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21 Oct 2009: 5) Prove the statement (b) of problem (4) above by completing the following. ... 6) Let f : B B be a continuous map from the open disc B = {(x, y) R2 : x2 y2 < 1} into it-self.

21 Jan 2009: B[p] := {bp : b B}. in B provided that B is reduced.Let t be a positive integer. ... Hence ι is surjective. Because b. p B1 for all b B, we see that ι is injective.In fact, it is easy to see that the inverse of ι is

9 Oct 2009: and B = (b : T B B), the function space [A, B] is given by T -algebra structure on the. ... r(x) = {b B | a M x. arb }. It is immediate that r preserves directed sups, that is, it is Scott continuous.

16 Oct 2009: of the form6 Z =. [b ca b. ]. Let us consider the map given by. ... 0X1t Et Xt dt. Write A :=[b ca b. ]. We have to prove that.

4 Aug 2009: Using this axiom and the laws of algebra (e.g. a b = b a), we can nowrefound the calculus. ... n1. bn1 g(b). Then G(0) = G′(0) = = G(n)(0) = 0 and.

13 Aug 2009: Theorem 12.5. Let B be a set of first category in T. ... iii) If A and B are of firs category so is A B.

12 Aug 2009: a(b1c22 b. 23c1) b(a2c. 21 a. 23c2) c(a. 22b3 a3b. 21). ... The only remaining possibility is (a,b) = (0, 0). 2. Example 4.5.