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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.

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You can pick for each $r$ a vector $v_r$ not in $V_p$ for $p<r$. The family $\{v_r\}$ is an uncountable independent set, contradicting the basis being countable. –  Michael Greinecker Jul 29 '12 at 19:36
    
@MichaelGreinecker: Obviously. But that's not what this question is about. It is about this specific argument I quoted. –  tomasz Jul 29 '12 at 19:52
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I'm not sure I understand how you define your space $V_r$ in the case of a countably infinite dimensional space. –  Joel Cohen Jul 29 '12 at 20:00
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@JoelCohen: Choose an arbitrary countable basis of $V$, index it by rationals, then $V_r$ is the span of all the basic vectors indexed by numbers less than or equal to $r$. Or the intersection of the old $V_r$ with the new $V$, if you'd rather see it that way. –  tomasz Jul 29 '12 at 20:01
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The problem is that we have $$V_r = \bigcup_{s > r} V_s$$. So it's not possible to find an element in $V_r$ that's not in the previous spaces (finding such an element for each $r$ would allow us to construct an uncountable independent set). –  Joel Cohen Jul 29 '12 at 20:20
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2 Answers 2

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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.

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The comment by Joel Cohen (appeared while I was typing) had the main point one minute before I posted the answer. –  Jyrki Lahtonen Jul 29 '12 at 20:32
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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.

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+1 This is exactly it. My comment was a bit sloppy. See Joel's comment or my answer for an explanation as to how the argument fails in the countable dimensional space, and how to fix it in the original case. –  Jyrki Lahtonen Jul 29 '12 at 20:22
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