p.59 - 5) If x gt 0, there exists 1/n lt x (corr 2.5.3), hence x not in [0, 1/n] for that n, hence not in the intersection. (It is easily shown that 0 is in the intersection, but no negative numbers are.) p.59 - 8) eta is the inf of the b_n's, hence is less than or equal to each b_n. If eta were lt some a_n*, then that a_n* would be a lb for the b_n's since a_n* leq a_n leq b_n for n geq n*, and b_n* leq b_n for n lt n*. This contradicts that eta is the glb (inf). so eta is geq each a_n hence in each interval, hence the intersection. Since xi and eta are in the interval and nothing lt xi or gt eta is in each interval, (by def of inf and sup) they are the endpoints of the intersection, and the intersection is closed. p.66 -6) f(x) = 2x-1 p.66 - 12) Let S be an infinite set, and A a denumerable (countably infinite) subset of S. Let f:A to Z+ be a bijection from A to the positive integers, which must exist by definiton of denumerable. Then let F be defined by F = identity {x in S\A F = g where g(x) = f^(-1)(f(x)+1) x in A F is well defined since f and f^(-1) are. f^(-1)(1) is not in the image of F, hence F is a bijection onto a subset of S. (An argument can be added that since f is 1-1, f^(-1) is 1-1, hence its composition g is 1-1; the identity map is 1-1 and since S and S\A are disjoint the union of id and g is 1-1. a 1-1 funtion is a bijection onto its image (range). p. 16 - 7) This problem is just to shrink and shift the interval. A linear map will work, so just get the enpoints to match up. f(x) = (x-a)/(b-a). (it is readily verified that the enpoints map appropriately).

Sun Sep 21 1997