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4. Topologies and Continuous Maps. In this section we introduce the concept of a topological space. Recall that a graph was defined to be a set together with a collection of subsets that are called ”edges”. A simplicial complex is a set together with a collection of subsets called ”simplices”. The notion of a topological space extends this idea vastly. A topological space is a set together with a collection of subsets called ”open subsets”. In this section we describe some efficient ways of generating a topology, we give some examples and we introduce some first properties of topological spaces. Topological Spaces. We say that a subset of R is an open subset of R if it is a union of open intervals (a, b) ⊂ R. Given this notion the usual ǫ − δ-definition for continuous maps takes a rather simple form. 4.1. Theorem. A map f : R → R is continuous if Klaus Johannson, Topology I 56 . Topology I either one of the following two conditions hold: (1) the preimage under f open. of every open subset is (2) for every point x ∈ R and every number ǫ > 0, there is a number δ = δ(x, ǫ) > 0 such that |z−x| < δ implies |f (z) − f (x)| < ǫ, for every z ∈ R. Proof. (1) ⇒ (2): Let us be given x ∈ R and ǫ > 0. Then the set Y := { y ∈ R | |y − f (x)| < ǫ } is an open interval and therefore an open set. Because of (1), the preimage f −1 (Y ) is also an open set. By definition, this means that that preimage is the union of open intervals. In particular, the point x must lie in some open interval (a, b) ⊂ Y . Set δ := min { |x − a|, |x − b| }. Then (x − δ, x + δ) ⊂ (a, b) and so (x − δ, x + δ) is mapped under f into Y . This means that |z −x| < δ implies f (z) ∈ Y , i.e., |f (z) − f (x)| < ǫ. In other words, f satisfies (2). (2) ⇒ (1): Let us be given an open subset, say W , of Y . Set V := f −1 W . Then we are asked to show that V is an open subset, i.e., we are asked to show that V is a union of open intervals. In other words we are Klaus Johannson, Topology I §4 Topologies 57 asked to show that every point of V lies in some open interval which in turn is contained in V . To do this let x ∈ V . Then there is an open interval in W which contains f (x) since W is supposed to be an open subset. More precisely, there is an ǫ > 0 such that the open interval (f (x) − ǫ, f (x) + ǫ) ⊂ W . Because of condition (1), there is a δ > 0 such that |z − x| < δ implies |f (z) − f (x)| < ǫ. It follows that the open interval (x − δ, x + δ) is mapped under f into the open interval (f (x) − ǫ, f (x) + ǫ) ⊂ W . In particular, this shows that the point x lies in the open interval (x−δ, x+δ) which in turn lies in f −1 V = W . ♦ Observe that the collection of all open sets in R forms a topology in the following sense. Definition. Let X be any set (finite or not). Then a collection X of subsets of X is called a topology for X if the following holds: (1) ∅ ∈ X and X ∈ X . (2) X is closed under taking finite intersections. (3) X is closed under taking arbitrary unions. A set together with a topology is a topological space. Klaus Johannson, Topology I 58 . Topology I The elements of the topology X of X are called open subsets of X. A subset A ⊂ X is called closed if its complement X − A is open. Remark. Strictly speaking, a topological space X is a pair (X, X ) of X with a topology X (and in this way a topological space is a generalization of a simplicial complex). However, one often does not list the topology X explicitly. Instead one often simply writes X instead of the more correct (X, X ) especially if it is clear from the context what X is. 4.2. Definition. Two topologies X1 , X2 of a set X are equal if X1 = X2 , i.e., if the open sets in one topology are also open in the other topology. The topology X1 is called finer (resp. coarser) than X2 if X1 ⊃ X2 (resp. X1 ⊂ X2 ). We now take a quantum leap forward and generalize the definition of continuous maps as follows. 4.3. Definition. Let (X, X ) and (Y, Y) be topological spaces. Then a map f : X → Y is a continuous map if the preimage of every open set is open, i.e., if f −1 U ∈ X , for every U ∈ Y. A homeomorphism Klaus Johannson, Topology I §4 Topologies 59 is a bijection h : X → Y such that h as well the inverse h−1 is a continuous map. Two spaces X, Y are homeomorphic (notation: X ∼ = Y ) if there is a homeomorphism h : X → Y . By what we have seen above this definition of continuity includes as special case the familiar continuous maps of the real line. However, we also have to be aware that the definition is so general that it has to be taken with a grain of salt. For instance, it may be surprising that not every continuous bijection is a homeomorphism. But it is actually quite easy to construct counterexamples. For this take any set X and equip it with the discrete or the indiscrete topology. Then the identity but not its inverse is continuous. Here the discrete topology of X is the power set of X, i.e., the collection of all subsets of X. The indiscrete topology of X is the topology which consists only of the empty set and the set X. A slightly more typical example is the map f : [0, 1) → S 1 , t 7→ (cos(2πt), sin(2πt)). It is clear that this map is a continuous bijection but not a homeomorphism (why?). Extending the range, we get the map f¯ : [0, 1) → R2 , t 7→ (cos(2πt), sin(2πt)). This is an example of a map which is an embedding but not a homeomorphism onto its image. Embeddings which Klaus Johannson, Topology I 60 . Topology I are homeomorphisms onto their image are also known as imbeddings. To illustrate the power of the definition on this level we give a topological proof of a well-known theorem in elementary number theory. The proof is due to Furstenberg [Amer. Math. Monthly 62 (1955)]. 4.4. Theorem. numbers. There are infinitely many prime Proof. We first introduce a topology P on the set Z of integers (the so called profinite topology) by setting U ∈ P :⇔ for every a ∈ U there is a t ∈ Z − 0 with a + tZ ⊂ U . (here tZ denotes the set tZ := { tx ∈ Z | x ∈ Z }. We claim P forms a topology of Z. For this we have to verify the three conditions of a topology. Now, it is clear that ∅, Z ∈ P. Moreover, if U1 , U2 ∈ P and if a ∈ U1 ∩ U2 , then there are t1 , t2 ∈ Z with a + t1 Z ⊂ U1 , a + t2 Z ⊂ U2 . It then follows that a + (t1 t2 )Z ⊂ U1 ∩ U2 , and so U1 ∩ U2 ∈ P. Finally, Klaus Johannson, Topology I §4 Topologies 61 it is obvious that arbitrary unions of sets from P are sets in P. To continue the proof observe that tZ, Z − tZ ∈ P, i.e,, tZ is open and closed in the profinite topology. Of course, tZ is open since (ta) + tZ S ⊂ tZ. To see that tZ is closed, note that Z − tZ = s6=t sZ, i.e., the complement is a union of open sets and so open. We are now ready to finish the proof. Assume the subset P ⊂ Z S of all prime numbers in Z is finite. S Then the union p∈P pZ S is closed. Hence Z − p∈P pZ is open. But Z − p∈P pZ = {±1} and {±1} is not open in the profinite topology. This contradiction finishes the proof. ♦ Remark. The construction used in the previous proof is much more general. Indeed, using normal subgroups N in groups, instead of tZ in Z, one can define a profinite topology for every group. Open and Closed Sets. Recall that a subset Y ⊂ X of a topological space (X, X ) is called an open subset iff it is an element of the topology of X, i.e., iff Y ∈ X . It is called a closed subset if its complement is an open subset. An open subset U ⊂ X is called an open neighborhood Klaus Johannson, Topology I 62 . Topology I for a set A ⊂ X if A ⊂ U (the subset A may be just a single point). Let Y ⊂ X be any subset of the topological space (X, X ), Then we denote by o [ int(Y ) := Y := {U | U ∈ X and U ⊂ Y } \ cl(Y ) := Ȳ := {X − U | U ∈ X and Y ⊂ X − U }. ∂Y := Ȳ − Y o We say Y o is the interior, Ȳ is the closure and ∂Y is the boundary of Y . Thus the interior of a set Y is the union of all open subsets contained in Y and the closure of Y is the intersection of all closed subsets containing Y . We say a point x ∈ X is a limit point (or an accumulation point) for Y if, for every open neighborhood U of x, the intersection U ∩ ( Y − {x} ) is non-empty. We denote Klaus Johannson, Topology I §4 Topologies 63 Y ′ := the set of all limit points of Y . Remark. Keep in mind that a point x need not be a limit point of a set Y if all its open neighborhoods meet Y . For this we have to make sure that they meet Y in points other than x ! 4.5. Theorem. Y o is an open set and Ȳ is a closed set. Moreover, Ȳ = Y ∪ Y ′ . Proof. Y o is an open set since any union of open sets is open. To show that ȲT is closed, it sufficies to show that any intersection A∈A A of closed T sets is closed. But this follows from the fact X − A∈A A = S A∈A (X − A) is open. Next, we show that Y ∪ Y ′ ⊂ Ȳ . So let x ∈ Y ∪ Y ′ . If x ∈ Y , then of course x ∈ Ȳ . Hence let x ∈ Y ′ . Then every open neighborhood U (x) of x intersects Y (in a point other than x). In particular, x cannot lie in X − Ȳ . For otherwise, there is at least one closed set A ⊂ X with Y ⊂ A and x 6∈ A. But X − A is open since A is closed. Hence there is an open neighborhood of x which does not meet A and so not Y . Contradiction. This proves Y ∪ Y ′ ⊂ Ȳ . Klaus Johannson, Topology I 64 . Topology I For the other direction let x ∈ Ȳ . We wish to show that x ∈ Y ∪ Y ′ . If x ∈ Y , then there is nothing to show. So let x ∈ Ȳ − Y and let U (x) be an open neighborhood of x in X. Then U (x) must meet Y , for otherwise the complement of U (x) is a closed set containing Y but not x. This is impossible since x ∈ Ȳ and since Ȳ is the intersection of all closed sets containing Y . It therefore follows that every x ∈ Ȳ − Y has an open neighborhood that meets Y in a different point than x. ♦ Remark. Note that Y o ⊂ Y ⊂ Ȳ . Note further that Y is open iff Y o = Y and it is closed iff Ȳ = Y . The closure of sets gives us the following useful characterization of continuous maps. 4.6. Theorem. Let f : X → Y be a map between topological spaces X, Y . Then f is continuous if and only if f (Ā) ⊂ f (A), for all subsets A ⊂ X. Klaus Johannson, Topology I §4 Topologies 65 Proof. Suppose f is continuous. Let x ∈ Ā. If f (x) ∈ f (A), we are done. So assume f (x) ∈ Y − f (A). Let U be an open neighborhood of f (x). Then f −1 U is an open neighborhood of x. Hence it contains a point, say z, of A different from x. Then f (z) is a point in f (A), different from f (x) (recall f (x) ∈ Y − f (A)), which lies in U ∩ f (A). This implies that f (x) lies in the closure of f (A). Suppose f (Ā) ⊂ f (A), for all subsets A ⊂ X. Let B ⊂ Y be any closed subset of Y . To show that f is continuous it suffices to show that the preimage A := f −1 B is closed. For this it is enough to show that Ā ⊂ A (because then Ā = A). Now, x ∈ Ā implies f (x) ∈ f (Ā) ⊂ f (A) ⊂ B̄ = B. Hence x ∈ f −1 B = A. ♦ The notion of a limit point allows some more terminology (and in this way some better questions that can be asked about topological spaces). 4.7. Definition. Let X be a topological space and let A ⊂ X be a subset. Then (1) A is dense in X if Ā = X, Klaus Johannson, Topology I 66 . Topology I (2) A is nowhere dense if (Ā)o = ∅, complement of Ā is dense in X, i.e., the (3) A is perfect if A = A′ , (4) a point x ∈ A is isolated if it is open, i.e., there is an open neighborhood U of x in X with U ∩ {x} = {x}. Remark. To see the equivalence in (2) recall that C( int(A) ) = cl( C(A) ) and C( cl(A) ) = int( C(A) ), where C(A) denotes the complement. Hence C(∅) = X = cl( C(cl(A)) ) = C( int( cl(A)) ) ) and the claim follows. Remark. The word ”limit point” is perhaps slightly misleading. One says that a sequence {xn }n∈N of points in a topological space X converges to a point x ∈ X if every open neighborhood U of x contains all but finitely many xn . But it is not true that every limit point x has a sequence {xn }n∈N converging to it. As an example consider R with the topology given by ∅ and all sets opne in R and containing 0. Then every point x ∈ R − {0} is a limit point of the set A = {0}, but no sequence in A converges to x. Klaus Johannson, Topology I §4 Topologies 67 Construction of Topologies. The definition of a topological space is rather abstract and not very easy to work with. In general, it is out of the question to list all open subsets of a topology although this is sometimes possible (e.g., for topologies of finite sets). In praxis, however, one describes a topology differently, namely through a basis or even a subbasis. In fact, we have given the topology of R (see above) through a basis. 4.8. Definiton. A basis of a topology is a collection B of subsets of a set X with (1) every open set of the topology is a union of basis elements. (2) if B1 , B2 are basis elements and if x ∈ B1 ∩ B2 , then there is a basis element B3 such that x ∈ B3 ⊂ B1 ∩ B2 . A collection of subsets of a set X is a subbasis if its union equals X. We turn a subbasis into a basis by taking all finite intersections. We generate a topology from a basis by taking all unions. So, for instance, the collection of all open intervals is a basis as well as a subbasis. Klaus Johannson, Topology I 68 . Topology I Remark. Note that the concept of subbasis makes it now a trivial matter to construct topologies for a set X. For this we simply give ourselves any collection A of subsets of X whose union is X. This already defines a unique topology, given by first taking all intersections of elements of A and then all unions of the resulting subsets. The topology obtained in this way is called the topology generated by the subbasis A. Here are some less obvious ways of constructing topologies. Product Topology. Let (X, X ), (Y, Y) be two topological spaces. Then notice that B := { U × V | U ∈ X , V ∈ Y } is not a topology. It satisfies all axioms of a topology except that it is not closed under taking unions. So it is a basis. The topology generated by this basis is called the product topology for X × Y . A set in the product topology is open if every point of it has an open neighborhood of the form U × V . Given two topological spaces X, Y , it is understood that X × Y denotes the topological space with the product topology. Klaus Johannson, Topology I §4 Topologies 69 Quotient Topology. Given a surjective map ϕ : X → Y between a topological space (X, X ) and a set Y , the quotient topology for Y (with respect to ϕ) is the topology given by Y := { U | ϕ−1 (U ) ∈ X }. The topological space (Y, Y) with the quotient topology is often called the quotient space and denoted by X/ϕ. Remark. Note that the map ϕ defines an equivalence relation on X by setting x ∼ y ⇔ ϕ(x) = ϕ(y). Vice versa, an equivalence relation ∼ on the space X gives rise to a surjective map ϕ : X → Y = X/ ∼, x 7→ [x] which maps every element x to its equivalence class. It is the quotient map induced by the equivalence relation ∼. The notation ”quotient space” is due to the fact that quotient spaces are usually defined via equivalence relations. Klaus Johannson, Topology I 70 . Topology I The quotient topology reflects the fact that points are being identified. Here are two typical examples. Example. Let R be the set of real numbers with the Euclidean topology. Define an equivelance relation ∼ on the unit interval I := [0, 1] = { x ∈ R | 0 ≤ x ≤ 1 } by setting x ∼ y ⇔ x = y, or x = 0 and y = 1. Let ϕ be the induced quotient map. Then I/ϕ is homeomorphic to the unit circle S 1 . Example. Let again R be the set of real numbers with the Euclidean topology. Let d : R → R, x 7→ x + 1, be the translation by 1. Then d induces an equivalence relation ∼ by setting x ∼ y ⇔ y = dn (x), for some n ∈ N Let ϕ be the induced quotient map. Then R/ϕ is again the unit circle S 1 . Klaus Johannson, Topology I §4 Topologies 71 Subspace Topology. Last but not least we define the subspace topology for a subset Y ⊂ X of a topological space (X, X ) to be the topology Y := { U ∩ Y | U ∈ X }. Note that Y is an open set in the topological space (Y, Y) even if Y is a closed set in (X, X ). Some Examples of Topologies. The mere definition of a topology is not very restrictive and therefore it covers a lot of territory. One basically only needs to check that it is closed under taking intersections and unions. So before one goes overboard inventing new topologies one should be aware of the fact that most of them are pretty useless unless we can say what we need them for. One aspect of a topology that one may want to keep in mind is that a topology decides which sequences of elements in a set converge. More precisely, we define 4.9. Definition. Let (X, X ) be a topological space. Let x ∈ X be a point. We say a sequence xn , xn ∈ X, converges to x if, for every open neighborhood U (x) of x there is an index N > 0 such that xn ∈ U (x), for all n > N. Klaus Johannson, Topology I 72 . Topology I Remark. Note that convergence is defined here without the use of any ǫ and δ. Example. In the indiscrete topology every sequence converges to every point. In the discrete topology only the eventually constant sequences converge. Here are some typical examples for topological spaces. Example. As a first example consider the finite product Rn = R×. . .×R. If n ≥ 2, there are actually two different ways of putting a topology on Rn . The first is to start with the Euclidean toplogy of R and iterate the construction of the product topology. More precisely, one defines recursively the topology for Rn+1 to be the product topology of Rn × R. The second way, uses the fact that an open interval cannot only be described by the order relation <, but also by a distance function. The Euclidean distance is given by p d(x, y) := (x1 − y1 )2 + . . . + (xn − yn )2 . Given the Euclidean distance function, one sets U (x, ǫ) := { y ∈ Rn | d(x, y) < ǫ }, Klaus Johannson, Topology I §4 Topologies 73 for every x ∈ Rn and every ǫ > 0. One says that U (x, ǫ) is the ǫ-neighborhood for x. It is easily verified that the set B := { U (x, ǫ) | x ∈ Rn , ǫ > 0 } of all ǫ-neighborhoods defines a basis. This basis in turn defines a topology for Rn . Which of the two topologies defined so far should be called the Euclidean topology of Rn ? Fortunately, we do not have to make a decision here. Both topologies are the same. For this one only has to show that every point in every box (a1 , b1 ) × . . . × (an , bn ) has an ǫ-neighborhood contained in the box, and that every point in every epsilon neighborhood lies in a box (a1 , b1 ) × . . . × (an , bn ) contained in the the ǫ-neighborhood. But this is easy to verify. Example. The Euclidean topology of Rn induces a topology on the set Matm (R) of all m × m-matrices with m2 = n. Indeed, the entries of an m×m matrix define a unique point in Rn and vice versa. So there is a bijection ϕ : Matm (R) → Rn if m2 = n. Define the topology X of Matm (R) by setting X := { ϕ−1 (U ) | U open in Rn }. Klaus Johannson, Topology I 74 . Topology I In this topology two matrices are ”close” if all their entries are close. Thus a sequence An ∈ Matm (R) of matrices in Matm (R) converge to a matrix A ∈ Matm (R) if and only if (aij )n → aij for the entries of the respective matrices. Example. The Euclidean topology of Rn induces a topology for finitely generated matrix groups. For instance, consider SLm (R) = { A ∈ Matm (R) | det(A) = 1 }. Then SLm (R) is not only a set of matrices but a space of matrices with respect to the topology induced by the topology of Matm (R). Moreover, SLm (R) is a group with respect to matrix multiplication, because of the product formula for determinants. Now, let G ⊂ SLm (R) be a finitely generated group of matrices. Let A1 , . . . , Ag ∈ G be the finitely many matrices that generate G. By definition, this means that every matrix in G is a product of the matrices A1 , . . . , Ag . The set Hom(G, SLm (R)) := { ϕ : G → SLm (R) | ϕ is a homomorphism } Klaus Johannson, Topology I §4 Topologies 75 can be viewed as a subset of R(ng) where n = m2 , because ϕ is determined by the g matrices ϕ(A1 ), . . . , ϕ(Ag ) which in turn can be identified with points in Rn . Thus Hom(G, SLm (R)) is a subspace of some Euclidean space and therefore is a topological space itself with respect to the subspace topology. Example. Let Map(X, Y ) be the set of all maps f : X → Y from X to Y . Given that Y is a topological space, the sets S(x, U ) := { f : X → Y | f (x) ∈ U }, for all open sets U ⊂ Y and all points x ∈ X, defines a subbasis. The topology of Map(X, Y ), generated by this subbasis, is called the topology of pointwise convergence. It is one of several useful topologies for function spaces. Example. Let R be a commutative ring (e.g., a polynomial ring). Then denote spec(R) := the set of all prime ideals in R. The Zariski topology of spec(R) is given by specifying closed sets (rather than open sets). More precisely, a set C ⊂ spec(R) is closed in the Zariski Klaus Johannson, Topology I 76 . Topology I topology if there is an ideal A ⊂ R such that C = {P | P is prime ideal with A ⊂ P }. The complements of the closed sets are the open sets in the Zariski topology. Properties of Topological Spaces. For future reference we need to list some more properties of topological spaces. 4.10. Definition. Let X Then be a topological space. (1) X is disconnected if it is the disjoint union X = A ∪ B of two open and non-empty subspaces A, B ⊂ X, (2) X is connected, if it is not disconnected. (3) X is totally disconnected, if the only connected subspaces are the one-point sets, (4) X is path-connected if, for any two points x, y ∈ X, there is a continuous map (path) w : [0, 1] → X with w(0) = x and w(1) = y. Klaus Johannson, Topology I §4 Topologies 77 A subspace C ⊂ X is a component of X if it is connected and if it equals every connected subspace containing it. Remark. Components are closed since the closure of a component is a component. However, in general, a component is not open. Components are of course open if X has only finitely many components since then the complement is a finite union of closed sets. 4.11. Example. Consider the following subspace K in R2 with the subspace topology. a b c The intervals [a, c) and (c, b] are components. They are closed but not open. Moreover, there is no separation K = U ∪ V with a ∈ U and b ∈ V and yet a, b Klaus Johannson, Topology I 78 . Topology I do not lie in a component of K. Here a separation of K is a disjoint union K = U ∪ V of non-empty, open subsets U, V . Next we list some separation axioms. We say that two disjoint subsets A, B ⊂ X in a topological space X can be separated if there are open neighborhoods U (A), U (B) of A, B, respectively, with U (A) ∩ U (B) = ∅. Given this notation we next define: T1: A topological space X is a T1-space if every subset with a single element is closed (short: all points are closed). T2: A topological space X is a T2-space if any two distinct points in X can be separated. A T2-space is also called a Hausdorff space. T3: A topological space X is a T3-space if any point and any disjoint, closed set can be separated. A T3-space is also called a regular space. T4: A topological space X is a T4-space if any two disjoint, closed sets can be separated. A T4-space is also called a normal space. Klaus Johannson, Topology I §4 Topologies Hausdorff regular 79 normal 4.12. Theorem. Let X be a T1-space. Then (1) X is a Hausdorff space if and only if the diagonal ∆ := {(x, x) | x ∈ X} is closed in the product space X × X. We also have the following sandwich properties: (2) X is a regular space if and only if, for every x ∈ X and every open neighborhood U (x), there is an open subset V ⊂ X with x ∈ V ⊂ V̄ ⊂ U (x). 3) X is a normal space if and only if, for every closed subset A ⊂ X and every open neighborhood U (A), there is an open subset V ⊂ X with A ⊂ V ⊂ V̄ ⊂ U (A). Remark. In general, V 6= V̄ . However, in the next section we will see a sandwich property with V = V̄ . Klaus Johannson, Topology I 80 . Topology I Proof. ad (1) Let x, y be two distinct points in X. Then (x, y) is a point in X × X which does not lie on the diagonal ∆ of X × X. If ∆ is closed, there is an open box U × V ⊂ X × X (with U, V ⊂ X open) such that (x, y) ∈ U × V and U × V ∩ ∆ = ∅. This means that x ∈ U, y ∈ V and that there is no point (u, v) ∈ U × V with u = v. Thus U, V are disjoint and open neighborhoods of x, y, respectively. Similarly, one shows that if U, V are disjoint and open neighborhoods of x, y, respectively, then U × V is an open neighborhood of (x, y) that does not meet the diagonal ∆. This proves (1) ad (2) Suppose X is regular and let x ∈ X. Recall that X is also supposed to be a T1-space. So the one-point set {x} forms a closed set. Moreover, the complement X − U (x) is a closed set. Hence there are open neighborhoods U, V with x ∈ U and X − U (x) ⊂ V such that U ∩ V = ∅. It follows that X − V ⊂ U (x). Hence we have x ∈ U ⊂ X − V ⊂ U (x). Moreover, Ū ⊂ X − V since X − V is closed. Thus, altogether, we have x ∈ U ⊂ Ū ⊂ U (x). This proves one direction of (2). Klaus Johannson, Topology I §4 Topologies 81 For the other direction of (2), let x ∈ X be a point and let A ⊂ X be a closed subset disjoint to x. Then U (x) := X − A is an open neighborhood of x. Hence there is an open set V ⊂ X with x ∈ V ⊂ V̄ ⊂ U (x). In particular, U := X − V̄ is an open neighborhood of A which does not meet the open neighborhood V of X. This proves that X is regular. ad (3) The proof for (3) follows the same line of argument. Simply replace the point x by the set A and argue as before. ♦ Appendix: The Cantor Set. The discussion of topological spaces has been rather dry so far. It is time to add some meat. For this let us consider a concrete example, the Cantor set. The Cantor set has some strange properties that help us to get a first idea of the intricacies of certain topological spaces that come up quite naturally. We begin with the definition of the Cantor set. Basically, the Cantor set is the set C := Y {0, 2} = {0, 2}N n∈N Klaus Johannson, Topology I 82 . Topology I of 02-sequences together with a certain topology. This topology is the subspace topology given by the embedding f : C → [0, 1], c1 c2 c3 . . . 7→ X cn . n∈N 3n Thus the topology of C is given by the sets f −1 (U ), U open in [0, 1]. Here is another view of the Cantor set which helps to visualize its topology. Denote by Cn := {c1 c2 . . . cn cn+1 . . . | c1 , . . . , cn ∈ {0, 2} and ci ∈ {0, 1, 2}, for i ≥ n + 1 }, i.e., Cn is the set of all infinite words in the alphabet {0, 1, 2} with theSfirst n letters different from 2. Then of course C = n Cn , i.e., the sets Cn ”converge” to C. The images of Cn under the embedding f are the following intervals: Klaus Johannson, Topology I §4 Topologies 83 C0 C1 C2 C3 4.13. Theorem. Let C be the Cantor space. Then (1) C is perfect. (2) C is totally disconnected. (3) C is homogeneous, i.e., for any two points x, y ∈ C there is a homeomorphism h : C → C with h(x) = y. (4) C is not homeomorphic to [0, 1]. Proof. ad (1). We are asked to show that C ′ = C. Let c = c1 c2 c3 c4 . . . ∈ C be a point in C. Then the sequence c1 000 . . . , c1 c2 00 . . . , c1 c2 c3 0 . . . , c1 c2 c3 c4 . . . Klaus Johannson, Topology I 84 . Topology I is a sequence of points in C different from c that converges to c. It follows that C ⊂ C ′ . To prove C ′ ⊂ C, let x be a limit point of C. Then x is a limit point of Cn , for every n ≥ 1. But Cn is closed as a finite T union of closed intervals. Thus x ∈ Cn and so x ∈ Cn = C. ad (2). Let U be a connected subset of C. Then U ⊂ [0, 1]. Let a = inf U and b = sup U be the greatest lower bound resp. the least upper bound of U . Then every point x ∈ (a, b) lies in U , for otherwise the connected set U can be written as a disjoint union (U ∩(a, x))∪(U ∩(x, b)) of non-trivial open sets which is impossible. Thus U is an interval. But every interval in C lies in Cn , for all n. Hence such an interval has length 3−n , for all n. It follows that U has length 0. This means that U is a set which consists of a single point. ad (3). We follow an argument given in [Rosemann, Elementary Topology]. Here is the idea. Suppose x, y ∈ C are two arbitrary points x = x1 x2 x3 . . . and y = y1 y2 y3 . . . . Klaus Johannson, Topology I §4 Topologies 85 Then define a homeomorphism hxy : C → C, p = p1 p2 p3 . . . 7→ hxy (p) = c1 c2 c3 . . . according to the rule ci = pi , if xi = yi pi + 2 (mod2), otherwise Here is an example x = 220202222 . . . y = 220022020 . . . p = 020202020 . . . hxy (p) = 020022222 . . . Check that h(x) = y. For instance, x = 220202222 . . . y = 220022020 . . . x = 220202222 . . . hxy (x) = 220022020 . . . For the general case note that hxy (X) = y, because digits common to x and y are left unchanged under Klaus Johannson, Topology I 86 . Topology I h, while digits of x different from the corresponding y-digit are changed (necessarily into the one for y). It remains to show that f is a homeomorphism (Exercise). ad (4). C cannot be an interval since C is totally disconnected and an interval is connected. ♦ We finish this section by showing that the Cantor set is not only bijective but also homeomorphic to the product C × C. The product C 2 = C × C is also known as the Sierpinski dust: Klaus Johannson, Topology I §4 Topologies 87 4.14. Theorem. Let C be the Cantor set. Then there is a homeomorphism h : C → C 2 . Proof. (We here follow an argument from [Sieradski, An introduction into topology and homotopy]). Define a map h : C → C 2, c1 c2 c3 c4 . . . 7→ (h1 (x), h2 (x)) = (c1 c3 c5 . . . , c2 c4 c6 ..). Klaus Johannson, Topology I 88 . Topology I We have the following for the distances of points x = x1 x2 x3 . . . , y = y1 y2 y3 . . . in C and image points h(x), h(y) in C 2 : dist(x, y) < 3−2n ⇔ x, y ∈ C2n (see above for the definition of Cn ) ⇔ xi = yi , for all 1 ≤ i ≤ 2n ⇔ h1 (x), h1 (y) as well as h2 (x), h2 (y) lie in a component of Cn ⇔ maximal distance of h(x) and h(y) is smaller than ( 13 )n . It follows that h and its inverse are continuous. ♦ Remark. In contrast to the previous theorem we have that R is not homeomorphic to the product R × R = R2 . To see this note that R−{0} is disconnected but R2 −{p} is connected, for every p ∈ R2 . On the other hand, we also have D ∼ = D × D, for every discrete space D. So the Cantor set is more like a discrete space than R. Is the Cantor set homeomorphic to a discrete space? For the answer see the next lecture. Klaus Johannson, Topology I