This is the wrong question. For Hilbert spaces (such as Hilbert space) there is a separate concept of Hilbert basis, whose cardinality is used to define the "Hilbert dimension" of such a space. (See http://en.wikipedia.org/wiki/Hilbert_space#Hilbert_dimension.) But a Hilbert space still has an underlying vector space structure and vector space dimension. At least for the standard Hilbert space H (take its incarnation as square-summable sequences of reals), the "Dedekind cut" method, that works to show that the vector space dimension of the direct product R^oo is c, works to show that the vector space dimension of H is also c. For, instead of the vectors {e_j} used in the proof for R^oo, use the vectors {e_j/j}. Again letting f: Z+ -> Q be a bijection to the rationals, let w_r denote the Hilbert space sum of all the e_j/j such that f(j) < r. (w_r is easily seen to be well-defined in H.) Then mimicking the argument for R^oo shows that the set {w_r | r in R} is a linearly independent set, so once again c = |{w_r | r in R}| <= dim(H) <= |H| = c, so dim(H) = c. Whereas, the Hilbert dimension of H is aleph_0. --Dan On 2012-08-27, at 12:49 PM, Fred lunnon wrote: << . . . . . . whether this definition of "dimension" leads to a sensible results in Hilbert space . . . . . . . . .