Topology Homework 3 Section 3.1 - Section 3.3 Samuel Otten 3.1 (1) Proposition. The intersection of finitely many open sets is open and the union of finitely many closed sets is closed. Proof. Note that S 1 S 2 S 3 S n = (((S 1 S 2 ) S 3 ) S n ) for any family of sets {S i }, i N, and any natural number n. Thus, for an intersection of finitely many open sets we can take the intersection pairwise and, by Definition 1(ii), each set along the way is open and the end result is open. Similarly, S 1 S 2 S 3 S n = (((S 1 S 2 ) S 3 ) S n ). Thus, for a union of finitely many closed sets we can take the union pairwise and, by Proposition 1(b), each set along the way is 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 topology is not induced by any metric. Proof. We shall employ Proposition 2 of the text to demonstrate that (N, F) is a topological space. First, by definition, F = N is closed and φ is closed because it contains a finite number of positive integers, namely zero. Second, let F 1 and F 2 be elements of F. If at least one of F 1 and F 2 is equal to all of N, then F 1 F 2 = N which is closed. If this is not the case, then both contain a finite number of positive integers. Thus, their union contains a finite number of positive integers and is closed by definition. Third, we consider the intersection of a family of elements of F. The intersection 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 defines a topology on N. Suppose a metric D on N did induce (N, F). Since (N, D) is a metric space, it must satisfy Proposition 12 of Chapter 2, which says that for any closed subset of the space and any point in its complement, there exist disjoint open subsets such that the closed subset is contained in one and the point is contained in the other. Consider F = {1, 2, 3} and x = 4. It is obviously the case that F is closed and x N F, so Proposition 12 must hold. Let U be an open set containing F and let V be an open set containing x such that U V = φ. Since U and V are disjoint, both cannot contain infinitely many elements. Therefore, at least one of them must contain finitely many elements, which leads us to two cases. Suppose U is finite. Then U C = N U has infinitely many elements. Moreover, U C N because it lacks the elements 1, 2, and 3. Hence, U C is 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 V C is infinite but lacks the element 3. So, in this case, V C is not closed and we also have a contradiction. This implies that this topology is not induced by any metric. 1
(7) Proposition. Define a family F of subsets of R 2 as follows: Let F F if and only if F = R 2. F = φ, or F is a set consisting of finitely many points together with the union of finitely many straight lines. Then the topology determined by F is the coarsest in which lines and points are closed sets. Proof. Suppose (R 2, F ) is a topological space such that F is strictly coarser than F. This means F F, so there must exist some F F such that F / F. By definition, F consists of finitely many points and lines. But since F is not an element of F, the points and lines from F are not closed sets in F. If they were individually closed, then their unions taken pairwise would have to be in F by Proposition 1(b). But this union is precisely F, which we just said was not in F. 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 the union of any two closed sets is closed. Since a one-point subset of the real plane corresponds to an intersection of lines, and lines are closed, the singleton set would also have to be closed. It is possible, however, to have a topology on the real plane in which every one-point set was closed but every line was not. This also follows from Proposition 1(b) which implies that the union of uncountably 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 that B 1 = {N(x, q) q Q, x X} and B 2 = {N(x, 1/n) x X and n a positive integer} are both bases for the topology induced on X = R 2 by the Pythagorean metric. We will verify this claim using our in-class definition of a basis. For B 1, let x be any element in X. It is clear that x N(x, q) for any rational q > 0 because a point is always in its own neighborhood. Choose x B 1 B 2 where B 1, B 2 B 1. This means that x is in two overlapping neighborhoods with rational radii. Set p equal to the lesser of the Pythagorean distances from x to the frontiers of the neighborhoods. If p is rational, then we have found a B 3 B 1 such that x B 3. If p is irrational, then we set q equal to a rational number that is less than p, which exists because the rationals are dense. Thus, B 1 is a basis for the real plane with the Pythagorean metric. For B 2, let x be any element in X. Again, x is guaranteed to be in its own neighborhood for any radius 1/n. Choose x B 1 B 2 where B 1, B 2 B 2. As above, set p equal to the lesser of the Pythagorean distances from x to the frontiers of the neighborhoods surrounding it. If p = 1/n for some positive integer n then we are done. If not, then we choose n so that 1/n < p, which exists because the sequence 1/n converges to zero. Thus, x B 3 B 2 and B 2 is also a basis. (6) We will list all the other open sets in the smallest topologies on N for which the following are collections 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. 2
3.3 (4) Proposition. For each x R 2, let N x be the set of interiors of all triangles which contain x in the interior. Then the collection N x forms a neighborhood system for a topology on R 2. Moreover, this topology is the same as the topology induced by the Pythagorean metric. Proof. We will use our in-class definition to show that this is a neighborhood system. (i) There are obviously triangle interiors that contain x for any point x R 2. Thus, N x φ. (ii) By definition, x is in the interior of the triangle and so x N for each N N x. (iii) Let N 1 and N 2 be two sets from N x. These sets both necessarily contain x. Since the interior of triangles (excluding the boundary) are open sets in the real plane, and the intersection of N 1 and N 2 is not empty, we know that infinitely many points belong to N 1 N 2. Choosing any three non-collinear points from this set will result in N 3 N x. (iv) Let N be any set from N x and let y be any point from N. Consider the points that lie halfway between y and each of the three vertices of the triangle at the frontier of N. These three points are non-collinear, and if we let N be the interior of the triangle formed by the three points then N contains y. Thus, N N y and N N. It is clear to see that this open neighborhood system determines the same topology as that induced by the Pythagorean metric because circles (D-neighborhoods) can be circumscribed about any triangular interior and a triangle can be described about any circle. (6) Proposition. Let (X, D) be a metric space and {a n } a sequence of positive real numbers which converges to 0. For each x X, set N x = {N(x, a n ) n N}. Then the collection of N x thus formed constitutes an open neighborhood system for the topology induced by D. Proof. (i) Since {a n } comprises positive real numbers, let p = a 1 > 0. Then N(x, p) N x for each x X, and N x φ. (ii) Since D(x, x) = 0 and a n > 0 for each n, we know that x N(x, a n ) for any n. Thus, x N for any N N x. (iii) Suppose N 1 and N 2 are in N x. Then N 1 = N(x, a j ) and N 2 = N(x, a k ) for some j, k N. Without loss of generality, suppose a j a k. Then N 1 N 2 = N(x, a k ). Since a n 0, we can find an a m that is arbitrarily close to 0. In this case, we choose a m < a k and define N 3 = N(x, a m ). Hence, N 3 N x and N 3 N 1 N 2. (iv) Let N = N(x, a p ) be a set from N x and let y be an element in N. If p = a p D(x, y) (note that 0 < p a p ), we know by the convergence of {a n } that there exists an a q such that a q < p. Therefore, N(y, a q ) N y is contained in N(x, a p ). (v) A subset U of X is clearly open if and only if for each x U there is N N x such that N U. This follows because open sets in the topology induced by D are defined in terms of neighborhoods, and it is neighborhoods that make up N x. Also, since a n 0, we will always be able to find N U for any x U. Therefore, N x satisfies the textbook definition of an open neighborhood system. (7) CLAIM: If N x is a neighborhood system of a set X with the property that each member of each N x is finite, then the topology τ on X determined by N x is the discrete topology. This claim is false. We shall consider the following counterexample. Let X = {1, 2, 3} and let N x have only one element for x = 1, 2, 3; specifically, N 1 = {{1, 2}}, N 2 = {{2}}, and N 3 = {{2, 3}}. Clearly, N x is never the empty set and x N for each N x (there is only one). It is true that N 3 N x always exists such that N 3 N 1 N 2 because, in fact, N 1 = N 2 = N 3 for each N x. Moreover, the only element y of an N N x that is distinct from x is 2 (in both N 1 and N 3 ). But N N y is the set containing 2, which is obviously a subset of N in either case. Therefore, N x is a neighborhood system by definition. Also, each member is finite. However, N x does not determine the discrete topology because {1}, {3}, and {1, 3} are subsets of X that are not the union or intersection of sets within N x. 3
3.G (1) Let B be a basis for a topology τ on X. a) We will show that τ is the intersection of all topologies containing B. Let T τ. Then, by the definition of a basis (p. 44), T is the union of members of B. Also, by part (iii) of the definition of a topology (p. 40), any topology containing B must also contain the union of any family of its members. This means T is in any topology containing B and T is in the intersection of all topologies containing B. Thus, τ is a subset of that intersection. Conversely, let T be any set from the intersection of all topologies containing B. Since τ is a topology containing B, T τ. Thus, the intersection of all topologies containing B is a subset of τ. Therefore, we can conclude that they are, in fact, equal. b) The conclusion from part (a) also holds when B is a subbasis. The argument is the same, except now we use the definition of subbasis (p. 46) which includes the intersection of finitely many members, and part (ii) of the definition of a basis which also deals with the intersection. (2) For the set of integers Z, let B be the set of all bi-infinite arithmetic sequences. Note that an arithmetic sequence is of the form an + b for n N and some fixed a, b Z. a) We will show that B is a basis for a topology τ on Z. First, observe that {..., 2, 1, 0, 1, 2,... } is an element of B. Thus, for any x Z there is a B B such that x B. Second, let B 1 and B 2 be arbitrary sets from B and let x B 1 B 2. We know that B 1 and B 2 contain elements of the form an + b and cn + d, respectively. Since x belongs to the intersection, x = an 1 + b and x = cn 2 + d for some n 1, n 2 N. But x + ac = a(n 1 + c) + b = c(n 2 + a) + d, so x + ac B 1 B 2. By adding x and all the multiples of ac, we will have generated B 3 B such that B 3 B 1 B 2. Therefore, B is a basis by definition. b) Let us consider some simple closed sets in this topology. By definition, the complement of a closed set is open; in particular, the complement of a closed set is a bi-infinite arithmetic sequence. So {..., 3, 1, 1, 3,... } is closed because its complement is the bi-infinite arithmetic sequence {..., 2, 0, 2,... }. (Note that both sets just mentioned are simultaneously open and closed.) Similarly, {..., 5, 4, 2, 1, 1, 2, 4, 5,... } is closed because its complement is the set of multiples of 3. Furthermore, there are more complicated closed sets whose complements are some union of bi-infinite arithmetic sequences, as we shall see below. c) Proposition. There are infinitely many prime numbers. Proof. Let τ be the topology from part (a). Define m to be the set of multiples of m N. Thus, 2 = {..., 4, 2, 0, 2, 4,... } 3 = {..., 6, 3, 0, 3, 6,... } 5 = {..., 10, 5, 0, 5, 10,... }. p = {..., 2p, p, 0, p, 2p,... }. The sets p for p a prime number are open because they are bi-infinite arithmetic sequences. However, it is important to note that the sets p are also closed. Consider the complement of 2, which is {..., 3, 1, 1, 3,... }. This is the set of all numbers congruent to 1(mod 2). Similarly, the complement of 3 is {..., 2, 1, 1, 2,... }, which is the union of the set of all numbers congruent to 1(mod 3) and the set of all numbers congruent to 2(mod 3). Since both of these complements are open, 2 and 3 themselves are closed. This reasoning can be generalized to p. The complement of p is the union of open sets; namely, the sets containing all numbers congruent to 1(mod p), 2(mod p),..., p 1(mod p). Since this union is open, p is closed. 4
Let us observe one more fact before moving on to the heart of the argument. By definition, a prime number is not a multiple of any natural number preceding it. That is, p / p for any p preceding p. Moreover, if q is a composite number, then q p for some p preceding q. Now, suppose to the contrary that the number of prime numbers is finite. Let p be the last prime. Then 2 3 p is the union of finitely many closed sets. We proved in Problem (1) of Section 3.1 that such a union is closed. This means that the complement of this union is open. Note that for any natural number q which is composite, q p for some prime p and, therefore, q is in the union. Certainly all primes are in the union. Since all integers are either prime or composite (other than 1 and -1), it follows that 2 3 p = X { 1, 1}. From this, we see that { 1, 1} is the complement of 2 3 p. But we said the complement has to be open, and { 1, 1} is not open because it is not bi-infinite. This is a contradiction. We conclude, then, that the number of prime numbers is not finite but is indeed infinite. 5