Alright, so we ran into a peculiarity in answering this question. Let R be the set of all functions f defined on the interval [0,1] such that - (1) f(t) is nonzero at no more than countably many points t1, t2, . . . (2) Σi = 1 to ∞ f2(ti) < ∞ . Define addition of elements and multiplication of elements by scalars in the ordinary way, i.e., (f + g)(t) = f(t) + g(t), (αf)(t) = αf(t). If f and g are two elements of R, nonzero only at the points t1, t2, . . . and t'1, t'2, . . . respectively, define the scalar product of f and g as (3) (f,g) = Σi,j = 1 to ∞ f(ti)g(t'j) . Prove that this scalar product makes R into a Euclidean space. By the looks of it, (3) is not referring to a sum across all pairs (i,j), since that may induce (absolute) convergence issues for certain elements in the set, where it might not be possible for them to have a finite norm. So we figured that the sum (3) is such that i and j run in parallel across Z+, like it would be in l2. The peculiarity we found next is in the ordering of the countable domain of points with non-zero image. It may not be well-ordered for some functions, and even if you were to assume only well-ordered domains, there can be several different countable orderings (given that the sums of functions are included). One possibility we considered is if f has smaller ordering than g, we can generalize the sum (3) so that it only goes up to the ordering of f (like we would if f had a finite domain and g had a countable domain). But even so, the list of plausible orderings goes a long way in [0,1], and then there is a problem with resolving one of the properties of the scalar product: (iv) (f , g+h) = (f , g) + (f , h) Typically, proving (iv) would involve showing that (f , g+h) remains absolutely convergent. But the problem is how to consider the domain of g+h in the left expression as opposed to the individual domains of g and h in the right expression. In any case, we're stuck.