Topology Homework 3Section 3.1 - Section 3.3Samuel Otten3.1(1)Proposition. The intersection of finitely many open sets is open and the union of finitely manyclosed sets is closed.Proof. Note that S1∩ S2∩ S3∩ · · · ∩ Sn= (((S1∩ S2) ∩ S3) · · · ∩ Sn) for any family of sets {Si},i ∈ N, and any natural number n. Thus, for an intersection of finitely many open sets we can takethe intersection pairwise and, by Definition 1(ii), each set along the way is open and the end resultis open.Similarly, S1∪ S2∪ S3∪ · · · ∪ Sn= (((S1∪ S2) ∪ S3) · · · ∪ Sn). Thus, for a union of finitelymany closed sets we can take the union pairwise and, by Proposition 1(b), each set along the wayis closed and the end result is closed. ¤(6)Proposition. Let F ⊂ N be closed if F contains a finite number of positive integers or F = N.Then the set F of all closed subsets of N thus defined forms a topology on N. Moreover, this topologyis not induced by any metric.Proof. We shall employ Proposition 2 of the text to demonstrate that (N, F) is a topologicalspace. First, by definition, F = N is closed and φ is closed b ecause it contains a finite number ofpositive integers, namely zero. Second, let F1and F2be elements of F. If at least one of F1andF2is equal to all of N, then F1∪ F2= N which is closed. If this is not the case, then both containa finite number of positive integers. Thus, their union contains a finite number of positive integersand is closed by definition. Third, we consider the intersection of a family of elements of F. Theintersection of sets containing finitely many elements must itself contain finitely many elements.The worst that the intersection could do is result in N itself, but this is closed. Therefore, F definesa topology on N.Suppose a metric D on N did induce (N, F). Since (N, D) is a metric space, it must satisfyProposition 12 of Chapter 2, which says that for any closed subset of the space and any point inits complement, there exist disjoint open subsets such that the closed subset is contained in oneand the point is contained in the other. Consider F = {1, 2, 3} and x = 4. It is obviously the casethat F is closed and x ∈ N − F, so Proposition 12 must hold. Let U be an open set containingF and let V be an open set containing x such that U ∩ V = φ. Since U and V are disjoint, bothcannot contain infinitely many elements. Therefore, at least one of them must contain finitely manyelements, which leads us to two cases. Suppose U is finite. Then UC= N − U has infinitely manyelements. Moreover, UC6= N because it lacks the elements 1, 2, and 3. Hence, UCis not closed.(Note that an open set in this topology is defined as the complement to some closed set from F.)Therefore, U is not open, a contradiction. Similarly, if V is finite, then VCis infinite but lacks theelement 3. So, in this case,VCis not closed and we also have a contradiction. This implies thatthis topology is not induced by any metric. ¤1(7)Proposition. Define a family F of subsets of R2as follows: Let F ∈ F if and only if F = R2.F = φ, or F is a set consisting of finitely many points together with the union of finitely manystraight lines. Then the topology determined by F is the coarsest in which lines and points areclosed sets.Proof. Suppose (R2, F0) is a topological space such that F0is strictly coarser than F. This meansF0⊂ F, so there must exist some F∗∈ F such that F∗/∈ F0. By definition, F∗consists of finitelymany points and lines. But since F∗is not an element of F0, the points and lines from F∗are notclosed sets in F0. If they were individually closed, then their unions taken pairwise would have tobe in F0by Proposition 1(b). But this union is precisely F∗, which we just said was not in F0.Therefore, any topology that is coarser than F does not have all lines and points as closed sets. ¤It is not possible to have a topology on the real plane in which every line was a closed set,but every one-point subset was not. This fact follows from Proposition 1(b) which states that theunion of any two closed sets is closed. Since a one-point subset of the real plane corresponds toan intersection of lines, and lines are closed, the singleton set would also have to be closed. It ispossible, however, to have a topology on the real plane in which every one-point set was closedbut every line was not. This also follows from Proposition 1(b) which implies that the union ofuncountably many closed sets may not be closed, and a line is the union of uncountably many(closed) one-point sets.3.2(5) Example 6 of the text made the claim thatB1= {N(x, q) | q ∈ Q, x ∈ X} andB2= {N(x, 1/n) | x ∈ X and n a positive integer}are both bases for the topology induced on X = R2by the Pythagorean metric. We will verify thisclaim using our in-class definition of a basis.For B1, let x be any element in X. It is clear that x ∈ N(x, q) for any rational q > 0 becausea point is always in its own neighborhood. Choose x ∈ B1∩ B2where B1, B2∈ B1. This meansthat x is in two overlapping neighborhoo ds with rational radii. Set p equal to the lesser of thePythagorean distances from x to the frontiers of the neighborhoods. If p is rational, then we havefound a B3∈ B1such that x ∈ B3. If p is irrational, then we set q equal to a rational number thatis less than p, which exists because the rationals are dense. Thus, B1is a basis for the real planewith the Pythagorean metric.For B2, let x be any element in X. Again, x is guaranteed to be in its own neighborhood forany radius 1/n. Choose x ∈ B1∩ B2where B1, B2∈ B2. As above, set p equal to the lesser of thePythagorean distances from x to the frontiers of the neighborhoods surrounding it. If p = 1/n forsome positive integer n then we are done. If not, then we choose n so that 1/n < p, which existsbecause the sequence 1/n converges to zero. Thus, x ∈ B3∈ B2and B2is also a basis.(6) We will list all the other open sets in the smallest topologies on N for which the following arecollections of open sets.• N, φ are the only open sets. (trivial topology)• N, {1, 2}, {3, 4, 5} and {1, 2, 3, 4, 5}, φ.• N, {1, 2}, {3, 4, 5}, {1, 4, 7} and {1, 2, 3, 4, 5, 7}, {1, 2 , 3, 4 , 5}, {3, 4, 5, 7}, {1, 2, 4, 7}, {1}, {4}, φ.23.3(4)Proposition. For each x ∈ R2, let Nxbe the set of interiors of all triangles which contain x inthe interior. Then the collection Nxforms a neighborhood system for a topology on R2. Moreover,this topology is