M55: Exercise sheet 3

1. Let A be a closed subset of a topological space X and F ⊂ A. Show that F is closed in A if and only if F is closed in X.

2. Let B ⊂ A ⊂ X. Let C be the closure of B in X and C_{A} the closure of B in A. Show that C_{A} = A ∩ C.

3. (Another patching type theorem) Let A, B ⊂ X be closed subsets of a topological space X such that X = A ∪ B. Equip both A and B with the subspace topology. Let f : X → Y be a map into another topological space Y. Show that f is continuous if and only if both f_{|A} : A → Y and f_{|B} : B → Y are continuous (with respect to the subspace topologies). Use this to prove that f : [−1, 1] → ℝ defined by

f(x) = \begin{cases}−x & if x ∈ [−1, 0]; \\ x & if x ∈ [0, 1]; \end{cases}

is continuous.

4. Let X be a topological space with base ℬ and A ⊂ X. Let ℬ_{A} = {A ∩ B : B ∈ ℬ}. (a) Show that ℬ_{A} is a base for a topology on A.

(b) What topology on A is it?

5. (The Moore plane) Let Γ denote the closed upper half plane {(x, y) ∈ ℝ^{2} : y ≥ 0}. Let ℬ denote all ε-balls lying in the region {(x, y) : y > 0} along with all sets of the form {z} ∪ A with z a point on the x-axis and A an open ball in the upper half plane tangent to the x-axis at z.

(a) Show that ℬ is the base for a topology on Γ.

(b) What is the induced topology on the set {(x, y) : y > 0}?

(c) What is the induced topology on the half-line {(0, y) : y ≥ 0}?

(d) What is the induced topology on the x-axis?

March 7, 2003

Home page: http://www.maths.bath.ac.uk/~feb/math0055.html

M55: Exercise sheet 3—Solutions

1. If F is closed in A then F = A ∩ G with G closed in X. Thus F is the intersection of two closed subsets of X and so is itself a closed subset of X. Conversely if F is closed in X and F ⊂ A, then F = A ∩ F is closed in A (note that closedness of A is not needed for this bit).

2. First note that A ∩ C is closed in A and contains B whence C_{A} ⊂ A ∩ C. For the converse, we use a characterisation of C_{A} given in lectures. Let x ∈ A ∩ C and x ∈ U with U open in A. We need to show that U ∩ B ≠ ∅. Now U = A ∩ V with V open in X and since x ∈ C, we know that B ∩ V ≠ ∅. Now, since B ⊂ A, B ∩ V = B ∩ (V ∩ A) = B ∩ U so that B ∩ U ≠ ∅ and x ∈ C_{A}.

3. If f is continuous then certainly f_{|A} and f_{|B} are continuous being the composition of f and the (continuous) inclusions of A, B into X. For the converse, it is a good idea to use the closed set formulation of continuity: so let C ⊂ Y be closed and try to show that f^{−1}(C) is closed in X. Now f^{−1}(C) = (f^{−1}(C) ∩ A) ∪ (f^{−1}(C) ∩ B) since X = A ∪ B. Also f^{−1}(C) ∩ A = f_{|A}^{−1}(C) which is closed in A (since f_{|A} is continuous) and so closed in X by question 1. Similarly, f^{−1}(C) ∩ B is closed in X whence f^{−1}(C) is closed in X being a union of two sets both closed in X. Thus f is continuous. For f : [−1, 1] → ℝ defined as above, f_{|[−1,0]} is clearly continuous—it is the restriction of the notoriously continuous x ↦ −x and similarly f_{|[0,1]} is continuous whence, by what we have just proved, f is continuous since [−1, 0] and [0, 1] are closed in ℝ and so in [−1, 1].

4. (a) We check the axioms for a base: first, since ℬ is a base, any x ∈ A lies in some B ∈ ℬ and so in A ∩ B ∈ ℬ_{A}. Second, let A ∩ B_{1}, A ∩ B_{2} ∈ ℬ_{A} with the B_{i} ∈ ℬ. If x ∈ (A ∩ B_{1}) ∩ (A ∩ B_{2}) then x ∈ B_{1} ∩ B_{2} and there is B_{3} ∈ ℬ with x ∈ B_{3} ⊂ B_{1} ∩ B_{2}. Thus x ∈ A ∩ B_{3} ⊂ (A ∩ B_{1}) ∩ (A ∩ B_{2}) and A ∩ B_{3} ∈ ℬ_{A} so that ℬ_{A} is a base for a topology on A.

(b) The topology generated by ℬ_{A} is the subspace topology. Indeed, a subset of A is open in the subspace topology if and only if it is of the form A ∩ V with V open in X, that is, of the form A ∩ (⋃_{i∈I} B_{i}), with B_{i} ∈ ℬ, or, equivalently, of the form ⋃_{i∈I}(A ∩ B_{i}), that is, a union of elements of ℬ_{A}.

5. (a) We check the axioms: first any point in Γ clearly lies in some element of ℬ. Second the intersections of two elements of ℬ come in two flavours: either one gets the intersection of two ε-balls in {(x, y) ∈ ℝ^{2} : y ≥ 0} (this also happens when an ε-ball intersects a basic set A ∪ {z} or when basic open sets A ∪ {z}, B ∪ {z′} with z ≠ z′ intersect) or we have two sets A ∪ {z} and A′ ∪ {z} where A and A′ are two balls tangent to the x-axis at z. In the first case, the intersection is a metric open set and thus certainly a union of ε-balls in {(x, y) ∈ ℝ^{2} : y ≥ 0}. In particular, such an intersection is a union of elements of ℬ. In the second case, one of the balls contains the other (draw a picture) and so the intersection is itself in ℬ.

(b) The induced topology on A = {(x, y) : y > 0} is the usual metric topology. Indeed, by question 4, a base for A is given by the intersections of A with the elements of ℬ. But these intersections are precisely the ε-balls in A which generate the metric topology.

(c) Again we get the metric topology and by the same argument: the intersections of the basic open sets with the y-axis are the intervals of the form [0, ε) and (a, b) with a > 0. These are precisely the ε-balls of the induced metric.

(d) The induced topology on the x-axis is an entirely different kettle of fish but the same argument applies: the only basic open sets that intersect the x-axis are those of the form {z} ∪ A with z on the x-axis and A an ε-ball tangent to the axis. Such a set intersects the X-axis in the singleton set {z}. Thus the base for the induced topology consists of all the singleton sets and so the induced topology is the discrete topology.