Nice and general, Michael! My solution just determines dim(R^oo). Here of course oo denotes aleph_0. In what follows we identify aleph_0 with the first countable ordinal, w, and so we denote R^oo by R^w. Lemma: ------ |R^w| = c. Proof: ------ Since |R| = 2^ω = c, it follows that |R^w| = (2^w)^w = 2^(wxw) = 2^w = c. Hence, |R^w| = c is an upper bound for any basis of R^w, i.e., dim(R^w) <= c. Definition: ----------- B = the set of vectors {e_j in R^w | j in Z+} such that the jth coordinate of e_j is 1 and the rest are 0. Let f: Z+ -> Q (the rationals) be any fixed bijection. Definition: ----------- For each real number r, define vector v_r in R^w via v_r satisfies pi_n(v_r) is 1 if f(n) < r, and is 0 otherwise. (I.e., each v_r corresponds to the indicator function of the Dedekind cut for r.) Lemma: ------ No nontrivial linear combination of the v_r's can be the 0 vector. Proof: ------ In any linear combination of the v_r's with nonzero coefficients, the largest r, among the v_r's occurring in it, contains coordinates that are 1, but which are 0 in all the other v_r's occurring in it. Hence the set of all v_r's is linearly independent Since |{v_r | r in R}| = c, we have c <= dim(R^w) <= |R^w| = c, so |R^w| = c. --Dan _____________________________________________ Michael Reid wrote: << I 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 .)