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Math 440, Spring 2012, Solution to HW 1 (1) Page 83, 1. Let X be a topological space; let A ⊂ X. Suppose that for each x ∈ A there is an open set U containing x such that U ⊂ A. Show that A is open in X. Suppose x ∈ A. From the given hypothesis, we have an open set, say Ux such that x ∈ Ux ⊂ A. It is [ easy to see that A = Ux : x∈A As x0 ∈ A =⇒ x0 ∈ Ux0 ⊂ [ Ux , we have A ⊂ U. x∈A Conversely, since each Ux ⊂ A, we have [ Ux ⊂ A. x∈A Since each Ux is open and arbitrary unions of open sets are open, A is an open set. (2) Page 83, 3. Let X be a set; let Tc be the collection of all subsets U f X such that X − U is either countable or all of X. We will show that Tc is a topology on X. First of all, note that if X is a countable set, then Tc is the Discrete topology on X, as every subset has a countable complement. So this example is more interesting in the case of uncountably infinite X. (i) Since X − φ = X, by definition, φ ∈ Tc and X − X = φ, which has zero elements, so φ ∈ Tc . (ii) Finite intersections: If U1 , U2 are nonempty elements of Tc , then X − Ui is countable for each i. By de Morgan’s laws, we have X − (U1 ∩ U2 ) = (X − U1 ) ∪ (X − U2 ) and finite union of countable sets is countable. So U1 ∩ U2 ∈ Tc . If either of them is empty, then the intersection is empty, which is also in Tc . (iii) Arbitrary unions: Let Ui be elements of Tc ,for i ∈ I. If for some i0 , Ui0 is nonempty, then X − Ui0 is countable. By de Morgan’s laws, we have [ \ X − ( Ui ) = (X − Ui ) ⊂ X − Ui0 . i∈I As the last set is countable, it follows that i∈I [ Ui inTc . Otherwise, if each Ui is empty, then the union is i∈I empty as well and therefore it is in Tc . Part II. T∞ = { U | X − U is infinite or φ or X }. Is this a topology on X? Note that if X is finite, then this defines the Indiscrete topology on X. However, in general (for an infinite set X), this is not a topology. For example, let X = N, U = { 2n | n ∈ N } and V = { 2n+1 | n ∈ N }. Then U, V ∈ T∞ , but U ∪ V 6∈ T∞ , since X − (U ∪ V ) = {1}, which is neither X, φ or infinite. (3) Page 83, 8. (a) Apply Lemma 13.2 to show that countable collection B = { (a, b) | a < b, a, b ∈ Q }. 1 is a basis that generates the standard topology on R. Statement of Lemma 13.2: Let X be a topological space. Suppose that C is a collection of open sets of X such that for each open set U of X and each x in U , there is an element C of C such that x ∈ C ⊂ U. Then C is a basis for the topology of X. Note that the standard topology on R is generated by open intervals (a, b) where a, b ∈ R. So by definition of a topology generated by a basis, for any open set U and any x ∈ U , we have x ∈ (a, b) ⊂ U, for some a, b ∈ R. Now, by density of Q in R, there exist c, d ∈ Q such that a ≤ c < x and x < d ≤ b, or, in other words: x ∈ (c, d) ⊂ (a, b) ⊂ U and c, d ∈ Q. By Lemma 13.2, the above B is a basis for the standard topology on R. (b) Show that the collection C = { [a, b) | a < b, a, b ∈ Q } is a basis that generates a topology different from the lower limit topology on R. First we need to check that C satisfies the conditions to be a basis. (i) For each x ∈ R, there are rationals a, b such that a ≤ x < b.. So there is a member [a, b) of C such that x ∈ [a, b). (ii) Condition (ii) follows from the stronger statement that if two members of C have a nontrivial intersection, then the intersection is also a member of C. [a, b) ∩ [c, d) = [max{a, c}, min{b, d}). So C is a basis. Note that the lower limit topology is generated by intervals [x, y) whre x, y ∈ R. So, [π, 10) is an open set the lower limit topology, and clearly π ∈ [π, 10). However, since for no rational numbers a, b we have π ∈ [a, b) ⊂ [π, 10), by the definition of the topology generated by a basis, [π, 10) is not an open set in the topology generated by the basis C. (4) Page 83, 5. Show that if A is a basis for a topology on X, then the topology, say T , generated by A equals the intersection of all topologies that contain A. Prove the same if A is a subbasis. First of all, note that, T is a member of the collection { Tα }, where each Tα is a topology containing T A. Therefore Tα ⊂ T . (Note that this part of the argument works even if A is a subbasis.) Next, recall from Lemma 13.1 that sets in T are unions of sets in A. Since each of the Tα contains A as a subcollection and being a topology each Tα is closed under taking arbitrary unions, it follows that T every member of T is also a member of each Tα . Therefore T ⊂ Tα . T Thus, T = Tα . If A is a subbasis, from the definition of a subbasis we know that sets in T are unions of finite intersections of sets in A. Since each of the Tα contains A as a subcollection and being a topology each Tα is closed under taking finite intersections as well arbitrary unions, it follows that every member of T T is also a member of each Tα . Therefore T ⊂ Tα . 2