Why does a countably infinite dimensional space not have an uncountable chain of subspaces?

Question

This question is motivated by Jyrki Lahtonen's comment in this question. It was an attempt at a short proof that for arbitrary field $k$, we have that $k^\mathbf N$ (as the full set-theoretic product with obvious linear structure) is not of countable dimension over $k$.

Don't know about one-liners. You could also argue that the subspaces $V_r$ defined for all real numbers $r$ as $$V_r:=\{ s:\mathbf Q \to k \lvert s(q)=0\forall q>r\}$$ form an uncountable chain of subspaces (without repetitions): $V_{r_1}\subsetneq V_{r_2}$ whenever $r_1<r_2$.

At first glance, it seems like a really nice and elegant argument, but when you look a little closer, there's no obvious (to me) reason for it not to work in the case of a countably infinite dimensional space. Sure, individual $V_r$ may be smaller, but they still seem to form an uncountable strictly increasing chain of subspaces, contradicting countable dimension.

This obviously can't be right, so where does the problem lie?

Countability is probably immaterial here, you could make a very similar argument using some linear orders larger than rationals, dense in yet larger linear orders to obtain pretty much the same "result" for arbitrarily large cardinals.

Answer 1

The subspace $V'$ tomasz is interested in consists of functions $s:\mathbb{Q}\rightarrow k$ with finite support. It seems to me that $V'$ also has an uncountable chain of subspaces. However, this does not lead to an uncountable set of linearly independent vectors. This is because, for example, $$ V'\cap V_{\sqrt2}=\bigcup_{r<\sqrt2}(V'\cap V_r). $$ This follows from the fact that any function on the l.h.s. has a finite support in $\mathbb{Q}\cap (-\infty,\sqrt2]$ and hence also belongsto the r.h.s. union.

The same does not work in the case of the original space, where we don't require the functions to have a finite support. I repeat the argument from my comment to the linked question. The characteristic function $\chi_r$ of the set $\mathbb{Q}\cap(-\infty,r]$ is in the difference: $$ \chi_r\in V_r,\qquad \chi_r\notin \bigcup_{r'<r}V_{r'} $$ for all real numbers $r$. Therefore the set $\{\chi_r\mid r\in\mathbb{R}\}$ is an uncountable linearly independent set.

So the answer is that the existence of an uncountable chain of subspaces does not really prove that the space has uncountable dimension, unless we can, for each subspace in the chain, give a vector that does not belong to the union of the preceding subspaces.

Answer 2

I think that you're right. There can exist uncountable chains of subspaces in a vector space of countable dimension. Thus Jyrki was wrong when he suggested that the existence of such a chain constituted an argument that the entire space must be uncountable-dimensional.

For finite chains of subspaces, the length of the chain gives a lower bound on the dimension of the entire space. But that is because every finite total order is a well order. That is not the case for countably infinite total orders.