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Munkres SolutionsAmit RajaramanLast updated March 31, 2021Contents2 Topological Spaces and Continuous2.13 Basis for a Topology . . . . . . . .2.16 The Subspace Topology . . . . . .2.17 Closed Sets and Limit Points . . .2.18 Continuous Functions . . . . . . .2.19 The Product Topology . . . . . . .2.20 The Metric Topology . . . . . . . .2.21 The Metric Topology (continued) .2.22 The Quotient Topology . . . . . .Functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .3 Connectedness and Compactness3.23 Connected Spaces . . . . . . . . . . . .3.24 Connected Subspaces of the Real Line3.25 Components and Local Connectedness3.26 Compact Spaces . . . . . . . . . . . .222355678.9. 9. 10. 10. 104 Countability and Separation Axioms124.30 The Countability Axioms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

Munkres Solutions2 / 13Amit Rajaraman§2. Topological Spaces and Continuous Functions2.13. Basis for a TopologyExercise 2.13.1. Let X be a topological space and A X. Suppose that for each x A, there is an open set Ucontaining x such that U A. Show that A is open in X.SolutionFor each x A, denote by Ux an open subset of A that contains A. Then A union of open sets is open and thus, so is A.Sx AUx . However, an arbitraryExercise 2.13.5. Show that if A is a basis for a topology on X, the topology generated by A equals the intersectionof all topologies that contain A. Prove the same if A is a subbasis.SolutionSLet T be the topology generated by A and T 0 be a topology that contains A. Let U T . Then U i I Bifor some (Bi )i I in A. However, each Bi is also in T 0 . Since an arbitrary union of open sets is open, U T 0 aswell. Therefore, T T 0 , proving the result. The solution for the case where A is a subbasis is very similar andso omitted.2.16. The Subspace TopologyExercise 2.16.1. Show that if Y is a subspace of X and A is a subset of Y , then the topology A inherits as asubspace of Y is the same as the topology it inherits as a subspace of X.SolutionThe topology A inherits as a subspace of X isT {U A : U open in X} {(U Y ) A : U open in X} {V A : V open in Y },which is the topology it inherits as a subspace of Y .Exercise 2.16.2. If T and T 0 are topologies on X and T 0 is strictly finer than T , what can you say about thecorresponding subspace topologies on the subset Y of X.SolutionIt is easily seen that TY0 is finer than TY . We further see that it need not be strictly finer by considering theexample X {a, b, c}, Y {a, b}, T { , X, {a}, {b}, {a, b}}, and T 0 as the discrete topology on X.Exercise 2.16.3. Consider Y [ 1, 1] as a subspace of R. Which of the following is open in Y ? Which are openin R? 1A x: x 12 1B x: x 12 1C x: x 12 1D x: x 12 E x : 0 x 1 and 1/x 6 Z

Munkres Solutions3 / 13Amit RajaramanSolutionA and B are open in Y and only A is open in R. This is reasonably straightforward to prove.C is not open in Y (and so not R either) because there is no basis element U of the order topology such that1/2 U C. A similar argument holds for D as well.E is open in both R and Y because it can be written as a union of basis elements \ 11E ,.n 1 n n ZExercise 2.16.4. A map f : X Y is said to be an open map if for every open U of X, f (U ) is open in Y . Showthat π1 : X Y X and π2 : X Y Y are open.SolutionWe shall only show that π1 is open, the other case is nearly identical. Let[U Ui Vii Ibe open in X Y for some indexing set I, where each Ui and Vi are open in X and Y respectively. Then,![[[π1 (U ) π1Ui Vi π1 (Ui Vi ) Uii Ii Ii Iis open in X.2.17. Closed Sets and Limit PointsExercise 2.17.1. Let C be a collection of subsets of set X. Suppose that and X are in C and that finite unionsand arbitrary intersections of elements of C are in C. Show that the collection T {X \ C : C C} is a topology onX.SolutionLet (Ui )i I be in T with Ui X \ Ci for each i. Then[\Ui X \Ci X \ C Ti Ii Ifor some C C. Closure under finite intersections is shown similarly. We trivially have , X T becauseX, C.Exercise 2.17.2. Show that if A is closed in Y and Y is closed in X, then A is closed in X.SolutionLet U be open in X such that Y \ A U Y . Then, we can write A as X \ ((X \ Y ) U ). Since X \ Y and Uare open in X, A is closed in X.Exercise 2.17.3. Show that if A is closed in X and B is closed in Y , A B is closed in X Y .Solution

Munkres Solutions4 / 13Amit RajaramanObserve that(X Y ) \ (A B) ((X \ A) (Y \ B)) ((X \ A) Y ) (A (Y \ B)).Since each of the sets on the right are open in X Y , A B is closed.Exercise 2.17.4. Show that if U is open in X and A is closed in X, U \ A is open in X and A \ U is closed in Y .SolutionThis is easily seen on writing U \ A U (X \ A) and A \ U A (X \ U ).Exercise 2.17.13. Show that if X is Hausdorff iff the diagonal {x x : x X} is closed in X X.SolutionLet us first show the forward direction. Let (x, y) (X X) \ . Then since X is Hausdorff, there are someopen U1 3 x, U2 3 y such that U1 U2 . Since the open set U1 U2 3 (x, y) does not intersect , isclosed (no other point is in its closure).For the backward direction, this means that for any (x, y) X X with x 6 y, there exists an open set U 3 (x, y)such that U . We may further assume that U is a basis element of the product topology and can bewritten as U1 U2 . However, (U1 U2 ) iff U1 U2 , so the required follows.Exercise 2.17.14. In the finite complement topology on R, what point or points does the sequence xn 1/nconverge to?SolutionWe claim that xn converges to any point in R. Let x R and U 3 x be open. U contains all but finitely manyxi since we are in the finite complement topology and therefore, xn converges to x.Exercise 2.17.19. If A X, define the boundary of A byBd A A X \ A. (a) Show that A and Bd A are disjoint, and A A Bd A.(b) Show that Bd A iff A is both open and closed.(c) Show that U is open iff Bd U U \ U . (d) If U is open, is it true that U U ? Justify your answer.Solution(a) Let x A \ A . Then for any open U 3 x, U 6 A (otherwise, A U ) A is open and contained in A).That is, U (X \ A) 6 . However, this implies that x X \ A, that is, A \ A X \ A. Therefore,A \ A (A \ A) (A \ A ) X \ AA A X \ A A (A X \ A) A (A X \ A) A Bd A.(b) If A is not closed, A ) A intersects X \ A X \ A, contradicting Bd A . Similarly, X \ A is closed aswell, so A is both open and closed.The other direction is similarly straightforward.(c) If U is open, X \ U is closed so Bd U U (X \ U ) U \ U .On the other hand, if U (X \ U ) U X \ U , X \ U must be closed. Indeed, otherwise, X \ U \ (X \ U )

Munkres Solutions5 / 13Amit RajaramanU U , contradicting the equality. (d) No, this is not the case. Consider the open set U (1, 2) (2, 3) R. Then U (1, 3).2.18. Continuous FunctionsExercise 2.18.1. Suppose that f : X Y is continuous. If x is a limit point of A X, is f (x) necessarily a limitpoint of f (A)?SolutionNo. Let f : R R be the zero function, A {1/n : n N}, and x 0.Exercise 2.18.8. Let Y be an ordered set in the order topology. Let f, g : X Y be continuous.(a) Show that the set {x : f (x) g(x)} is closed in X.(b) Let h : X Y be the functionh(x) min{f (x), g(x)}.Show that h is continuous.Solution(a) It suffices to show that V {x : f (x) g(x)} is open in X. Let x V . Since Y is Hausdorff, there areopen sets U1 , U2 such that f (x) U1 , g(x) U2 , and U1 U2 .We can consider a basis element B1 U1 that contains f (x) in lieu of U1 . That is, we may suppose wlogthat U1 and U2 are disjoint basis elements. Further, we may assume that U1 is of the form (a, ) (if it is ofthe form (a, c) instead, we can replace it with (a, )). Due to the disjointness assumption, this means thatwe can consider U1 to be of the form (a, ) and U2 of the form ( , b) such that for any c U1 , d U2 ,c dNow, let U f 1 (U1 ) g 1 (U2 ) 3 x. f and g are continuous so U is open. Further, for any y U ,f (y) g(y) (by the above assumption), that is, U V .It follows that V is open (for any x V , there is an open V U such that x V ).(b) Let U1 {x : f (x) g(x)} and U2 {x : f (x) g(x)}. By (a), U1 and U2 are both closed. Since g iscontinuous on U1 , f is continuous on U2 , and f g on U1 U2 , we can use the pasting lemma to concludethat h is continuous (h(x) g(x) on U1 and f (x) on U2 ).2.19. The Product TopologyExercise 2.19.4. Show that (X1 · · · Xn 1 ) Xn is homeomorphic to X1 · · · Xn .SolutionLet the two topological spaces above be denoted by X and Y respectively. Consider the map f : X Y withf (x) (π1 (π1 (x)), . . . , πn 1 (π1 (x)), π2 (x)).We claim that f is a homomorphism.If Ui is open in Xi for each i, then f 1 (U1 · · · Un ) (U1 · · · Un 1 ) Un is open in X (U1 · · · Un 1is open in X1 · · · Xn 1 and Un is open in Xn ). Therefore, f is continuous.On the other hand, if U is open in X1 · · · Xn 1 and Un is open in Xn , then f (U Un ) U1 · · · Un ,where each Ui is open in Xi by Exercise 2.16.4.QExercise 2.19.6. Let x1 , x2 , . . . be a sequence of points in the product spaceXα . Show that this sequenceconverges to the point x iff the sequence πα (x1 ), πα (x2 ), . . . converges to πα (x) for each α. Is this true if we use thebox topology instead of the product topology?

Munkres Solutions6 / 13Amit RajaramanSolutionWe first show the backward direction. Suppose (πα (xi )) converges to πα (x). LetQU be an open set containing xand B U be a basis element of the product topology containing x. Let B α Uα where Uα 6 Xα for thefinite set {α1 , . . . , αn }. Since (παj (xi )) converges to παj (x) for each j, each Uαj contains all but finitely manyπα (xi ). As a result, B contains all but finitely many xi and therefore, (xi ) converges to x.On the other hand, let (xi ) converge to x. Let Uα beQ an open set in Xα containing πα (x). We wish toQ show thatit contains all but finitely many πα (xi ). Let U 0 Uβ0 be a basis element containing x and V Vβ , whereVβ Uβ0 for β 6 α and Vα Uα Uα0 . Since V is an open set containing x, it contains all but finitely many xi .In particular, Vα contains all but finitely many πα (xi ). The required follows.Observe that the forward proof works even if we use the box topology instead, but the backward direction breaks.To see that the result need not hold for the box, let the product space be Rω , πn (xi ) n/i and πn (x) 0 foreach i, n.Exercise 2.19.7. Let R be the subset of Rω consisting of all sequences that are eventually 0 (xi 6 0 for finitelymany i). What is the closure of R in Rω in the box and product topologies?SolutionQWe claim that the closure under the product topology is Rω . Let x Rω and U Un be a basis elementcontaining x. We wish to determine when U R 6 . Consider y R such that yn xn if Un 6 R and 0otherwise. Then y U R , thus proving the result.QFor the box topology, we claim that R is closed. Indeed, for any x Rω \ R , consider the open set U Un ,where Un (xn /2, 3xn /2) if xn 6 0 and R otherwise. Then U R is empty, completing the proof.2.20. The Metric TopologyExercise 2.20.3. Let X be a metric space with metric d.(a) Show that d : X X R is continuous.(b) Let X 0 denote a space having the same underlying set as X. Show that if d : X 0 X 0 R is continuous, thetopology of X 0 is finer than the topology of X.This means that if X has a metric d, the metric topology induced by it is the coarsest topology with respect to whichd is continuous.Solution(a) Let x (x1 , x2 ) X X and for some ε 0, U B(f (x), ε) be a basis element of R containing f (x).Consider the open sets U1 Bd (x1 , ε/4) and U2 Bd (x2 , ε/4). If y1 U1 and y2 U2 , thend(y1 , y2 ) d(x1 , x2 ) d(x1 , y1 ) d(x2 , y2 ) f (x) ε/2.Therefore, d(y1 , y2 ) U . As a result, U1 U2 is an open set such that x f (U1 U2 ) U and d iscontinuous.(b) Consider the continuous function g : X 0 X 0 X 0 given by y 7 (x, y) (we proved continuity in Exercise 2.16.4). Since the composition of continuous functions is continuous, so is d g : X 0 R. Note that forany ε 0, (d g) 1 ((0, ε)) Bd (x, ε). It follows that Bd (x, ε) is open in X 0 and therefore, the topology onX 0 is finer than that on X.Exercise 2.20.5. Let R be the subset of Rω consisting of all infinite sequences that are eventually 0. What is theclosure of R in Rω in the uniform topology?

Munkres Solutions7 / 13Amit RajaramanSolutionWe claim that the closure of R is the set of all sequences that converge to 0. Let x Rω . Case 1. The sequence x does not converge to 0. There is then some ε 0 such that for infinitely many n, xn ε. We may assume that ε 1. Consider the open set U Bρ (x, ε). We claim that U R .Indeed, for any y R , xn y ε for infinitely many n so y 6 Bd (x, ε), thus proving that x 6 R . Case 2. The sequence x converges to x. Let Bρ (x, ε) be an arbitrary basis element containing x. Thenthere exists N N such that for all n N , xn ε/2. Consider the element y R such that yn xnfor n N and yn 0 otherwise. Then y Bρ (x, ε), thus proving that x R .Exercise 2.20.6. Let ρ be the uniform metric on Rω . Given x (x1 , x2 , . . .) Rω and 0 ε 1, letU (x, ε) (x1 ε, x1 ε) · · · (xn ε, xn ε) · · · .(a) Sho