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Class Notes for Math 871: General Topology, Instructor Jamie Radcliffe

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Class Notes for Math 871: General Topology, Instructor Jamie Radcliffe University of Nebraska - Lincoln DigitalCommons@University of Nebraska - Lincoln Math Department: Class Notes and Learning Materials Mathematics, Department of 2010 Class Notes for Math 871: General Topology, Instructor Jamie Radcliffe Laura Lynch University of Nebraska-Lincoln, [email protected] Follow this and additional works at: https://digitalcommons.unl.edu/mathclass Part of the Science and Mathematics Education Commons Lynch, Laura, "Class Notes for Math 871: General Topology, Instructor Jamie Radcliffe" (2010). Math Department: Class Notes and Learning Materials. 10. https://digitalcommons.unl.edu/mathclass/10 This Article is brought to you for free and open access by the Mathematics, Department of at DigitalCommons@University of Nebraska - Lincoln. It has been accepted for inclusion in Math Department: Class Notes and Learning Materials by an authorized administrator of DigitalCommons@University of Nebraska - Lincoln. Class Notes for Math 871: General Topology, Instructor Jamie Radcliffe Topics include: Topological space and continuous functions (bases, the product topology, the box topology, the subspace topology, the quotient topology, the metric topology), connectedness (path connected, locally connected), compactness, completeness, countability, filters, and the fundamental group. Prepared by Laura Lynch, University of Nebraska-Lincoln August 2010 1 Topological Spaces and Continuous Functions Topology is the axiomatic study of continuity. We want to study the continuity of functions to and from the spaces C; Rn;C[0; 1] = ff : [0; 1] ! Rjf is continuousg; f0; 1gN; the collection of all in¯nite sequences of 0s and 1s, and H = 1 P 2 f(xi)1 : xi 2 R; xi < 1g; a Hilbert space. De¯nition 1.1. A subset A of R2 is open if for all x 2 A there exists ² > 0 such that jy ¡ xj < ² implies y 2 A: 2 In particular, the open balls B²(x) := fy 2 R : jy ¡ xj < ²g are open. This de¯nition of open lets us give a new de¯nition of continuity, other than the basic ²¡± de¯nition. To do this, ¯rst we need to de¯ne the preimage of a function: If f : X ! Y and A ⊆ Y; then the preimage is f ¡1(A) = fx 2 Xjf(x) 2 Ag: Lemma 1.2. A function f : R2 ! R2 is continuous if and only if the preimage of every open set is open. Proof. First suppose f is continuous. Consider an open A ⊆ R2: We want to show f ¡1(A) is open. Let x 2 f ¡1(A) (if ¡1 f (A) = ;; done). As A is open, there exists ² > 0 such that B²(f(x)) ⊆ A: By continuity, there exists ± > 0 such that ¡1 ¡1 jy ¡ xj < ± implies jf(y) ¡ f(x)j < ²; that is, f(y) 2 B²(f(x)) ⊆ A: Thus y 2 f (A); which implies B±(x) ⊆ f (A): Therefore, f ¡1(A) is open. 2 Now suppose the preimage of every open set is open. Note that for all x and for all ² that B²(x) is open. Let x 2 R and ¡1 ¡1 ² > 0 be given. By hypothesis, f (B²(f(x)) is open. So there exists ± such that B±(x) ½ f (B²(f(x))); that is, for all y with jy ¡ xj < ±; we have f(y) 2 B²(f(x)); which says jf(y) ¡ f(x)j < ²: Properties of Open Sets in R2: ² Arbitrary unions of open sets are open. ² Finite intersections of open sets are open. We want to abstract this notion to a more general setting. To do so, consider the following de¯nition. De¯nition 1.3. Given a set X; a topology on X is a collection ¿ 2 P(X) such that ² For all A ⊆ ¿; we have [O2AO 2 ¿: ² For all O1;O2 2 ¿; we have O1 \ O2 2 ¿: ² ;;X 2 ¿: Given a topology ¿ on X; we call the sets in ¿ open or ¿ ¡ open and we call the pair (X; ¿) a topological space. Examples. The following are topologies. 1. The usual topology on Rn. 2. The discrete topology on X; where ¿ = P(X): 3. The indiscrete topology on X; where ¿ = f;;Xg: 4. Let X = C[0; 1]: Then we can de¯ne ¿ by saying A is open if and only if for all f 2 A there exists ² > 0 such that for all g 2 X with supx2[0;1] jf(x) ¡ g(x)j < ² we have g 2 A: 2 2 5. The Zariski topology on R ; where A ½ R is open if there exists polynomials f1; :::; fn 2 C[x; y] such that A = C f(x; y)jfi(x; y) = 0; i = 1; :::; ng : 6. The co¯nite topology on X; where ¿ = fA : jAC j < 1g [ f;g: Examples. The following are continuous functions. 1. If R is the topological space of the set R with the usual topology, then f : R ! R is continuous if and only if for all x 2 R and for all ² > 0 there exists ± > 0 such that jy ¡ xj < ± implies jf(y) ¡ f(x)j < ²: 2. If X is any topological space, then idX : X ! X de¯ned by x 7! x is continuous. 3. If X; Y are arbitrary topological spaces and f : X ! Y is constant (that is, for all x; x 7! y0), then f is continuous. Lemma 1.4. If f : X ! Y; g : Y ! Z are continuous, then so is g ± f : X ! Z: ¡1 ¡1 ¡1 ¡1 Proof. Given A 2 ¿Z ; (g ± f) (A) = f (g (A)): By continuity of g; we have g (A) 2 ¿Y and by continuity of f; we have ¡1 ¡1 f (g (A)) 2 ¿X : De¯nition 1.5. A homeomorphism from X to Y is a bijection f : X ! Y such that both f and f ¡1 are continuous. 1.1 Bases De¯nition 1.6. If B ½ P(X) is a collection of sets satisfying X = [B2BB and for all A; B 2 B and all x 2 A \ B; there exists C 2 B such that x 2 C ⊆ A \ B (that is, for all A; B 2 B;A \ B 2 ¿B), then the topology generated by B is ¿B = fA ⊆ Xj for all x 2 A; there exists B 2 B such that x 2 B ⊆ Ag = f[B2CB : C ½ Bg: In particular, B ⊆ ¿B: We call B a basis for ¿B: Theorem 1.7. ¿B is a topology. Examples. ² The usual topology on R is generated by the basis f(x ¡ ²; x + ²)jx 2 R; ² > 0g = f(a; b): a < bg: ² The discrete topology on X is generated by ffxg : x 2 Xg: ² Bases are NOT unique: If ¿ is a topology, then ¿ = ¿¿ : Theorem 1.8. If f : X ! Y is a function and the topology on Y is generated by B; then f is continuous if and only if f ¡1(B) is open for all B 2 B: ¡1 Proof. We need only to prove the backward direction. So assume f (B) is open for all B 2 B: Consider A 2 ¿B: Then ¡1 ¡1 ¡1 ¡1 A = [B2AB for some A ⊆ B: Now, f (A) = f ([B2AB) = [B2Af (B): As f (B) is open, the union is. De¯nition 1.9. If S ⊆ P(X); then the topology generated by X as a subbasis is the topology farbitrary unions of ¯nite intersections of sets in Sg with basis fS1 \ ¢ ¢ ¢ \ Snjn ¸ 0;Si 2 S; i = 1; ::; ng: [Note: This is a topology, if we consider \; = X]. Theorem 1.10. If f : X ! Y and a topology on Y is generated by a subbasis S; then f is continuous if and only if ¡1 f (S) 2 ¿X for all S 2 S: ¡1 Proof. We need only prove the backward direction. So assume f (S) 2 ¿X for all S 2 S: If A = S1 \ ¢ ¢ ¢ \ Sn is a basic ¡1 n ¡1 ¡1 open set, then f (A) = \i=1f (Si) 2 ¿X : Thus f (A) 2 ¿X for all basic open sets A: Thus f is continuous. 1.2 Product Topology Suppose X1 and X2 are topological spaces. Then X1 £ X2 = f(x; y)jx 2 X1; y 2 X2g: The product topology on X1 £ X2 has basis BX1£X2 = fU £ V : U 2 ¿X1 ;V 2 ¿X2 g: 2 2 Example. In R ; we have two bases: B1 = fB²(x): x 2 R ; ² > 0g and B2 = fU £ V : U; V are open in Rg: In fact, ¿B1 = ¿B2 : To prove this fact, note that B1 ⊆ ¿B2 and vice versa. Theorem 1.11. BX1£X2 is a basis. Proof. X1 £ X2 = X1 £ X2; were X1 2 ¿X1 ;X2 2 ¿X2 : Thus X1 £ X2 2 BX1£X2 : Note that (U1 £ V1) \ (U2 £ V2) = (U1 \ U2) £ (V1 \ V2) 2 BX1£X2 : Lemma 1.12. If ¿ is a topology on X and B ⊆ ¿; then B is a basis for ¿ if and only if 1. [B2BB ¶ X 2. For all A 2 ¿ and for all x 2 A; there exists B 2 B with x 2 B ⊆ A: (Equivalently, for all A 2 ¿; A = [B2CB where C ⊆ B: Proof. Note that for B; B0 2 B;B \ B0 2 ¿: Hence, for all x 2 B \ B0; there exists C 2 B with x 2 C ⊆ B \ B0: So B is a basis for some topology. Since B ⊆ ¿; any union of a subfamily of B belongs to ¿: Thus ¿B ⊆ ¿: Now, condition (2) says A ⊆ ¿ implies A ⊆ ¿B: Thus ¿ = ¿B: Lemma 1.13. If B1; B2 are bases for the topologies on X1;X2; then B1£2 = fB1 £ B2jBi 2 Big is a basis for ¿X1£X2 : 0 0 0 Proof. Certainly B1£2 ⊆ BX1£X2 ⊆ ¿X1£X2 and X1 £ X2 = ([B2B1 B) £ ([B 2B2 B = [B £ B = [B2B1£2 B: Now suppose x 2 A 2 ¿X1£X2 : Then if x = (x1; x2); we see there exists Ui 2 ¿Xi such that (x1; x2) 2 U1 £ U2 ⊆ A: Since Bi are bases, there exists Bi 2 Bi such that xi 2 Bi ⊆ Ui: Thus x 2 B1 £ B2 ⊆ U1 £ U2 ⊆ A: De¯nition 1.14.
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