Can a (possibly infinite-dimensional) vector space ever be a finite union of proper subspaces?
If the ground field is finite, then any finite-dimensional vector space is finite as a set, so there are a finite number of 1-dimensional subspaces, and it is the union of those. So let's assume the ground field is infinite.
Let $V$ be the union $\cup_{i=1}^n V_i$, where the $V_i$ are proper subspaces and the ground field $k$ is infinite. Pick a non-zero vector $x\in V_1$. Pick $y\in V-V_1$, and note that there are infinitely many vectors of the form $x+\alpha y$, with $\alpha\in k^\ast$. Now $x+\alpha y$ is never in $V_1$, and so there is some $V_j$, $j\neq 1$, with infinitely many of these vectors, so it contains $y$, and thus contains $x$. Since $x$ was arbitrary, we see $V_1$ is contained in $\cup_{i=2}^n V_i$; clearly this process can be repeated to find a contradiction.
Steve
You can prove by induction on n that:
An affine space over an infinite field $F$ is not the union of $n$ proper affine subspaces.
The inductive step goes like this: Pick one of the affine subspaces $V$. Pick an affine subspace of codimension one which contains it, $W$. Look at all the translates of $W$. Since $F$ is infinite, some translate $W'$ of $W$ is not on your list. Now restrict all other subspaces down to $W'$ and apply the inductive hypothesis.
This gives the tight bound that an $F$ affine space is not the union of $n$ proper subspaces if $|F|>n$. For vector spaces, one can get the tight bound $|F|\geq n$ by doing the first step and then applying the affine bound.
I recently completed a short expository note on this subject, Covering Numbers in Linear Algebra. See:
Here's a reduction to the finite dimensional case. Let $F$ be a finite set of subspaces of $X$. For each finite dimensional subspace $Y$ of $X$, let $u(Y)$ be the set of elements $Z$ of $F$ such that $Y$ is contained in $Z$. By assumption, $u(Y)$ is non-empty for every $Y$. Since any two finite dimensional subspaces are contained in a third, the intersection of the sets $u(Y)$, as $Y$ runs among all finite dimensional subspaces of $X$, is non-empty. Hence there is at least one set in $F$ that contains every finite dimensional subspace of $X$, hence contains $X$.
For the finite dimensional case, let $F$ be a finite set of subspaces of $X$. By induction, every codimension 1 subspace of $X$ is contained in some $Y$ from $F$. But there are infinitely many codimension $1$ subspaces, so some $Y$ in $F$ contains more than one such subspace. Any two distinct codimension 1 subspaces $\operatorname{span} X$ (if $\dim X > 1$) so $Y = X$.
Anton Geraschenko's comment prompted me to write a new version of this short answer. I'm leaving the old version to make Anton's comment clearer (and also to increase the probability of having at least one correct answer).
NEW VERSION. Let $A$ be an affine space over an infinite field $K$, and let $f_1,\dots,f_n$ be nonzero $K$-valued functions on $A$ which are polynomial on each (affine) line. Then the product of the $f_i$ is nonzero. In particular the $f_i^{-1}(0)$ do not cover $A$.
Indeed, as pointed out by Anton, the $K$-valued functions on $A$ which are polynomial on each line form obviously a ring $R$. This ring is a domain, because if $f$ and $g$ are nonzero elements of $R$, then there is a line on which none of them is zero, and their product is nonzero on this line.
OLD VERSION. Let $A$ be an affine space over an infinite field $K$, and let $f_1,\dots,f_n$ be nonzero $K$-valued functions on $A$ which are polynomial on each finite dimensional affine subspace. Then the product of the $f_i$ is nonzero. In particular the $f_i^{-1}(0)$ do not cover $A$.
Indeed, we can assume that $A$ is finite dimensional, in which case the result is easy and well known.
I needed this result for a paper I wrote with David Leep ten years ago. Bruce Reznick came up with a nice proof which we included in the paper (Marriage, Magic, and Solitaire, published in the American Math Monthly). I don't think the proof was any better than the ones already given here, and I seriously doubt this was the first time a proof had ever appeared in print, but I wonder if anyone knows an earlier citation.
For a slightly worse answer for the fin dim case - prove the following - if k is an infinite field then if f is a polynomial in n variables over k there exists a point of k^n x such that f(x) is non zero (proving this really isn't much easier than the actual problem though - I told you it was a worse answer.) Each subspace is mapped to zero by some poly over k, multiplying the polys gives a contradiction.
This is a late response to the post, but I noticed that the question was not answered in general.
No vector space is the finite union of proper subspaces.
EDIT: In response to my false solution, Phil Hartwig pointed out that $\mathbb{F}$$_{2}^2$ is a vector space that is the union of three proper subspaces. Indeed, the "routine" induction was less routine and more nonsensical. I had fixed my proof, only to realize that my solution was much less elegant than Halmos' solution found in his Linear Algebra Problem Book. You can view the page here.
In the class of Banach spaces there is a stronger result:
If $B$ is a Banach space, then $B$ is not the countable union of proper subspaces.
This relies on the fact that a proper subspace of a topological vector space has empty interior. To appeal to your intuition in $\mathbb{R}^3$, every proper subspace (a plane or line through the origin) cannot completely contain an open ball (an open set in the usual norm topology).
Since $B$ is complete (by definition), by Baire's Theorem it is not the countable union of nowhere dense sets. Since proper subspaces are nowhere dense, $B$ is not the countable union of proper subspaces.
Take a Hamel basis $e_\alpha$ . Each vector can be written $v = \sum_\alpha x_\alpha e_\alpha$.
Let $V_\alpha$ the subspace of those vectors for which $x_\alpha = 0$
It is clear that $V$ is the union of any sequence $V_{\alpha_n}$ for $n\in \N$ with $\alpha_n$ differents.
– juan May 22 '12 at 21:21Baire theorem is no problem since an hyperplane may be of second Baire category.
In fact there was a conjecture by Klee and Wilansky that first category hyperplane are the same as closed hyperplanes. I proved this conjecture is not true
Dense Hyperplanes of First Category. Mathematische Annalen. Vol. 249. 1980. Pag. 111-114.
– juan May 23 '12 at 15:20Functional analysts use the word subspace in infinite dimensional space synonymously with the term "closed subspace". The term "linear manifold" is used for a not-necessarily closed subspace.
For example, the set of all sequences in $\ell^2$ with finitely many non-zero coordinates is a dense linear manifold in $\ell^2$, but not a dense subspace. It's semantics, really, but I wanted to clear up what I meant in the original post.
– Adam Azzam May 24 '12 at 23:09It is certainly true that no proper closed subspace has interior in the norm topology. For if it did contain an open ball, one could shift it to the origin and take successive dilations (which would be elements of the subspace) to exhaust the entire space.
– Adam Azzam May 24 '12 at 23:14Your remarked phrase is false. You may put it as
If B is a Banach space, then B is not the countable union of proper closed subspaces.
This is true and you must correct. In the proof you also must put the adjective closed where this is your meaning.
– juan May 25 '12 at 08:00The finite dimensional case cannot happen by dimension counting (just view everything as affine spaces).
