INTRODUCTION TO TOPOLOGY (MA30055) SEMESTER SEMESTER 2 MATHEMA MATHEMATICS: TICS: PROBLEM SHEET 1: SOLUTIONS SOLUTIONS 1. True or false? (a) In the discrete topology Y = Y , Y , for all Y ⊆ X . X . X , for all Y ⊆ X . X . [Hint: is your answer always (b) In the trivial topology Y = X , answer always true?] A or (c) If a topological space X space X = A ∪ B , with A with A,, B disjoint open subspaces, then one of A B is empty. f : X → Y be a map between (d) Let f : between topologica topologicall spaces spaces.. Suppose Suppose that, that, whenev whenever er F is F ) is closed in Y . Y . Then, U is open in X , f ∗ (U ) is closed in X , the image f ∗ (F ) Then, whenev whenever er U Y . open in Y . Solution: (a) True: every subspace is closed. (b)False:
∅
= ∅, but true for non-empty subspaces.
(c) False: e.g. X = { 0, 1} with the discrete topology, A = { 0}, B = { 1}. (d) False: remember that f that f ∗ (U c ) U ))c (it’s f f ∗ (U c ) = (f ! (U )) U ))c ). This statement does = (f ∗ (U )) hold whenever f is f is bijective. Options for a counterexample include: ( •) any map between cofinite topological spaces preserves closed, i.e. finite, subspaces under direct image, but not necessarily open ones. ( •) Let Σ be the Sierpi´ nski nski space and define f : f : Σ → Σ by f ( f (x) = 0 for all x ∈ Σ. Σ . Note that the converse converse is not true either: a map that preserves preserves open subspaces under direct direct image does not necessarily preserve closed subspaces. X , then X \ \ Y ◦ = X \ \ Y and X \ \ Y = 2. Show that, if X if X is is a topological space and Y ⊆ X , \ Y ) Y )◦ . (X \ Solution: X \ Y ◦ is closed and contains X contains X \ Y (because Y (because Y Y ◦ is open and Y and Y ◦ ⊆ Y ). Y ). Therefore \ Y . \ Y X \ Y ◦ ⊇ X \ Y . On the the other other hand, hand, X \ (X Y is open and ( X \ Y ◦ ) = Y ◦ , and X \ X \ \ X \ \ Y ⊆ X \ \ (X \ \ Y ◦ ), so, taking contained in Y , Y , so it is contained in Y ◦ . Therefore X \ \ Y ⊇ X \ \ Y ◦ . complements, X \ N is a neighbourhood of x X if there is U ∈ T X with x ∈ U ⊆ N . N . Prove 3. Recall that N is of x ∈ X if the following properties. N is a neighbourhood of x (a) If N is of x and N ⊆ N , then N is a neighbourhood of x of x. (b) If N 1 and N 2 are neighbourhoods of x, then so is N 1 ∩ N 2 . Y if and only if every neighbourhood of x Y . (c) x ∈ Y if of x intersects Y . Solution: (a) If x If x ∈ U ⊆ N N with U with U open, open, then x then x ∈ U ⊆ N . Thus N is a neighbourhood of x.
(b) If x ∈ U 1 ⊆ N 1 and x ∈ U 2 ⊆ N 2 with U 1 , U 2 open, then x ∈ U 1 ∩ U 2 ⊆ N 1 ∩ N 2 and U 1 ∩ U 2 is open because it is th intersection of two open sets. Thus N 1 ∩ N 2 is a neighbourhood of x. (c) One option is to prove the contrapositive: x ∈ / Y if and only if x ∈ X \ Y = (X \ Y )◦ , i.e. there is an open U ⊆ X such that x ∈ U and U ⊆ X \ Y . Alternatively, tackle the problem head-on. For this, x ∈ Y if and only if Y ⊆ Z implies ∅ for each closed set x ∈ Z for each closed set Z ⊆ X . But that says x ∈ U ⇒ Y ∩ U = U ⊆ X , i.e. every open neighbourhood of x intersects Y . However, every neighbourhood of x contains an open neighbourhood of x, so every neighbourhood of x meets Y if and only if every open neighbourhood of x does. 4. Let Y be a subspace of a topological space X . Show that subspace F ⊆ Y is closed in Y if and only F = Y ∩ E for some closed subspace E ⊆ X . Deduce that the closure of any subspace Z ⊆ Y in Y is the intersection of Y with the closure of Z in X . Solution: Firstly, Z closed in Y ⇐⇒ Y \ Z open in Y ⇐⇒ Y \ Z = Y ∩ U for some U open in X ⇐⇒ Z = Y ∩ (X \ U ) for some U open in X ⇐⇒ Z = Y ∩ E , for some E closed in X . Z ⇐⇒ y ∈ U ) is equivalent to The third equivalence is justified because ∀ y ∈ Y (y ∈ ∀ y ∈ Y (y ∈ Z ⇐⇒ y ∈ U ). But then, for any Z ⊆ Y , note that Z ⊆ Y ∩ W for W ⊆ X if and only if Z ⊆ W . Therefore Z (in Y ) =
Z ⊆W ∈CY
W =
(Y ∩ W ) = Y ∩
Z ⊆W ∈CX
Z ⊆W ∈CX
W = Y ∩ Z (in X ),
where C X and C Y are the collections of all closed subsets in X and Y . 5. Let f : X → Y be a map of topological spaces. Show that f is continuous if and only if for every W ⊆ X , f ∗ (W ) ⊆ f ∗ (W ). [Hint: use the equivalent definition of continuity that for any Z ⊆ Y , f ∗ (Z ) ⊆ f ∗ (Z ), and a relationship between inverse and direct image from Sheet 0.] Solution: First, suppose that f is continuous, so that for Z ⊆ Y , f ∗ (Z ) ⊆ f ∗ (Z ), i.e., f ∗ (f ∗ (Z )) ⊆ Z , and put Z = f ∗ (W ), noting that R ⊆ f ∗ (f ∗ (W )), so that f ∗ (W ) ⊆ f ∗ (f ∗ (f ∗ (W )) ⊆ f ∗ (W ). Conversely if the stated property holds for all W ⊆ X , consider W = f ∗ (Z ), and use f ∗ (f ∗ (Z )) ⊆ Z .
Can you formulate a characterisation of continuity using f ! (N 0 )? 6. (Kuratowski’s closure axioms.) Let X be a topological space with closure operation Y . Y → (a) Show that Y = Y if and only if Y is closed. (b) Show that, for all A, B ⊆ X : (K1) A ⊆ A;
(K2) A = A; (K3) A ∪ B = A ∪ B; (K4)
∅
= ∅;
(K5) if A ⊆ B then A ⊆ B. (c) Show that (K3) implies (K5). [Hint: A ⊆ B ⇐⇒ A ∪ B = B.]
Y satisfying axioms (K1)–(K4) is the closure operation (d) Show that any operation Y → of some topology on X . Solution: (a) Y is closed, so if Y = Y then Y is closed; conversely, if Y is closed then it is certainly the smallest closed subspace containing Y , so Y = Y . (b) (K1) A is an intersection of sets containing A, so it contains A. (K2) if Y = A, then Y is closed and so Y = Y , by (a). (K3) A ∪ B is closed as it is the intersection of closed sets. If F is any closed set with A ∪ B ⊆ F then A ⊆ F , so A ⊆ F and B ⊆ F , so B ⊆ F . Hence A ∪ B ⊆ F and so A ∪ B is the smallest closed set containing A ∪ B. (K4)
∅
is closed, so this follows from (a).
(K5) A ⊆ B ⊆ B. Hence B is a closed set which contains A, so A ⊆ B. c) If A ⊆ B then B = A ∪ B. Hence, by (K3), B = A ∪ B and so A ⊆ B. (d) Given the operation Y → Y , define a set C = {F ⊆ X : F = F }. We show first that this satisfies the properties of a collection of closed sets and so defines a topology on X . (K4) implies that ∅ ∈ C , while (K1) implies that X = X since this is the only set which contains X : hence X ∈ C . If A, B ∈ C , then A ∪ B = A ∪ B by (K3), but that is = A ∪ B so A ∪ B ∈ C .
If Aα ∈ C for α ∈ I and F = α∈I Aα , then for all α ∈ I we have F ⊆ A α so F ⊆ A α = A α . Hence F ⊆ α∈I A α = F . But F ⊆ F , by (K1), so F = F and so F ∈ C .
It remains to show that the topology defined by C has the original operation as its closure operation. To see this observe that (K2) implies that Y ∈ C for all Y , while for any other
set F ∈ C with Y ⊆ F , we have Y ⊆ F = F by (K5). Hence Y is the smallest set in C that contains Y and so is the closure in this topology. Moral: a closure operation is another way of defining the notion of a topological space. Can you formulate yet another definition, using the idea of an “interior operation” N → N ◦ ? GKS, 8/2/16