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Extra info for Linear functional analysis; introduction to Lebesgue integration and infinite-dimensional problems

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Suppose that a <> b = (' <> a = O. Then clearly b = c, and so, if a is quasi-invertible, there is a unique dement b E A such that a <> b = b <> a = 0; b is the quasi-mverse of a, and is denoted bya'l. \Ve write q-Inv A for the set of quasi-invertible elements of A. Clearly 0, <> b = 0 if and only if (eA - a)(cA - b) = eA in A#, and so Inv A# = eA + q-Inv A. Let A be an algebra. and let Sand T be non-empty subsets of A. 2, S . T = {ab : a E S, Ii E T}. We define ST = lin S . T, so that ST = {t aiaibi : 0'1, ...

Let A be an algebra. Then MIn(A) is an algebra for the product given by (ai])(bij ) = (L~=l aikbkj): it is the full matrix algebra of order n over A. We write MIn for MIn(C), so that MIn is a complex algebra of dimension 11. 2 . ) for Inv MIn, the general {meaT group of order n. Let A be a commutative algebra, and take (ai]) E MIn(A). Then the determinant of (aij) is ° ° ° det(aij) = 2) (-1)10'Ia1,O'(l) ... O'(n) : IY E Sn} , where IIYI is the sign of IY. The following result is standard. 9 Let A be a commutative algebra, and let n EN, (i) deteST) = (det S)(det T) (S, T E Mn(A», (ii) Let T E Mn(A).

Again Vr = U, and so l' E U, a contradiction. Thus U = S, and the result is proved. 0 \Ve now give a construction that will show that each cancellative, abelian semigroup can be embedded in a group. Let S be an abelian semigroup, and let T he a subsemigroup of S. We say that (SI' tt) '" (S2' t2) in S x T if there exists l' E T such that r8 l t 2 = rS2tl' Routine verifications show that the relation", is an equivalence relation which is compatible with S x T. = (S x T)/"" and let 7r : S x T ----- S[T- I ] be T, set 1/J(s) = 7r«(8t, t») (8 E S).