A vector space over $R$ is not a countable union of proper subspaces

Solution 1:

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 $n$-dimensional 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 dependency 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 $\kappa$ (assuming WLOG that the $W_k$ are pairwise distinct). 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.

Solution 2:

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.