – Eric Wofsey Oct 4 '16 at 9:16
  • $\begingroup$ Thanks a lot for the answer! The solution is a lot more convoluted than I thought. It's a nice interplay between linear algebra and topology, but as topology is used as an auxiliary tool, I wonder if there is a more elementary proof somewhere. $\endgroup$ – Daniel Robert-Nicoud Oct 4 '16 at 12:15
  • $\begingroup$ Also, a minor remark: in the proof of Lemma 2, it should be "Then given any finite set $T\supseteq S$ such that $x_1,\ldots,x_{n+1}$ are linearly independent over $T$ (...)" I think. $\endgroup$ – Daniel Robert-Nicoud Oct 4 '16 at 12:17
  • $\begingroup$ (1) Yeah, the proof was more complicated than I wanted it to be too...I think that this result really is a "compactness" result in an essential way, so even if you don't use the topological setup you will still be making similar sorts of arguments. But I would expect/hope that there is a simpler way to set it all up (in particular, I was surprised I needed to go dig up Lemma 1 instead of just using Tychonoff's theorem to get the needed compactness). $\endgroup$ – Eric Wofsey Oct 4 '16 at 15:52
  • $\begingroup$ (2) No, the point is that we are assuming for a contradiction that $x_1,\dots,x_{n+1}$ is linearly dependent over $T$. Since $T\supseteq S$, $x_1,\dots,x_n$ are linearly independent over $T$, so the only way this can happen is if $x_{n+1}$ is a linear combination of them. I'll edit to make this a bit clearer. $\endgroup$ – Eric Wofsey Oct 4 '16 at 15:52
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ @EricWofsey I found this alternative proof while working on some other (kind of related) things. I think it works, it's shorter, and it's purely algebraic. What do you think? Please, leave your answer where it is, I still like it! $\endgroup$ – Daniel Robert-Nicoud Oct 25 '16 at 4:53
  • $\begingroup$ Looks good (and I would suggest you accept it over my answer)! You might just additionally mention that dualizing turns colimits into limits because the dual functor $Vect\to Vect^{op}$ is a left adjoint (its adjoint being the dual functor $Vect^{op}\to Vect$). $\endgroup$ – Eric Wofsey Oct 25 '16 at 5:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.