Michael Reid's example of functions f(n) = x^n is illuminating. In particular, to support my doubt about whether this definition of "dimension" leads to a sensible results in Hilbert space, I should need to answer affirmatively the following question: Given an arbitrary real sequence X = [x_i] with |x_i| < 1 , does there exist a real sequence A = [a_i] such that for all natural n \sum_i a_i (x_i)^n = 0 over natural i ? Consider for example specialising to x_i = 2^(-i) : we require A such that \sum_i a_i = \sum_i a_i/2^i = \sum_i a_i/4^i = ... = 0. Restricting 0 <= n < m, 0 <= i <= m yields homogeneous simultaneous linear equations; do their solutions A^m converge as m -> oo ? Resulting A^m for m = 1..8, fixing a_0 = 1 : [1,-1] [1, -3, 2] [1, -7, 14, -8] [1, -15, 70, -120, 64] [1, -31, 310, -1240, 1984, -1024] [1, -63, 1302, -11160, 41664, -64512, 32768] [1, -127, 5334, -94488, 755904, -2731008, 4161536, -2097152] [1, -255, 21590, -777240, 12850368, -99486720, 353730560, -534773760, 268435456] Apparently (-1)^m A^m is unimodal at |a_(m-1)| = 2^(m(m-1)/2) - 2^((m-1)(m-2)/2) ; so normalising to \sum (a_i)^2 = 1, as m -> oo all initial coefficients approach zero: in the limit, there is after all no (nonzero, infinite) linear relation! Tradition 1, Lunnon 0. Often happens. WFL On 8/27/12, Michael Reid <reid@gauss.math.ucf.edu> wrote:
Let R^oo denote the real vector space that is the countable direct product of copies of the reals.
I.e., all countable-tuples of reals with componentwise addition.
Puzzle: What is the dimension of the real vector space R^oo ???
I thought about the same question, in a slightly more general setting a few years ago. Specifically, let k be a field, and let Map(N, k) be the k-vector space of all functions from N (the set of natural numbers) to k . The question I considered, was "what is the dimension of Map(N, k) over k ?"
I found it very counterintuitive that the answer depends upon the field k !
I did not solve it in all cases, but I can do Dan's case, in which case the dimension is c = 2^Aleph_0 .
In general, the dimension is at least max(|k|, c) . I do not know if this is the right answer if k has cardinality strictly between Aleph_0 and c (in which case CH is false).
For x in k , consider the function n |--> x^n . These functions are linearly independent over k , so the dimension is at least |k| . On the other hand, the vector space V = Map(N, k) has cardinality at least c . If k is an infinite field, and B is a basis of V , then |V| = |B| |k| = max(|B|, |k|) . This shows that if |k| < c , then |B| >= c . If k is finite, then it's easy to see that |V| = c , and B must be infinite, in which case |V| = |B| , so the dimension is c .
(When k = R , the vector space has cardinality c , so a basis can't have cardinality larger than c .)
Michael Reid
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