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By S. M. Srivastava (auth.)

ISBN-10: 3642854737

ISBN-13: 9783642854736

ISBN-10: 3642854753

ISBN-13: 9783642854750

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So in this case, v(a, (-yn»(k) :I: v(a', (-y~»(k). Now assume that for some i, "Yi :I: "Y~. Choose j such that "Yi(j) :I: --I;(j). Let k = u(i,j). So I(k) = i and r(k) = j. Then, as u is one-to-one, v(a, (-yn»(k) = u(a(k), "Yi(j» '" u(a'(k), "Y;(j» = v(a', (1n»(k). Thus v is one-to-one in this case too. We now show that v is onto. Towards this, let /3 E NN. Define a ~ NN by . a(k) = 1(/3(k», kEN. For any n, define "Yn by "Yn(m) = r(/3(u(n, m))), mEN. Fix kEN. We have v(a, ("Yn»(k) = = = = This shows that v(a, ("Yn» u(a(k), "YI(k) (r(k))) u(I(/3(k», r(/3(u(l(k), r(k»))) u(I(/3(k», r(/3(k))) /3(k).

Symbolically, we -:an express this as follows. (Po & 'v'n(Pn ~ Pn+d) ~ 'v'nPn . The proof of this proposition uses two basic properties of the set of natural numbers. First, it is well-ordered by the usual order, and second, every nonzero element in it is a successor. 1 gives us the following. 2 (Definition by induction) Let X be any nonempty set. Suppose Xo is a fixed point of X and 9 : X - - X any map. Then there is a unique rna, I : N - - X such that 1(0) = Xo and I(n + 1) = g(f(n)) for all n.

Thus a subspace of a complete metric space need not be complete. However, a closed subspace of a complete metric space is easily seen to be complete. For A ~ X we define diameter(A) = sup{d(x,y): x,y E A}. 28 Let (X, d) be a metric space. Show that for any A ~ X, diameter(A) = diameter(cl(A». 29 (Cantor intersection theorem) A metric space (X,d) is complete i/ and only if/or every decreasing sequence Fo ;2 Fl ;2 F2 ~ ... 0/ nonempty closed subsets 0/ X with diameter(Fn) - 0, the intersection Fn is a singleton.

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A Course on Borel Sets by S. M. Srivastava (auth.)

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