Can a vector space over an infinite field be a finite union of proper subspaces? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T10:24:43Z http://mathoverflow.net/feeds/question/26 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/26/can-a-vector-space-over-an-infinite-field-be-a-finite-union-of-proper-subspaces Can a vector space over an infinite field be a finite union of proper subspaces? Anton Geraschenko 2009-09-29T14:22:21Z 2012-05-22T18:55:21Z <p>Can a (possibly infinite-dimensional) vector space ever be a finite union of proper subspaces?</p> <p>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.</p> http://mathoverflow.net/questions/26/can-a-vector-space-over-an-infinite-field-be-a-finite-union-of-proper-subspaces/35#35 Answer by David Zureick-Brown for Can a vector space over an infinite field be a finite union of proper subspaces? David Zureick-Brown 2009-09-29T18:40:10Z 2009-09-29T18:40:10Z <p>The finite dimensional case cannot happen by dimension counting (just view everything as affine spaces). </p> http://mathoverflow.net/questions/26/can-a-vector-space-over-an-infinite-field-be-a-finite-union-of-proper-subspaces/36#36 Answer by Richard Dore for Can a vector space over an infinite field be a finite union of proper subspaces? Richard Dore 2009-09-29T18:50:40Z 2010-10-05T16:25:26Z <p>You can prove by induction on n that:</p> <p>An affine space over an infinite field F is not the union of n proper <em>affine</em> subspaces.</p> <p>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.</p> <p>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|&#8805;n by doing the first step and then applying the affine bound.</p> http://mathoverflow.net/questions/26/can-a-vector-space-over-an-infinite-field-be-a-finite-union-of-proper-subspaces/47#47 Answer by Jonathan Wise for Can a vector space over an infinite field be a finite union of proper subspaces? Jonathan Wise 2009-10-01T04:03:42Z 2009-10-01T04:03:42Z <p>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.</p> <p>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 span X (if dim X > 1) so Y = X.</p> http://mathoverflow.net/questions/26/can-a-vector-space-over-an-infinite-field-be-a-finite-union-of-proper-subspaces/666#666 Answer by Emile Bouaziz for Can a vector space over an infinite field be a finite union of proper subspaces? Emile Bouaziz 2009-10-15T21:38:46Z 2009-10-15T21:38:46Z <p>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. </p> http://mathoverflow.net/questions/26/can-a-vector-space-over-an-infinite-field-be-a-finite-union-of-proper-subspaces/14236#14236 Answer by Pete L. Clark for Can a vector space over an infinite field be a finite union of proper subspaces? Pete L. Clark 2010-02-05T06:11:31Z 2010-03-08T06:37:30Z <p>I recently completed a short expository note on this subject, <em>Covering Numbers in Linear Algebra</em>. See:</p> <p><a href="http://math.uga.edu/~pete/coveringnumbersv2.pdf" rel="nofollow">http://math.uga.edu/~pete/coveringnumbersv2.pdf</a></p> http://mathoverflow.net/questions/26/can-a-vector-space-over-an-infinite-field-be-a-finite-union-of-proper-subspaces/14241#14241 Answer by Steve D for Can a vector space over an infinite field be a finite union of proper subspaces? Steve D 2010-02-05T07:00:55Z 2010-02-05T07:00:55Z <p>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 $\alpha\in k^{*}$. 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.</p> <p>Steve</p> http://mathoverflow.net/questions/26/can-a-vector-space-over-an-infinite-field-be-a-finite-union-of-proper-subspaces/14256#14256 Answer by Gerry Myerson for Can a vector space over an infinite field be a finite union of proper subspaces? Gerry Myerson 2010-02-05T11:40:25Z 2010-02-05T11:40:25Z <p>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. </p> http://mathoverflow.net/questions/26/can-a-vector-space-over-an-infinite-field-be-a-finite-union-of-proper-subspaces/69094#69094 Answer by Adam Azzam for Can a vector space over an infinite field be a finite union of proper subspaces? Adam Azzam 2011-06-29T05:51:12Z 2012-05-22T18:55:21Z <p>This is a late response to the post, but I noticed that the question was not answered in general. </p> <blockquote> <p>No vector space is the finite union of proper subspaces. </p> </blockquote> <p>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 spaces. 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 <a href="http://books.google.com/books?id=DQP3AIlrCP0C&amp;lpg=PA211&amp;dq=No%2520vector%2520space%2520over%2520an%2520infinite%2520field%2520is%2520a%2520finite%2520union%2520of%2520proper%2520subspaces&amp;pg=PA211#v=onepage&amp;q&amp;f=false" rel="nofollow">here</a>. </p> <p>In the class of Banach spaces there is a stronger result:</p> <blockquote> <p>If $B$ is a Banach space, then $B$ is not the <em>countable</em> union of proper subspaces.</p> </blockquote> <p>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). </p> <p>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.</p> http://mathoverflow.net/questions/26/can-a-vector-space-over-an-infinite-field-be-a-finite-union-of-proper-subspaces/69227#69227 Answer by Pierre-Yves Gaillard for Can a vector space over an infinite field be a finite union of proper subspaces? Pierre-Yves Gaillard 2011-07-01T03:44:57Z 2011-07-01T17:25:13Z <p><a href="http://mathoverflow.net/users/1/anton-geraschenko" rel="nofollow">Anton Geraschenko</a>'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).</p> <p><strong>NEW VERSION.</strong> 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 <strong>not</strong> cover $A$.</p> <p>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.</p> <p><strong>OLD VERSION.</strong> 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 <strong>not</strong> cover $A$.</p> <p>Indeed, we can assume that $A$ is finite dimensional, in which case the result is easy and well known.</p>