Download PDF by I Kaplansky: Rings of operators

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A) Show that AO = O. ) = 0 implies x = O. 2. ISOMETRIES OF THE PLANE R2 Show that any inner product space isomorphism of £2 (2) onto itself may be represented by a matrix multiplication y = Ax where A is of the form ( 8 sin 8 COS - sin 8 ) ( cos 8 cos 8 or sin 8 sin 8 ) - cos 8 . Deduce from this that the only linear isometries of £2 (2) are rotations through an angle 8, reflection about a line making an angle 8/2 with the x-axis, or compositions of these. If we drop "linear," there are other distance-preserving maps-translations, for example.

A point x is called an ADHERENCE POINT of S if every open ball B(x, r), r> 0, meets (has nonempty intersection with) S. • The collection cl S, or 5, of all adherence points of S is the CLOSURE of S. If S = cl S then S is called CLOSED. • If cl S =X then S is DENSE in X . 0 As there are no points adherent to 0, the null set is closed-how could any set have nonempty intersection with 0? It is also clear that the whole space X is closed. For any subset S of X, cl S is closed, since an open ball B of positive radius about any point x of cl( cl S) must contain a point of cl S; therefore, B must also contain a point of S, and it follows that x E cl S.

6( e) is continuous on the irrationals. 6. PROJECTIONS ARE CONTINUOUS Let (Xi, di ), i = 1,2, ... 2). 6(d), we define the projection map Pi (1 ~ j ~ k) on Xl x X 2 X ••. X Xk as the map that sends (Xl, X2, ... , Xk) into xi. Show that each projection Pi, 1 ~ j ~ k, is continuous. 7. MAPS ON PRODUCTS Let X, Y, and Z be metric spaces. 2). (a) If! (x,Yo) is continuous at Xo. (b) Show that the converse is false. 2. Analysis 47 8. UNIQUE REPRESENTATION Let (en) be an orthonormal sequence in the inner product space X and let (an) be a sequence of scalars.

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Rings of operators by I Kaplansky

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