# Can a vector space over an infinite field be a finite union of proper subspaces?

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.

• Very interesting question.If I remember correctly a related problem is given in the book of Herstein, Topics in algebra. – Ali Taghavi Feb 24 '18 at 11:54
• @AliTaghavi It is a question in M. Artin's Algebra. – Akshay Hegde Mar 21 '18 at 15:07
• Related thread in Math.SE. – Jyrki Lahtonen Jan 18 '19 at 22:07
• Fun fact: From this, it easily follows that for $|k| = \infty$ and $k$-algebras $R$, the Prime Avoidance Lemma holds for arbitrary ideals. I.e. for ideals $J, I_1, \dots, I_n \subsetneq R$ with $J \subseteq \bigcup_{i=1}^n I_i$ we get $J \subseteq I_i$ for some $i$. – Qi Zhu Jul 21 '19 at 18:41

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.

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

• +1. Short and self-contained. I really liked it! – user2734 Feb 5 '10 at 10:50
• best proof and shortest one – Koushik Jan 21 '13 at 11:55
• Similar technique can be found in the proof of the fact that 'A finite extension is simple iff there exist only finitely many immediate subfields' – booksee Mar 11 '14 at 16:50
• With this proof we can even get the sufficient bound $|k|\geq n$: If this condition is met, then by the pigeonhole principle some $V_i$ with $i\neq 1$ must contain two different linear combinations of $x$ and $y$, which assures that $x\in V_i$. – Jose Brox Aug 28 '17 at 18:28
• Should be $|k|>n$ in my previous comment (to take into account the case $\alpha=0$). – Jose Brox Aug 28 '17 at 19:24

I recently completed a short expository note on this subject, Covering Numbers in Linear Algebra. See:

http://math.uga.edu/~pete/coveringnumbersv2.pdf

• +1 Thanks for sharing this. It was a treat to read. – kjo Jan 18 '14 at 20:56

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 $Y = X$.

• Very nice. For any future reader who gets confused in the same way I did, the assumption in the sentence "By assumption, u(Y) is non-empty for every Y" is the finite-dimensional case. – Anton Geraschenko Oct 1 '09 at 4:12

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.

• Ahhh +1. It took me a while to understand what's going on here. The confusing statement was "the product of the $f_i$ is nonzero". Here's my expansion of that statement. (1) The set of functions which are polynomial on each finite-dimensional subspace is clearly a ring. (2) Given two non-zero functions of this form, there is a finite-dimensional subspace on which they are both non-zero (a 2-dimensional subspace spanned by two points witnessing the non-zero-ness of the two functions will do). (3) By (2) and the finite-dimensional case, the ring is an integral domain. – Anton Geraschenko Jul 1 '11 at 16:21
• @Anton Geraschenko - Thanks for your vote and your comment, so much clearer than my answer! – Pierre-Yves Gaillard Jul 1 '11 at 16:29

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.

• Well, this is Thm 1.2 in Roman's book Advanced Linear Algebra. I have the 3rd edition from 2008. The 1st edition is from 1993; I don't know if it was in there. He doesn't cite anything. – Michael Benfield Feb 5 '10 at 12:31
• I believe it is one of those problems that turns up again and again over the years. I have seen it posed at least once as a problem in the American Mathematical Monthly (unfortunately I don't remember when, even up to a decade, and I don't know of a good way to search for it), but my guess is that its provenance dates back from well before 1993. – Pete L. Clark Feb 5 '10 at 18:20

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.

• The "routine induction" doesn't work. As Anton already noted in his initial post $(\mathbb{F}_2)^2=\{(0,0),(1,0)\}\cup \{(0,0),(0,1)\}\cup \{(0,0),(1,1)\}$. – Philipp Hartwig Jun 29 '11 at 6:18
• @Philipp: Under the assumption the OP gave that the field is infinite, your counterexample does not hold. – Asaf Karagila Jun 29 '11 at 13:35
• A vector space $V$ of dimension infinite is always union of a countable set 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 21:21
• @Adam: proper subspaces of a Banach space can be dense. – George Lowther May 22 '12 at 22:39
• What I said is true for any infinite dimension. I speak about a Hamel basis. Each vector has only a finite number of non null coordinates. – juan May 23 '12 at 11:49

The finite dimensional case cannot happen by dimension counting (just view everything as affine spaces).

• This argument certainly doesn't suffice by itself; it doesn't explain where the infinitude of the field comes into play. – Qiaochu Yuan May 22 '12 at 20:01