Solution 4

1. In a discrete metric space (where d(x, y) = 1 if x y) a ¹/₂ -neighbourhood of a point p is the singleton set {p}. Thus since every singleton is open and any subset A is the union of all the singleton sets of points in A we get the result that every subset is open.
   Since all the complements are open too, every set is also closed.
   
   Since all inverse images are open, every function from a discrete space is continuous.
   
   If f: R X is continuous and X is discrete, this means that the inverse image of every singleton set is open. But then we cannot have any interval (a, b) which is mapped to more than one point. Hence the only continuous functions are constant ones.
   

2. The map x x² maps (-1, 1) to [0, 1) and so open sets are not mapped to open sets necessarily.
   
   For any continuous map f: X Y between metric (or topological) spaces we have f ^-1(Y - B) = X - f ^-1(B). So if B is open in Y, Y - B is closed and its inverse image is the complement of the open set f ^-1(B) and hence is closed.
   

3. Suppose x V(p). Then d(x, p) < and if we take - d(x, p) = a -neighbourhood of x is in V(p).
   If A X is open then A = V_(a)(a) where the union is over all a A and (a) is chosen so that V_(a)(a) is inside A. It is easy to verify the double inclusion to prove equality.
   
   Not every open set can be written as a union of countably many -neighbourhoods. For example, take R with the discrete topology. Then any -neighbourhood is either R or a singleton set and so no proper uncountable set can be written as a countable union of these.
   However, R with its usual metric does have this property.

4. { (x, y) R² | f(x, y) > 0 } = f ^-1((0, )) and so is an open set.
   
   For the unit disc take f(x, y) = 1 - x² - y²
   For the unit square: a function like x(1 - x) gives a vertical strip as an open set. Take the intersection of this with a similar horizontal strip.
   f ^-1([0, )) is a closed set by Question 2 above.

5. If were limit points of {a_i} then there would have to be subsequences converging to and to . But every subsequence of a convergent sequence converges to the same limit.

   Take a union of two convergent sequences.

   Take a union of countably many sequences converging to limits a finite distance apart. e.g. {m + 1/n | m, n N}

6. (i) and (iii) are not topologies, (ii) and (iv) are.
    
   

7. (q₁ , ) (q₂ , ) = (min{q₁ , q₂}, ) and (q₁ , ) (q₂ , ) = (max{q₁ , q₂}, ) and so is closed under finite unions and intersections.
   Take a monotonic decreasing sequence of rationals (r_i) converging to an irrational . Then (r_i, ) = (, ) and so is not in and is not a topology.

8. As in the last question, is closed under finite unions and intersections. This time, however, {n_i , n_i+1 , ... } is in the set since the minimum min{n_i | i N} exists since N is bounded below.
   
   
   The closed sets are the finite sets {1, 2, 3, ... , n} for each n N
   
   
   

9. It is easiest to check the corresponding topological properties for the closed sets which are the countable ones.
   The union of two countable sets is countable and since the intersection of (arbitrarily many!) countable sets is a subset of any one of them, it too is countable.