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I was looking for alternate proofs of the theorem that "a vector space $V$ of dimension greater than $1$ over an infinite field is not a union of fewer than $|\mathbf{F}|$ proper subspaces" and possible generalizations.

A simple measure-theoretic proof over $\mathbb{R}$ is as follows: By countable additivity the sum of the measures of any collection of subspaces is zero since the measure of each subspace is zero, which is a contradiction.

I would like to look at proofs over arbitrary infinite fields and would like to know if similar statements hold for say modules (finitely generated or otherwise) over infinite rings.

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@Timothy: The title says "countable union" but the first paragraph does not. Of course every vector space is a union of (enough) proper subspaces... And you need $\dim(\mathbf{V})\gt 0$. –  Arturo Magidin Nov 17 '10 at 22:02
@Arturo: Thanks, modified. –  Timothy Wagner Nov 17 '10 at 22:11
@Jason De Vito: In general, you need at least $|\mathbf{F}|$ proper subspaces. –  Arturo Magidin Nov 17 '10 at 22:11
@Jason: I fixed the statement; just to nitpick, the counterexample requires $\dim(V)\gt 1$. –  Arturo Magidin Nov 17 '10 at 22:18
@Timothy: I fixed the statement to what I believe is what you intended (perhaps you were thinking "real numbers" instead of "field" –  Arturo Magidin Nov 17 '10 at 22:19

3 Answers 3

up vote 10 down vote accepted

Here's a proof of the finite dimensional case over any field (finite or infinite):

Theorem. Let $\mathbf{V}$ be a nonzero finite dimensional vector space over $\mathbf{F}$. If $\mathbf{V}$ is a union of $\kappa$ proper subspaces, then $\kappa\geq|\mathbf{F}|$.

Proof. Write $\mathbf{V}=\bigcup\limits_{k\in\kappa} W_{k}$, with $W_k$ a proper subspace of $\mathbf{V}$. By enlarging the $W_k$ as necessary, we may assume that $\dim(\mathbf{V}/W_i) = 1$ for all $i$.

The proof is by induction on $\dim(\mathbf{V})$. The result is trivially true if $\dim(\mathbf{V})=1$, because $\mathbf{V}$ is never the union of proper subspaces in this case.

For the case of $\dim(\mathbf{V})=2$, let $\{w_1\}$ be a basis for $W_1$, and let $v\notin W_1$. For each $\alpha\in \mathbf{F}$ there exists $j_{\alpha}\in\kappa$ such that $w_1+\alpha v\in W_{j_{\alpha}}$. Moreover, if $\alpha\neq\beta$, then $w_1+\alpha v$ and $w_1+\beta v$ are linearly independent, since $\{w_1,v\}$ are a basis for $\mathbf{V}$. Thus, no $W_k$ contains more than one $w_1+\alpha v$. This gives an injection from $\mathbf{F}$ to $\kappa$, proving that $\kappa\geq|\mathbf{F}|$, as required.

Assume the result holds for $n$-dimensional vector spaces, and let $\mathbf{V}$ be $(n+1)$-dimensional. Let $\{w_1,\ldots,w_n\}$ be a basis for $W_1$, and $v\notin W_1$. For each $\alpha\in\mathbf{F}$, consider the subspace $W_{\alpha}=\mathrm{span}(w_1+\alpha v,w_2,\ldots,w_n)$. If $W_{\alpha}$ is contained in some $W_k$, then $W_{\alpha}=W_k$ by dimension considerations; and if $W_{\alpha}=W_{\beta}$, then $\alpha=\beta$, for otherwise we would be able to find a nontrivial linear combination involving $w_1,\ldots,w_n,v$. So there is again an injection from the set $$S=\{\alpha\in\mathbf{F} \mid W_{\alpha}=W_k\text{ for some }k\in \kappa\}$$ to $\kappa$. If the set has cardinality $|\mathbf{F}|$ we are done. Otherwise, let $\alpha_0\in\mathbf{F}\setminus S$, and look at $W_{\alpha_0}$. For each $k\in \kappa$, $W_{\alpha_0}\cap W_k\neq W_{\alpha_0}$; since $$W_{\alpha_0} = W_{\alpha_0}\cap\mathbf{V} = W_{\alpha_0}\cap\left(\bigcup_{k\in\kappa}W_k\right) = \bigcup_{k\in\kappa}(W_{\alpha_0}\cap W_k),$$ then by the induction hypothesis we have that $\kappa\geq|\mathbf{F}|$. QED

I suspect a similar argument can be made in the infinite dimensional case; certainly, we can construct the analogous set to $S$, and if $|S|=|\mathbf{F}|$ then we are done. But if not, then $\dim(W_{\alpha})=\dim(\mathbf{V})$, so we cannot really use a reduction argument. But I think there may be a way to tweak this.

As Pete L. Clark has noted, the result does not hold in the infinite dimensional case.

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That's a lot of effort. Thank you very much. I will read it carefully and get back if I have any questions. Thanks again. –  Timothy Wagner Nov 18 '10 at 3:06
Technically the requirement that $V$ be nonzero is redundant, since the zero vector space can't be written as the union of proper subspaces—it doesn't have any. –  kahen Oct 31 '11 at 3:45
@kahen: Fair enough. –  Arturo Magidin Oct 31 '11 at 3:48

I wrote a short note on exactly this problem. It will appear in the Monthly one of these years.

Note in particular that your quoted statement is not necessarily true for infinite-dimensional vector spaces: an infinite-dimensional vector space over any field can be covered by $\aleph_0$-many proper subspaces.

You ask also about modules over rings. Yes, there has been work on that. I believe that Apoorva Khare, a mathematician who independently proved most of the results in my note, also has some work on the case of modules over various rings. If you google for "covering numbers", you'll find that many, many papers on this topic have been written.

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Aha! That would explain why (i) I couldn't reduce to irredundant coverings in general; and (ii) I couldn't quite close the deal with the infinite dimensional case. But I feel silly for not thinking of $\aleph_0\leq\lambda\lt|\mathbf{F}|$-dimensional vector spaces. Thanks. –  Arturo Magidin Nov 18 '10 at 5:42
Thanks, this is great. –  Timothy Wagner Nov 19 '10 at 14:52
Very nice note. –  copper.hat Jan 31 at 8:40

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