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6: A = n , n-dimensional Euclidean space, with 1/2 n−1 n n 2 |ai − bi | d(a , b ) = , i=0 the Euclidean distance. | and distance d(a, b) = |a − b|. A vector space or linear space A is a space consisting of points called vectors and two operations — one called addition which associates with each pair of vectors a, b a new vector a + b ∈ A and one called multiplication which associates with each vector a and each number r ∈ (called a scalar) a vector ra ∈ A — for which the usual rules of arithmetic hold, that is, if a, b, c ∈ A and r, s ∈ , then a+b (a + b) + c r(a + b) (r + s)a (rs)a 1a = = = = = = b + a (commutative law) a + (b + c) (associative law) ra + rb (left distributive law) ra + sa (right distributive law) r(sa) (associative law for multiplication a.

Clearly F = n=1 Fn since F contains all of the elements of all of the Fn and each element in F is in some Fn . Let An be a nonincreasing sequence of nonempty atoms in Fn as in the ∞ hypothesis for a basis. If the An are all equal for large n, then clearly n=1 An = ∅. If this is not true, then there is a strictly decreasing subsequence Ank and hence from (b’) ∞ ∞ Ank = ∅. An = n=1 k=1 This means that the Fn form a basis for F and hence F is standard. Lastly, suppose that F is standard with basis Fn , n = 1, 2, .

Proof: The basic idea is that if you take a standard space with a basis and remove some atoms, the original basis with these atoms removed still works since decreasing sequences of atoms must still be empty for some finite n or have a nonempty intersection. Given a basis Fn for (A, B), form a candidate basis Gn = Fn ∩ F c . This is essentially the original basis with all of the atoms in the sets in F removed. ” From the good sets principle, the Gn asymptotically generate B ∩ F c . Suppose that Gn ∈ Gn is a decreasing sequence of nonempty atoms.