TOPOLOGY: NOTES AND PROBLEMS

TOPOLOGY: NOTES AND PROBLEMS7expressed as the union a x b {x} (c, d) of open sets {x} (c, d) in theproduct topology Ω.6. Subspace TopologyLet (X, Ω) be a topological space and let Y be a subset of X. ThenΩY : {U Y : U Ω} is a topology on Y, called the subspace topology.Remark 6.1 : If U Ω and U Y then U is open in the subspace topology.Remark 6.2 : If V ΩY and Y Ω then V Ω.Example 6.3 : Consider the real numbers R with usual topology and letY : [ 1, 1]. Then(1) (1/2, 1) is open in the subspace topology: (1/2, 1) Y .(2) (1/2, 1] is open in the subspace topology: (1/2, 1] (1/2, 2) Y.(3) [1/2, 1) is not open in the subspace topology: If [1/2, 1) U Y forsome open subset U of R then 1/2 U. Thus (1/2 , 1/2 ) iscontained in U Y [1/2, 1) for some 0, which is not possible.(4) {x (0, 1) : 1/x 6 1, 2, · · · } is open in the subspace topology: Thisset is open in R. Now apply Remark 6.1.Remark 6.4 : Let (X, Ω) be a topological space with subsets Y, Z suchthat Z Y X. Then (Y, ΩY ) is a subspace of (X, Ω) and (Z, ΩYZ ) is asubspace of (Y, ΩY ). Also, (Z, ΩZ ) is a subspace of (X, Ω). Then ΩZ ΩYZ .Proposition 6.5. If B is basis for the topological space X, and Y is asubspace of X then the basis BY for the subspace topology on Y is given byBY {B Y : B B}.Example 6.6 : Consider the topological space Rl R and let L denote astraight line in the plane. Then the basis elements for the subspace topologyon L are of the form{(x, y) L : a x b, c y d} or {(x, y) L : a x b, c y d}provided L is neither X-axis nor Y -axis.Let X be a simply ordered set with order topology Ωo and let Y be asubset of X. Then the order relation on X makes Y into an ordered set.This makes Y a topological space with order topology ΩY,o . Also, Y has thesubspace topology ΩY . We will see in class an example in which ΩY,o andΩY are not same [4, Pg 90, Example 3].Lemma 6.7. ΩY,o ΩY .Proof. The basis elements for the order topology on Y are of the form B1 : {x Y : a x b}, B2 : {x Y : ã0 x b}, B3 : {x Y : a x

8TOPOLOGY: NOTES AND PROBLEMSb̃0 }, where ã0 , b̃0 denote the minimal and maximal elements of Y (if exists)respectively. Clearly, a0 ã0 , b̃0 b0 . Note the following:(1) Clearly, B1 (a, b) Y.(2) If a0 ã0 then B2 [a0 , b) Y.(3) If a0 ã0 then B2 (a0 , b) Y.(4) If b0 b̃0 then B3 (a, b0 ] Y.(5) If b0 b̃0 then B3 (a, b0 ) Y.Thus B1 , B2 , B3 belong to the basis for the subspace topology. That is,BY,o BY , and hence ΩY,o ΩY . Let X be a simply ordered set and let Y be a subset of X. We say thatY is convex if for any pair a, b Y, the interval (a, b) is contained in Y.Example 6.8 : Any interval (x, y) in a simply ordered set is convex: Ifa, b (x, y) then x a b y and hence (a, b) (x, y).Example 6.9 : Consider the subset [0, 1] [0, 1] of R2 with dictionary order.Then 0 0, 1 1 belong to [0, 1] [0, 1], however, 1/2 2 (0 0, 1 1)does not belong to [0, 1] [0, 1].The first quadrant is not a convex subset of R2 . However, the right halfplane is convex.Lemma 6.10. Let Y be a convex subset of topological space X with ordertopology. For any open ray R in X, R Y is open in order topology on Y.Proof. We prove the lemma only for the ray (a, ). Consider the cases:(1) a y for all y Y : Then Z Y.(2) y a for all y Y : Then Z .(3) There exists y1 , y2 Y such that y1 a y2 : Then a Y . Theconvexity of Y implies that a (y1 , y2 ) Y .If a / Y then Z Y or Z . If a Y then the open set Z : (a, ) Yin the subspace topology is actually an open ray in Y. Theorem 6.11. Let Y be a convex subset of the topological space X withorder topology. Then the order topology on Y is same as subspace topologyon Y, that is, ΩY,o ΩY .Proof. By the preceding lemma, for any open ray R in X, R Y is open inorder topology on Y. Note the following:(1) If a b then (a, b) Y (( , b) Y ) ((a, ) Y ).(2) If X has minimal elment a0 then [a0 , b) Y ( , b) Y .(3) If X has maximal element b0 then (a, b0 ] Y (a, ) Y .Thus every basis element for the subspace topology is open in order topologyon Y. The fact that every basis element for the order topology is open in thesubspace topology is established in Lemma 6.7. Remark 6.12 : If may happen that for a non-convex subset Y of X, ΩY,o ΩY . Consider for example X : R and Y : R \ {0}.

TOPOLOGY: NOTES AND PROBLEMS97. Closed Sets, Hausdorff Spaces, and Closure of a SetA topological space (X, Ω) is Hausdorff if for any pair x, y X withx 6 y, there exist neighbourhoods Nx and Ny of x and y respectively suchthat Nx Ny .Any metric space is Hausdorff. In particular, the real line R with usualmetric topology is Hausdorff.Exercise 7.1 : If X is Hausdorff then show that the complement of anyfinite set is open.Hint. Let x0 X then let x X \ {x0 } and apply the definition ofHausdorff set to the pair x, x0 .Recall that a subset of X is closed if its complement in X is open in X.Remark 7.2 : Arbitrary intersection of closed sets is closed. Finite unionof closed sets is closed.Example 7.3 : Consider the real line R with usual topology. Then the setQ of rationals is not closed. This follows since any neighbourhood of anirrational number contains rationals. If we enumerate Q as a sequence {rn }then Q n {rn }, which shows that countable union of closed sets need notbe closed.Exercise 7.4 : Show that X is Hausdorff iff the diagonal : {(x, x) X X} is closed in X X.Hint. Note that x 6 y iff (x, y) belongs to the complement of in X X.Check that X is Hausdorff iff X X \ is open in X X.Exercise 7.5 : Show that every topological space with order topology isHausdorff.Hint. Given x1 6 x2 , choose xi (ai , bi ) possibly (a1 , b1 ) (a2 , b2 ) 6 .We may assume that a2 b1 . If there is c (a1 , b1 ) (a2 , b2 ) then try (a1 , c)and (c, b2 ). Otherwise try (a1 , b1 ) and (a2 , b2 ).Exercise 7.6 : What are all closed subsets of R with VIP topology (resp.outcast topology) ?Example 7.7 : Consider the subset Y : ( , 0) [1, ) of R. Clearly, Yis not closed in the usual topology on R as R \ Y [0, 1) is not open in R.However, Y is closed in Rl .Exercise 7.8 : Show that the only non-empty subset of R which is open aswell as closed is R.Hint. Write R A B for open and disjoint sets A and B. If a Aand b B with a b then [a, b] A0 B0 , where A0 A [a, b] and

10TOPOLOGY: NOTES AND PROBLEMSB0 B [a, b] are disjoint. Let c : sup A0 [a, b]. Then either c A0 orc B0 . If c A0 then either c a or a c b. Since A0 is open in thesubspace topology on [a, b], there is d such that [c, d) A0 . This contradictsthat c sup A0 . Similarly, prove that c can not belong to B0 .Exercise 7.9 : Given an example of a non-empty subset of Y : R \ {0}which is open as well as closed in the subspace topology on Y .Exercise 7.10 : Let f : Rn R be a continuous function in the variablesx1 , · · · , xn . Show that the zero set Z(f ) : {x Rn : f (x) 0} is a closedsubset of Rn .2Exercise 7.11 : Identify Mn (C) with Cn with usual topology. Show thatGLn (C) is an open subset of Mn (C).Let (X, Ω) be a topological space with topology Ω. If A is a subset of Xthen the closure Ā of A in X is defined as the intersection of all closed setscontaining A:Ā \closedB.B ARemark 7.12 : Clearly, A Ā. Also, Ā A iff A is closed.Example 7.13 : The closure of rationals in the usual topology on R is R.To see this, let B be a closed subset of R such that Q B. If B 6 R thenthe complement of B in R is non-empty, and hence contains an interval. Butthen R \ B contains a rational, which is not possible since Q B.Example 7.14 : Let X be an ordered set with order topology. Note thatthe complement of [a, b] in X is open. Thus (a, b) [a, b].Example 7.15 : Considerthe topologicalspace Rl . Then the closure of (non-closed set) (0, 2) equals [0, 2). This follows from the observationthat R \ [0, 2) ( , 0) [ 2, ) is open in Rl . If one replaces thelower limit topology by the topology Ω generated by the basis {[a, b) : a, b Q} then the closure of (0, 2) in Ω equals [0, 2]. This provides anotherverification of the fact that Ω and the lower limit topology are different.Example 7.16 : Let X R with usual topology. Then the closure of (0, 1)in R equals [0, 1] as [0, 1] is the smallest closed set containing (0, 1). However,the closure of (0, 1) in the subspace topology on [0, 1) equals [0, 1) as [0, 1)is the smallest closed in the subspace topology that contains (0, 1).Theorem 7.17. Let (X, Ω) be a topological space with topology Ω. Let A, Ybe subsets of X such that A Y. Then

TOPOLOGY: NOTES AND PROBLEMS11(1) A is closed in the subspace topology on Y iff A B Y for someclosed subset B of X.(2) The closure of A in the subspace topology on Y equals Ā Y.Proof. (1) If A B Y for some closed subset B of X then (X \ B) Y Y \ A, and hence A is closed in the subspace topology on Y. If A is closedin the subspace topology on Y then Y \ A U Y for some open set U inX. Then A Y \ (U Y ) B Y for closed set B : X \ U in X.(2) Let B denote the closure of A in the subspace topology. Since Ā Yis a closed set containing A, and B is the smallest closed set in the subspacetopology that contains A, we have B Ā Y. Also, since B C Y for someclosed set C in X that contains A, Ā C. But then Ā Y C Y B. Corollary 7.18. Let Y be a subspace of X. If A is closed in Y and Y isclosed in X, then A is closed in X.Proof. Let Ā and ĀY denote the closures of A in X and Y respectively.By the last theorem, ĀY Ā Y. Suppose that A is closed in Y , that is,ĀY A. But then A ĀY Ā Y, which is the intersection of closed setsĀ and Y in X. Hence A is closed in X. Example 7.19 : Consider the topological space R R with dictionary ordertopology and consider the unit square [0, 1] [0, 1] with subspace topology.Note that A : {(1/n) 0 : n N} is closed in R R, and hence in thesubspace topology on [0, 1] [0, 1]. Note that the closure of B : {0 (1/n) :n N} in R R equals B {0 0}. This also shows that the closure of Bin the subspace topology equals B {0 0}.By a neighbourhood of a point, we mean an open set containing that point.By a deleted neighbourhood of a point x, we mean U \ {x}, where U is aneighbourhood of x. If A is a subset of the topological space X and if x Xthen we say that x is a cluster point of A if every deleted neighbourhood ofx intersects A.Example 7.20 : Consider the subset A : [0, 1) {2} of the real line withusual topology. Note that 1 is a cluster point of A but 2 is not a clusterpoint of A.Exercise 7.21 : What are all cluster points of {q} in VIP topology withbase point p on R ?Lemma 7.22. Every cluster point of A belongs to Ā.Proof. If possible, suppose that x / Ā. But then U : X \ Ā is a neighbourhood of x. Since x is a cluster point of x, the deleted neighbourhood U \ {x}must intersect A. This is not possible since U is contained in X \ A. Lemma 7.23. If x Ā (X \ A) then x is a cluster point of A.

12TOPOLOGY: NOTES AND PROBLEMSProof. If x is not a cluster point of A then there is a neighbourhood U of xsuch that (U \ {x}) A . Since x / A, we have U A , and henceA is contained in the closed set X \ U. By the definition of closure of a set,Ā X \ U. Since x Ā, we arrive at the contradiction that x / U. We combine the preceding two lemmas to get the following:Theorem 7.24. x Ā iff either x A or x is a cluster point of A.Theorem 7.25. If X is a Hausdorff space then every neighbourhood of acluster point of A contains infinitely many points from A.Proof. Suppose that X is Hausdorff and let x be a cluster point of a subsetA of X. Suppose, if possible, there is a neighbourhood U of x such thatU A F is a finite set. Since X is Hausdorff, F is closed in X. But thenthe deleted neighbourhood U (X \ F ) of x does not intersect at all. Thatcontradicts that x is a cluster point. 8. Continuous FunctionsLet (Xi , Ωi ) (i 1, 2) be two topological spaces and let f : X1 X2 bea function. We say that f is continuous if for any V Ω2 , the pre-imagef 1 (V ) of V under f belongs to Ω1 .Remark 8.1 : Note that f is continuous iff for any closed set B in X2 ,the pre-image f 1 (B) of B under f is closed in X1 . This follows from theobservation f 1 (X2 \ B) X1 \ f 1 (B).Exercise 8.2 : Let (X, Ωi ) (i 1, 2) be two topological spaces and considerthe identity function id from X onto itself. Show that Ω1 is finer than Ω2iff the identity function id is continuous.Example 8.3 : Let X1 , X2 , X3 denote the set R with usual topology, lowerlimit topology, K-topology respectively. The identity mapping id from X3onto X1 is continuous. However, neither the identity mapping id from X2onto X3 nor the identity mapping id from X3 onto X2 is continuous.Exercise 8.4 : Let X be a topological space with discrete topology and Ybe any topological space. Show that any function f : X Y is continuous.Exercise 8.5 : Let (Xi , Ωi ) be two topological spaces with topology Ωi (i 1, 2) and let x0 X2 . If X2 is Hausdorff then show that for any continuousfunction f : X1 X2 , the set C : {x X1 : f (x) x0 } is closed in X1 .Hint. Note that f 1 (X2 \ {x0 }) X1 \ C.Example 8.6 : Consider the metric space C[0, 1] with sup metric and Cwith usual topology. For fixed t0 [0, 1], consider the evaluation functionalEt0 : C[0, 1] C given by Et0 (f ) f (t0 ). Since f (t0 ) g(t0 ) d (f, g),the convergence in sup metric implies point-wise convergence. It follows

TOPOLOGY: NOTES AND PROBLEMS13that Et0 is continuous. Now {f C[0, 1] : f (t0 ) 6 0} is the complement ofthe null-space of Et0 , and hence by the last exercise, it is open in C[0, 1].Exercise 8.7 : If f : X Y is continuous and a sequence {xn } in Xconverges to x X, show that the sequence {f (xn )} in Y converges to f (x).Proposition 8.8. Let (Xi , Ωi ) (i 1, 2) be two topological spaces and let B2be a basis for Ω2 . A function f : X1 X2 is continuous iff for any B2 B2 ,the pre-image f 1 (B2 ) of V under f belongs to Ω1 .Proof. Suppose that for any B2 B2 , the pre-image f 1 (B2 ) of V underf belongs to Ω1 . Let V be an open subset of Ω2 . Then V α Bα . Sincef 1 (V ) α f 1 (Bα ), f 1 (V ) belongs to Ω1 . Example 8.9 : Consider the topological space X1 : (0, ) with the subspace topology inherited from R. Then the function f (x) x2 from X1 ontoitself is continuous. In view of the last proposition, we need to check thatthe pre-image of any open interval in (0, ) is open. However, note that for 0 a b , f 1 (a, b) ( a, b). If X2 denotes the topological space(0, ) with lower limit topology then by the same argument, the functionf : X1 X2 given by f (x) x2 is not continuous. Finally, note that thefunction f : X2 X1 given by f (x) x2 is continuous.Exercise 8.10 : Let E be a non-empty proper subset of the topologicalspace X. Consider the characteristic function χE of E. Show that χE iscontinuous iff E is a closed and open subset of X.Exercise 8.11 : Consider topological spaces Xi (X, Ωi ), i 1, 2 such thatΩ1 is finer than Ω2 . Show that f : X1 Y is continuous if so is f : X2 Y.The next result says that continuity is a local property.Proposition 8.12. Let (Xi , Ωi ) (i 1, 2) be two topological spaces. A function f : X1 X2 is continuous iff for each x X and each neighbourhoodVx of f (x), there is a neighbourhood Ux of x such that f (Ux ) Vx .Proof. If f is continuous then Ux : f 1 (Vx ) satisfies f (Ux ) Vx .Conversely, suppose that for each x X and each neighbourhood Vx off (x), there is a neighbourhood Ux of x such that f (Ux ) Vx . Let V bean open subset of X2 and let x f 1 (V ). Then f (x) V, and hence byhypothesis, there is a neighbourhood Ux of x such that f (Ux ) V. But thenUx f 1 (V ), and hence f 1 (V ) is open. We say that f : X1 X2 is continuous at x X1 if for each neighbourhood Vx of f (x), there is a neighbourhood Ux of x such that f (Ux ) Vx .By the last result, f is continuous iff f is continuous at every point.Example 8.13 : Consider the function f : R R given by f (x) x ifx Q and f (x) x if x R \ Q. We check that f is continuous only at 0.

14TOPOLOGY: NOTES AND PROBLEMSTo see that, first note that any neighbourhood V0 of f (0) 0 contains aninterval ( , ) for some 0. Then U0 : ( , ) satisfies f (U0 ) ( , ) V0 . Thus f is continuous at 0. If x 0 then for Vx : (x/2, 3x/2) (0, ),there is no neighbourhood Ux of x such that f (Ux ) Vx (since f (Ux ) alwayscontains negative numbers). Similarly, one can see that f is not continuousat x 0.Exercise 8.14 : Let X be a topological space and Y be a topological spacewith ordered topology. Let f, g : X Y be continuous functions. Showthat the set U : {x X : f (x) g(x)} is open in X.Hint. Let x0 U. If (g(x0 ), f (x0 )) then g 1 ( , f (x0 )) f 1 (g(x0 ), )is a neighbourhood of x0 contained in U. Otherwise, for any y (g(x0 ), f (x0 )),g 1 ( , y) f 1 (y, ) is a neighbourhood of x0 contained in U.Exercise 8.15 : Prove: Composition of continuous functions is continuous.Hint. Use Proposition 8.12.Corollary 8.16. Let (Xi , Ωi ) (i 1, 2) be topological spaces and let A be asubspace of X1 . If f : X1 X2 is continuous then so is map f A : A X2 .Proof. We apply the last proposition. Let a A and let Va be a neighbourhood of f (a) in X2 . Since A X, a X. Since the f is continuous, there is a neighbourhood Ua of a such that f (Ua ) Va . But thenf (A Ua ) f (Ua ) Va , where A Ua is a neighbourhood of a in thesubspace topology. Remark 8.17 : The inclusion map i : A , X is always continuous.One may wish to know whether or not the converse of Corollary 8.16is true: For X α Aα and f : X A, if each f Aα is continuous thenwhether f is continuous ? (Restriction Problem). The answer is No even ifeach Aα is closed.Example 8.18 : Let X : [0, 1] and A0 : {0}, An : [1/n, 1]. Then thefunction f : [0, 1] R given by f (0) 1, f (x) 1/x (0 x 1) isdiscontinuous at 0. However, f An is continuous for every n 0.The answer to the Restriction Problem is yes if each Aα is open.Corollary 8.19. For X α Aα with open Aα and f : X A, if eachf Aα is continuous then f is continuous.Proof. We apply Proposition 8.12. Let x X and let Vx be a neighbourhoodof x. Then x Aα for some α. Since f Aα is continuous, there exists aneighbourhood Ux,α of x such that f (Ux,α Aα ) Vx , where Ux : Ux,α Aαis the desired neighbourhood of x. An indexed family {Aα } of sub

TOPOLOGY: NOTES AND PROBLEMS Abstract. These are the notes prepared for the course MTH 304 to be o ered to undergraduate students at IIT Kanpur. Contents 1. Topology of Metric Spaces 1 2. Topological Spaces 3 3. Basis for a Topology 4 4. Topology Generated by a Basis 4 4.1. In nitude of Prime Numbers 6 5. Product Topo