354  5- We may cite an earlier result that every convergent sequence is 
        eventually constant, hence coverges to the integer at which it is
        constant (Z+ contains all its limi points).  Or we may show that its
        complemement is a union of open sets, hence open:
        (-infty, 1) U (!,2) U (2,3) U (3,4) U ...

    16- A point in A is in every set whic contains A, hence in the intersection
        of any such collection of sets.
        If x is not in A-, there is a neighborhood around it outside A-
        since A- is closed (A-' is open).  Therefore x is not a boundary
        point of A-, hence not in A--
        If x is in (AUB)-, it is the limit of a sequence of points in A and B;
        hence of a subsequence in A or B, hence in A- or B-.  If x is in A-
        (or B-), it is the limit of a sequence of points in A (or B), which 
        sequence is in AUB.
        If x is in (A /int B)-, it is the limit of a sequence which is in both 
        A and B, hence it is in both A- and B-; consider A=Q and B=(R\Q)
        The intersection (hence closure of the intersection) is the null set, 
        but the closure of each is R.
   
    17- Q 

359  1- {(1+(1/n), 2+(1/n))} Note that what is happening at the left endpoint 
        is what is important, I could use any number greater than 2 for the  
        right endpoint.

     4- Take any open cover of F, and add to it the open set F' (the complement
        of F).  The open cover of F with F' provides an open cover of K, hence
        there is a finite subcover of K, and removing F' leaves a finite
        subcover of F.

     9- Choose x_n in K_n for each n.  Choose a convergent subsequence (such 
        exists since K_1 is bounded, hence all K_i are bounded by the same
        bound).  Let x be the limit of the convergent subsequence, it is in
        each K_n since the tail of the subsequence is in each K_n (perhaps
        a later tail for larger n) and each K_n is closed.