Solutions to Homework 4

Home , Closure (topology), Interior (topology)

1. Let X be a topological space and A ⊂ X. The closure of A, denoted A¯, is the intersection of all closed sets containing A.

(a) Show that A¯ is the smallest closed subset of X containing A, in the following sense: if A ⊂ F ⊂ X and F is closed, then A¯ ⊂ F . Solution: By deﬁnition, A¯ is the intersection of all closed sets containing A: that is, x ∈ A¯ if and only if x ∈ F for all closed sets F ⊂ X with A ⊂ F . Thus if F ⊂ X is some closed set with A ⊂ F , then for all x ∈ A¯ we have x ∈ F ; thus A¯ ⊂ F . It may be worth remarking that A¯ is a closed set containing A: it’s an intersection of closed sets, hence is closed, and it’s an intersection of sets containing A, hence contains A. (If x ∈ A then x ∈ F for all closed sets F ⊂ X with A ⊂ F , so x is in the intersection of all such F , which is A¯.)

(b) Show that if A ⊂ B ⊂ X then A¯ ⊂ B¯. Solution: We have A ⊂ B ⊂ B¯, so B¯ is a closed set containing A, so by part (a) we get A¯ ⊂ B¯.

(c) Show that A ∪ B = A¯ ∪ B¯. Solution: We have A ⊂ A ∪ B, so A¯ ⊂ A ∪ B by part (b), and similarly B¯ ⊂ A ∪ B, so A¯ ∪ B¯ ⊂ A ∪ B. For the reverse inclusion, we have A ⊂ A¯ ⊂ A¯ ∪ B¯, and B ⊂ B¯ ⊂ A¯ ∪ B¯, so A ∪ B ⊂ A¯ ∪ B¯. But A¯ ∪ B¯ is a union of two closed sets, hence is closed, so by part (a) we get A ∪ B ⊂ A¯ ∪ B¯.

1 (d) Show that A ∩ B ⊂ A¯ ∩ B¯. Give an example where the inclusion is strict. Solution: We have A ∩ B ⊂ A ⊂ A¯, and A ∩ B ⊂ B ⊂ B¯, so A ∩ B ⊂ A¯ ∩ B¯. The latter is an intersection of closed sets, hence is closed, so by part (a) we get A ∩ B ⊂ A¯ ∩ B¯. For the counterexample, let X = R with the usual topology, let A = (0, 1), and let B = (1, 2). Then A ∩ B = ∅, which is closed, so A ∩ B = ∅. On the other hand, A¯ = [0, 1], and B¯ = [1, 2], so A¯ ∩ B¯ = {1}.

10. Let X be a topological space and A ⊂ X. The interior of A, denoted int A, or sometimes A◦, is the union of all open sets contained in A.

(a) Show that int A is the biggest open subset of A, in the following sense: if U ⊂ A and U is open, then U ⊂ int A. Solution: By deﬁnition, int A is the union of all open sets contained A: that is, x ∈ int A if and only if x ∈ U for some open set U ⊂ A. Thus if U ⊂ X is some open set with U ⊂ A, then for all x ∈ U we have x ∈ int A; thus U ⊂ int A. It may be worth remarking that int A is an open set contained in A: it’s a union of open sets, hence is open, and it’s a union of sets contained in A, hence is contained in A. (If x ∈ int A then x ∈ U for some open set U ⊂ A, so x ∈ A.) (b) Show that if A ⊂ B ⊂ X then int A ⊂ int B. Solution: We have int A ⊂ A ⊂ B, so int A is an open set contained in B, so by part (a) we get int A ⊂ int B. (c) Show that int A ∩ int B = int(A ∩ B). Solution: We have A∩B ⊂ A, so int(A∩B) ⊂ int A by part (b), and similarly int(A ∩ B) ⊂ int B; thus int(A ∩ B) ⊂ int A ∩ int B.

For the reverse inclusion, we have int A ∩ int B ⊂ int A ⊂ A, and int A ∩ int B ⊂ int B ⊂ B, so int A ∩ int B ⊂ A ∩ B. But int A ∩ int B is an intersection of two open sets, hence is open, so by part (a) we get int A ∩ int B ⊂ int(A ∩ B).

2 (d) Show that int A ∪ int B ⊂ int(A ∪ B). Give an example where the inclusion is strict. Solution: We have int A ⊂ A ⊂ A ∪ B, and int B ⊂ B ⊂ A ∪ B, so int A∪int B ⊂ A∪B. Now int A∪int B is a union of open sets, hence is open, so by part (a) we get int A ∪ int B ⊂ int(A ∪ B). For the counterexample, let X = R with the usual topology, let A = (0, 1], and let B = [1, 2). Then int A = (0, 1), and int B = (1, 2), so int A∪int B = (0, 1)∪(1, 2). On the other hand, A∪B = (0, 2) is open, so int(A ∪ B) = (0, 2).

2. Let X be a topological space and A ⊂ X.

(a) Show that X \ A¯ = int(X \ A), and X \ int A = X \ A. Solution: It may be clearer to use the notation Ac := X \ A for the complement. We ﬁrst remark that for any subsets B,C ⊂ X, we have B ⊂ C if and only if Cc ⊂ Bc; and (Bc)c = B. You can prove these if you want. Next, we have A ⊂ A¯, so (A¯)c ⊂ Ac; because A¯ is closed, (A¯)c is open, so (A¯)c ⊂ int(Ac) by 10(a). Replacing A with Ac, we get (Ac)c ⊂ int((Ac)c) = int A, so (int A)c ⊂ Ac. Thus we’ve proved one inclusion for each of the two desired equalities. For the reverse inclusions, we have int A ⊂ A, so Ac ⊂ (int A)c; because int A is open, (int A)c is closed, so Ac ⊂ (int A)c. Replacing A with Ac, we get A¯ ⊂ (int(Ac))c, so int(Ac) ⊂ (A¯)c by 1(a). Thus both equalities are proved.

4. Let X be a set, and let T be the set of subsets U ⊂ X that such that X \ U is ﬁnite, together with the empty set.

(a) Show that T is a topology. (It is called the “ﬁnite complement topology.”)

Solution: We have ∅ ∈ T by deﬁnition. To see that X ∈ T , note that X \ X = ∅ is ﬁnite.

Suppose that U1,...,Un ∈ T . If some Ui = ∅ then U1 ∩· · ·∩Un = ∅ ∈ T . Otherwise X \ Ui is finite for all i, so X \ (U1 ∩ · · · ∩ Un) = (X \ U1) ∪ · · · ∪ (X \ Un) is a finite union of finite sets, hence is finite, so U1 ∩ · · · ∩ Un ∈ T . Given an arbitrary subset S ⊂ T , we want to show that S S ∈ T . S If S = ∅ or S = {∅} then S = ∅ ∈ T . Otherwise there is some non-empty U ∈ S, so X \ U is finite. Then U ⊂ S S, so X \ T S ⊂ X ∩ U is a subset of a finite set, hence is finite, so S S ∈ T . (Or it might be nicer to show that the closed sets do what they should.)

(b) Find the closure, interior, and boundary of Z as a subset of R in the finite complement topology. Solution: A set is open if and only if its complement is finite, or it is empty. Thus a set is closed if and only if it is finite, or is the whole space R. Now Z is infinite, so the only closed set containing Z is R, so Z¯ = R. Similarly, R \ Z is infinite, so R \ Z = R, so int Z = ∅ by 3(a). Finally, ∂Z = Z¯ \ int Z = R \ ∅ = R.

5 5. Let T be the set of subsets U ⊂ R such that U contains 0, together with the empty set.

(a) Show that T is a topology.

Solution: We have ∅ ∈ T by deﬁnition, and R ∈ T because 0 ∈ R..

Suppose that U1,...,Un ∈ T . If some Ui = ∅ then U1 ∩· · ·∩Un = ∅ ∈ T . Otherwise 0 ∈ Ui for all i, so 0 ∈ U1 ∩ · · · ∩ Un, so U1 ∩ · · · ∩ Un ∈ T . (Indeed, we can see that T is closed under arbitrary intersections, not just ﬁnite intersections.) Given an arbitrary subset S ⊂ T , we want to show that S S ∈ T . S If S = ∅ or S = {∅} then S = ∅ ∈ T . Otherwise there is some non-empty U ∈ S, so 0 ∈ U, so 0 ∈ S S, so S S ∈ T .

(b) Find the closure, interior, and boundary of the one-point subsets {1} and {0}. Solution: A set is open if and only if it either contains 0, or is empty. Thus a set is closed if and only if it either does not contain 0, or is the whole space R. Thus {1} is closed, and it contains no non-empty open set, so its interior is ∅, its closure is {1}, and its boundary is {1}, just as in the usual topology. On the other hand, {0} is open, and it is not contained in any closed set apart from the whole space R, so its interior is {0}, its closure is R, and its boundary is (−∞, 0) ∪ (0, ∞), quite unlike the usual topology.

6 6. Let T be the subsets of R of the form (a, ∞) for some a ∈ R, together with the empty set and the whole set R. (a) Show that T is a topology. (It is called the “lower semi-continuous topology” and we discussed it in lecture on Friday.)

Solution: We have ∅ ∈ T and R ∈ T by deﬁnition. For ﬁnite intersections, by induction it is enough to prove it for two: if U, V ∈ T then U ∩ V ∈ T . Observe moreover that for any two subsets we either have U ⊂ V , in which case U ∩ V = U ∈ T , or V ⊂ U, in which case U ∩ V = V ∈ T . Given an arbitrary subset S ⊂ T , we want to show that S S ∈ T . S If R ∈ S then S = R ∈ T . If not, write S = {(a, ∞): a ∈ A}

or S = {(a, ∞): a ∈ A} ∪ {∅} for some set A of real numbers. If A is non-empty and bounded below then S S = (inf A, ∞) ∈ T . If A is non-empty and unbounded S S below then S = R ∈ T . If A is empty then S = ∅ ∈ T . (b) Find the closure, interior, and boundary of the interval (0, 1) as a subset of R in this topology. Solution: The closed sets are of the form (−∞, a] for a ∈ R, together with the whole set R and the empty set. A closed set containing (0, 1) is either all of R, or of the form (−∞, a] for some a ≥ 1. The closure of (0, 1) is the intersection of all these, which is (−∞, 1]. The only open set contained in (0, 1) is ∅, so its interior is ∅. Thus its boundary is (−∞, 1].