From: "G. A. Edgar" Subject: Re: basis of dual space Date: Mon, 24 Jul 2000 08:45:42 -0400 Newsgroups: sci.math Summary: [missing] In article <397af298.88167898@nntp.sprynet.com>, David C. Ullrich wrote: > I wonder whether in some sense the set of all > linear functionals on V1 which can be "written down" is > V2. That's a little vague - one precise version would be to > ask whether it's consistent with ZF (minus AC) that the > dual of V1 is V2. (Or that in general if V is the space > of all F-valued functions on a set B then the dual is > the space of all functions on B vanishing except at > finitely many points.) Yes... V1 (=countable product of copies of R) is naturally a topological space. (Polish space.) The only CONTINUOUS linear functionals are the members of V2 (=finite linear combinations of coordinate projections). Or the only BOREL MEASURABLE linear functionals. Or the only linear funcionals WITH THE PROPARTY OF BAIRE. Solovay showed that it is consistent with ZF (plus dependent choice) that every subset of a Polish space has the property of Baire (and thus that every linear functional on a Polish vector space has the property of Baire). This is one precise version of the assertion that the only linear functionals on V1 that we can actually "write down" are the members of V2. -- Gerald A. Edgar edgar@math.ohio-state.edu ============================================================================== From: dlrenfro@gateway.net (Dave L. Renfro) Subject: Re: basis of dual space Date: 24 Jul 2000 00:21:51 -0400 Newsgroups: sci.math Siu Lok Shun [sci.math 21 Jul 2000 23:55:51 GMT] wrote > Who can write down explicitly a basis for the dual space of a > infinitely dimensional space, for example F^\mathbb{N}? [The following is formatted for fixed font size.] I assume you mean "Hamel basis" (vector spaces), and not "Schauder basis" (normed spaces). I'll denote the cardinality of a set E by |E|. By 'c', I mean the cardinal number 2^{Aleph_0}. R denotes the set (or field, depending on context) of real numbers and N denotes the set of positive integers. THEOREM 1: If V is a vector space over a field F and dim(V) is infinite, then dim(V*) = |F|^[dim(V)]. This is proved on page 247 of Nathan Jacobson's book "Lectures in Abstract Algebra: Volume II--Linear Algebra", Van Nostrand, 1953 [QA266 .J158L] [This was reprinted as Springer-Verlag's Graduate Texts in Math. #31, QA162 .J3 1975.] In particular, if V is infinite dimensional, then dim(V*) > dim(V) (indeed, dim(V*) will have cardinality at least c), which implies that V* and V are not isomorphic. [Recall that any field has cardinality at least 2.] I don't know if by F^N you mean the direct sum (also called the subdirect product) of Aleph_0 many copies of F or the direct product of Aleph_0 many copies of F. The latter can be realized as the set of all infinite sequences of elements from F (with component-wise operations), whereas the former can be realized as the set of all infinite sequences of elements from F which contain only finitely many nonzero terms. You probably mean direct product, since one commonly writes F^(N) if a direct sum is intended. However, since the dual of F^(N) is the direct product of Aleph_0 copies of F [This follows from the more general fact that Hom[SUM(V_i), W] is isomorphic to Hom[PRODUCT(V_i), W], where the V_i's (not necessarily a countable collection) and W are vector spaces and 'SUM', 'PRODUCT' represent direct sum and direct product.], just finding a basis for F^N (direct product) is essentially equivalent to finding a basis for the dual space of F^(N) (direct sum). I'll pass on exhibiting a basis for the dual space of F^N (direct product), and focus on finding a basis for the dual space of F^(N) (direct sum). In fact, I'll only consider the problem of exhibiting a linearly independent set in F^(N) (direct sum) having cardinality c. There is a short argument given in Example 6.4 on page 46 of William C. Brown's book "A Second Course in Linear Algebra", John Wiley & Sons, 1988 [QA184 .B765 1987], for the fact that the dual space of F^(N) (direct sum) has uncountable cardinality. However, I don't believe the issue is as straightforward as Brown says. [He writes "A simple counting exercise will convince the reader that ...".] Below I'll prove a result that implies the dual space of F^(N) (direct sum) has cardinality at least c (equal to c when |F| is at most c) in a way that will roughly allow you to exhibit a linearly independent subset of (F^(N))* that has cardinality c. A number of years ago I took a linear algebra class, and in this class the instructor made a comment while we were in the middle of all the usual dual space results one proves in such a class (specifically, the last few sections of Chapter 3 from Hoffman/Kunze's text "Linear Algebra") that V and V* are NOT isomorphic when V is infinite dimensional. He said, however, that the proof of this was beyond the level of our course and so we would not prove it. At this time I had been learning some set theory on my own and had just learned that there exists a collection of c many almost disjoint subsets (definition is below) of a countably infinite set. It struck me that there must be some way of utilizing this result to prove dim(V*) > dim(V) when dim(V) is infinite, since both the notion of a Hamal basis and the notion of a collection of almost disjoint subsets involve the idea of equality up to a finite set. I was able to write up a proof during the rest of the lecture (whose notes I wound up having to get from someone else), which of course I proudly gave to the instructor after the class. [If only more of my guesses were correct and I was so easily able to prove them!] He gave it back to me the next day, after having a chance to read over it, and said that it was neat. For a couple of years I left it alone with all my other various writings, and then when I became more aware of the idea of publishing math papers, it occurred to me that this might be suitable for a short American Mathematical Monthly article. Unfortunately, I saw essentially the same proof somewhere shortly afterwards [I can't seem to find the specific reference right now. (See ** below, however.)], and so there went that idea. However, now with the internet I have an option intermediate between "left in one's drawer" and "published"--post it on the internet! [This is where a lot of my longer sci.math "essay posts" have come from, by the way.] [** Added later] I looked though some more photocopied papers I have and found something that *might* have been what led me to not write the result up. The paper I found gives a short proof (the entire paper is only half a page) that any infinite dimensional separable Banach space has (Hamel) dimension at least c using the fact that there exists a collection of c many almost disjoint subsets of the positive integers: H. Elton Lacey, "The Hamel dimension of any infinite dimensional separable Banach space is c", Amer. Math. Monthly 80 (1973), 298. [This fact was overlooked by Willard when, on page 184 of his text "General Topology", exercise 24J(7) asks the reader to prove "If a Banach space is Aleph_0--dimensional, it is separable."] For those interested in other published proofs of this fact about Banach spaces, I mention these other papers: W. R. Brauer and R. H. Benner, "The nonexistence of a Banach space of countably infinite Hamel dimension", Amer. Math. Monthly 78 (1971), 895-896; Nam-Kiu Tsing, "Infinite-dimensional Banach spaces must have uncountable basis--an elementary proof", Amer. Math. Monthly 91 (1984), 505-506. Also, exercise 1 on page 53 of Rudin's text "Functional Analysis" (3'rd edition) states: "If X is an infinite- dimensional topological vector space which is the union of countably many finite-dimensional subspaces, prove that X is of the first category in itself. Prove that therefore no infinite- dimensional F-space has a countable Hamel basis." Finally, a proof (using the Hahn-Banach theorem) that the dimension of any infinite dimensional Banach space is at least c is given on page 75 of Goffman and Pedrick's text "First Course in Functional Analysis" (1983 2'nd edition published by Chelsea). Interestingly, I couldn't find a proof in volume 1 of Dunford and Schwartz's treatise, but that doesn't mean it's not there. (I'd be quite suprised if it wasn't in DS.) DEFINITION: Let B be a countable set and let C be a collection of subsets of B. We say that C is almost disjoint if the intersection of any two distinct members of C is finite and the cardinality of each element belonging to C is Aleph_0. [Yes, I know this notion has a more general formulation.] FACT: Let B be a countably infinite set. Then there exists a collection of c many almost disjoint subsets of B. PROOF: Two proofs are given later. THEOREM 2: Let V be an infinite dimensional vector space over a field F. Then the dimension of V* is at least c. PROOF: Let B = {v1, v2, ...} be a countably infinite linearly independent subset of V and let {B_r: r \in R} be a collection of c many almost disjoint subsets of B. For each real number r let f_r: V --> R be a linear extension (which will not be unique if B isn't a basis) of the function that equals 1 if x belongs to B_r and equals 0 if x belongs to B - B_r. Then {f_r: r \in R} is a subset of V*. I claim that {f_r: r \in R} is linearly independent in V*. To this end, suppose that (*) c1*f_r1 + c2*f_r2 + ... + cn*f_rn = 0 for some c1, c2, ..., cn in F and r1, r2, ..., rn in R. Choose an integer N sufficiently large so that k > N ==> (f_ri)(vk) not equal (f_rj)(vk) whenever i not equal j. [Take N to be any integer larger than the largest index of any basis element appearing in the finite set UNION{B_i intersect B_j: i not equal j and i,j are between 1 and n, inclusive}. (Note that n is a fixed positive integer, being fixed due to equation (*) above.)] Now fix any i between 1 and n (inclusive). Since B_ri is infinite for each i (part of the definition of being an almost disjoint collection), we can find k > N such that vk \in B_ri. Evaluating both sides of (*) (an equation in V*) above at the vector vk \in V gives c1*(f_r1)(vk) + c2*(f_r2)(vk) + ... + cn*(f_rn)(vk) = 0 ==> ci*1 = 0 [all terms except the i'th term drop out], and so ci = 0. Since i was arbitrary (prior to being fixed), we have c1 = c2 = ... = cn = 0. Hence, {f_r: r \in R} is linearly independent (we've shown that every finite subset of {f_r: r \in R} is linearly independent), which implies that V* has dimension at least c. REMARK: Using some elementary cardinal arithmetic it is not difficult to use theorem 2 to show the following, where Q = the field of rational numbers: The vector space Q^N (direct product) over the field Q has dimension c. The vector space R^N (direct product) over the field R has dimension c. By way of contrast, note that Q^(N) and R^(N) (direct sums) over the fields Q and R, respectively, each have dimension Aleph_0. TWO PROOFS THAT THERE EXISTS AN ALMOST DISJOINT COLLECTION OF c MANY SUBSETS OF A COUNTABLY INFINITE SET. Note that it suffices to prove the result for any specific countably infinite set (unless you're concerned with constructivist issues, which I'm certainly not). 1. For each irrational number between 0 and 1 we associate a subset of the natural numbers as follows. If x = .abcdefg..., where a, b, c, etc. are the digits of the decimal expansion of x, let N(x) be the set {a, ab, abc, abcd, ...}. For example, N(sqrt(1/2)) = {7, 70, 707, 7071, ...}. Then it is not difficult to check (i) N(x) is infinite for each irrational x in (0,1), (ii) the correspondence x <---> N(x) is one-to-one and onto (hence, there are c many N(x)’s), and (iii) N(x) has finite intersection with N(y) whenever x is not equal to y. 2. We'll find an almost disjoint collection in N x N with cardinality c. For each real number r between .9 and 1.1, let N(r) = {(m,n) in N x N : rm - 2 < n < rm + 2}. In other words, N(r) is the set of points in N x N that belong to the open angular sector in the first quadrant formed by the lines y = rx - 2 and y = rx + 2. Then it is not difficult to check (i) N(r) is infinite for each real number r between .9 and 1.1, (ii) the correspondence r <---> N(r) is one-to-one and onto (hence, there are c many N(r)’s), and (iii) N(r) has finite intersection with N(s) whenever r is not equal to s. Proof #2 is given in J. R. Buddenhagen, "Subsets of a countable set", Amer. Math. Monthly 78 (1971), 536-537. I don't know if this proof was previously known to anyone, or even if it had been published previously. I'd be interested in learning of an earlier published appearance of proof #2 if anyone knows of one. Dave L. Renfro ============================================================================== From: dlrenfro@gateway.net (Dave L. Renfro) Subject: Re: basis of dual space Date: 24 Jul 2000 15:21:45 -0400 Newsgroups: sci.math Dave L. Renfro [sci.math 23 Jul 00 11:08:08 -0400 (EDT)] wrote (in part) > There is a short argument given in Example 6.4 on page 46 of > William C. Brown's book "A Second Course in Linear Algebra", > John Wiley & Sons, 1988 [QA184 .B765 1987], for the fact that > the dual space of F^(N) (direct sum) has uncountable cardinality. > However, I don't believe the issue is as straightforward as Brown > says. [He writes "A simple counting exercise will convince the > reader that ...".] > > Below I'll prove a result that implies the dual space of > F^(N) (direct sum) has cardinality at least c (equal to c when > |F| is at most c) in a way that will roughly allow you to > exhibit a linearly independent subset of (F^(N))* that has > cardinality c. and David C. Ullrich wrote (in part) [Not yet at the URL given above, but his post can be found at as a July 24 response to Siu Lok Shun's July 23 post.] > I see a proof has appeared elsewhere in the thread. But it's > really not hard to figure out the cardinality: Let's say B is > a basis for V; now we may assume that V is the space of all > functions from B to F which vanish except at finitely many > points. It's easy to see that the dual of V is exactly the > space of all functions from V to F (if L is a linear functional > on V, x is an element of V, and x is the sum over B of f_b * b, > where the f_b are scalars, then Lx = the sum of f_b * L(b). > The map taking L to the function (b -> L(b)) is an isomorphism > from the dual of V to the space of all functions from B to F.) > So the cardinality of the dual is exactly |F|^|B|. Ooops! Brown claimed that the DIMENSION of the dual space of F^(N) is uncountable, not just that the dual space of F^(N)--as a set--is uncountable. Also, what I later proved (and at that later point stated) is that the DIMENSION of the dual space of F^(N) is at least c. As Ullrich points out, finding the cardinality of the dual space of F^(N) is almost immediate. Here's another proof--essentially the same proof, actually--the proof that I think Brown had in mind (and which leads me to suspect that in his example Brown overlooked the distinction between the cardinality of a space and the dimension of that space) when he wrote the stuff I quoted from his text: The set of all functions from a (fixed) basis B of F^(N) into F, by the definition of cardinal exponentiation, has cardinality |F|^|B|. Since any such function has an extension to a linear function from F^(N) into F (and different functions from B into F obviously give rise to different linear functions from F^(N) into F), the cardinality of the dual space of F^(N) is at least |F|^|B|. Moreover, since any linear function from F^(N) into F is determined by the values it takes on B, the cardinality of the dual space of F^(N) is at most |F|^|B|. Finally, any field has at least two elements, and so |F|^|B| is at least 2^|B| = 2^(Aleph_0), which is uncountable. [Regardless of the field F, |B| = dim[F^(N)] = Aleph_0.] Incidentally, if F is the field of real numbers, then F^(N) and the dual space of F^(N) have the same cardinality. [F^(N) can be identified with F union FxF union FxFxF union ..., which has cardinality c + c^2 + c^3 + ... = c, and |F|^|B| has cardinality c^(Aleph_0) = (2^Aleph_0)^(Aleph_0) = 2^(Aleph_0 * Aleph_0) = 2^(Aleph_0) = c.] However, the DIMENSIONS of these two spaces are different. Dave L. Renfro