By C. H. Edwards Jr.

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If S is an infinite subset of A, then S has a limit point be B, because B is compact. But be A also, because b is a limit point of A, and A is closed. Thus every infinite subset of A has a limit point in A, so A is compact. | In the next theorem and its proof we use the following notation. Given x = (xl9 . . , xj e 0Γ and y = (yl9 . . , yn) e mn9 write (x, y) = (xl9 . . , xm,yl9 . . , yn) e @m+n. If A c ^ m and 5 c @n9 then the Cartesian product A x B = {(a, b)e@m+n:*eA is a subset of @ and b e B} m+n .

D) (rA)B = r(AB) = A(rB). PROOF We prove (a) and (b), leaving (c) and (d) as exercises for the reader. Let the matrices A, B, C be of dimensions k x /, / x m, and m x n respectively. Then let f:&l^&k,g: Mm -► M\ /? : &n -► Mm be the linear maps such that Mf = A,Mg = B, Mh = C. Then (/o(0oA))(x) =f(goh(x)) = f(9(h)x)))=(fog)(h(x)) = ((fog)oh)(x) n for all x e M , so f° (g ° h) = (f ° g) ° h. 2 therefore implies that A(BC) = MfMgûh = Mfo(goh) = (MfMg)Mh = (AB)C, thereby verifying associativity. To prove (b), let A be an / x m matrix, and B, C m x n matrices.

Show that A is compact if and only if every sequence of points of A has a subsequence that converges to a point of A. If |b„ | > n for each n, show that the sequence {b„}T has no limit. Prove that the union or intersection of a finite number of compact sets is compact. Let {An}™ be a decreasing sequence of compact sets (that is, An + 1 <= A„ for all n). Prove that the intersection Π^°=ι ^« ^s compact and nonempty. Give an example of a decreasing sequence of closed sets whose intersection is empty.