p.363 1- a) if a,b .gt. (or =) 0, the inverse image of (a, b) is the union of (-b^{-.5}, -a^{-.5}) and (a^{-.5}, b^{-.5}) if a, b .lt. (or =) 0, there is no preimage. if a .lt. 0 , b .gt. 0 the preimage is (-b^{.5}, b^{.5}) b) If a .lt. 0, b .gt. 0; |a| .lt. b, (a, b) maps to [0, b^2) 3- Since SQRT .gt.or = 0, we are interested in the preimage of [0, epsilon). This is [1, 1+epsilon^2), which is the intersection of (e.g.) (0, 1+epsilon^2) with with I. 7- R\{k} is open, hence its preimage is open, hence the preimage of {k} is closed. Or, consider a convergent sequence in the preimage of {k}, then by continuity the value of f at its limit point is k by continuity, hence the limit point is in the preimage, and the preimage is closed. p.370 1- condition (a) is satisfied because absolute values are taken. condition (b) is satisfied because | | is zero only if its argument is 0, and both [components] must be 0 for d1 and d_infty condition (c) is satisfied because | | is symmetric. condition (d) is satisfied for d1 because it is satisfied componentwise. For d2 it is satisfied because it is satisfied for the component which provides the maximum on the LHS, and if the other component is larger on the RHS in either rem, it will strengthen the inequality. 6- Let epsilon = .5, then |x_n - x| .lt. epsilon only if x_n = x, hence for convergence there must be a K s.t. x_n = x for n .gt. K 8- a) interior of a square with corners at (0,0), (1,0), (-1.0), (0,-1) b) interior of a square with corners (1,1), (1,-1), (-1,1), (-1,-1)

Sun Feb 22 1998