Homework 5 (pdf)
... (3) Let X be a topological sapce. Show that X is a T1 -space if and only if each point of X is a closed set. (4) Give a direct proof that a metric space (X, d) is Hausdorff. (Do not for example use the fact that a metric space is a T3 -space and every T3 -space is a T2 -space.) (5) Let f : X → Y be ...
... (3) Let X be a topological sapce. Show that X is a T1 -space if and only if each point of X is a closed set. (4) Give a direct proof that a metric space (X, d) is Hausdorff. (Do not for example use the fact that a metric space is a T3 -space and every T3 -space is a T2 -space.) (5) Let f : X → Y be ...
LOYOLA COLLEGE (AUTONOMOUS), CHENNAI – 600 034
... b) State and prove Urysohn’s embedding theorem. OR c) State and prove Tietze extension theorem. IV a) i) Prove that the product of any non-empty class of connected spaces is connected. OR a) ii) Let X be a Hausdorff space. If X has an open base whose sets are also closed then prove that X is totally ...
... b) State and prove Urysohn’s embedding theorem. OR c) State and prove Tietze extension theorem. IV a) i) Prove that the product of any non-empty class of connected spaces is connected. OR a) ii) Let X be a Hausdorff space. If X has an open base whose sets are also closed then prove that X is totally ...
MATH4530–Topology. PrelimI Solutions
... Q1-3: Find all Hausdorff spaces. (11pts) Solution: (1) Not Hausdorff: Every two open sets would intersect non-trivially in an infinite set with f.c. topology. (2) Hausdorff: Any subspace of a Hausdorff space is Hausdoff. (3) Hausdorff: Any subspace of a Hausdorff space is Hausdoff. (4) Hausdorff: An ...
... Q1-3: Find all Hausdorff spaces. (11pts) Solution: (1) Not Hausdorff: Every two open sets would intersect non-trivially in an infinite set with f.c. topology. (2) Hausdorff: Any subspace of a Hausdorff space is Hausdoff. (3) Hausdorff: Any subspace of a Hausdorff space is Hausdoff. (4) Hausdorff: An ...
Local compactness - GMU Math 631 Spring 2011
... These notes discuss the same topic as Section 29 of Munkres’ book. Definition 1. A topological space X is called locally compact at a point x ∈ X if there exists a neighborhood Ux 3 x such that Ux is compact. X is called locally compact if X is compact at every point. An equivalent definition: X is ...
... These notes discuss the same topic as Section 29 of Munkres’ book. Definition 1. A topological space X is called locally compact at a point x ∈ X if there exists a neighborhood Ux 3 x such that Ux is compact. X is called locally compact if X is compact at every point. An equivalent definition: X is ...
VI. Weak topologies
... Then κ(X) is a subspace of (X ) separating points of X ∗ , hence the topology σ(X ∗ , κ(X)) is Hausdorff. It is called the weak* topology of X ∗ , it is denoted by σ(X ∗ , X) or by w∗ . (3) Let Γ be a noenmpty set and let the space FΓ be equipped by the product topology (cf. Example V.1(2)). The pro ...
... Then κ(X) is a subspace of (X ) separating points of X ∗ , hence the topology σ(X ∗ , κ(X)) is Hausdorff. It is called the weak* topology of X ∗ , it is denoted by σ(X ∗ , X) or by w∗ . (3) Let Γ be a noenmpty set and let the space FΓ be equipped by the product topology (cf. Example V.1(2)). The pro ...
Homework Set #2 Math 440 – Topology Topology by J. Munkres
... that same chain will be a chain from x to z, so that z ∈ O as well. Thus, UN ⊆ O, and every point of O is an interior point. Note that we do not even need the cover to be open, just that the interiors of the sets cover X. Closed: Suppose that y ∈ / O, and let U be an element from U that covers y. Su ...
... that same chain will be a chain from x to z, so that z ∈ O as well. Thus, UN ⊆ O, and every point of O is an interior point. Note that we do not even need the cover to be open, just that the interiors of the sets cover X. Closed: Suppose that y ∈ / O, and let U be an element from U that covers y. Su ...
Topology 640, Midterm exam
... 1. Let (X, T ) be a topological space. Recall that a subset A ⊆ X is called dense if A ∩ U 6= ∅ for every non-empty open set U ⊆ X. (a) Show that A is dense if and only if A = X. (b) Show that if A1 and A2 are open dense subsets of X then A1 ∩ A2 is also a dense subset of X. (c) Show by example that ...
... 1. Let (X, T ) be a topological space. Recall that a subset A ⊆ X is called dense if A ∩ U 6= ∅ for every non-empty open set U ⊆ X. (a) Show that A is dense if and only if A = X. (b) Show that if A1 and A2 are open dense subsets of X then A1 ∩ A2 is also a dense subset of X. (c) Show by example that ...
Topology (Maths 353)
... Theorem. Let f : X → Y be a map where X is a set, Y is a topological space. Denote f ∗ OY := {U ⊂ X | U = f −1 (V ) where V ∈ OY }. Then: (1) f ∗ OY is a topology on X, called the induced topology (more precisely, the topology induced by f ); (2) f : (X, f ∗ OY ) → (Y, OY ) is continuous; (3) f ∗ OY ...
... Theorem. Let f : X → Y be a map where X is a set, Y is a topological space. Denote f ∗ OY := {U ⊂ X | U = f −1 (V ) where V ∈ OY }. Then: (1) f ∗ OY is a topology on X, called the induced topology (more precisely, the topology induced by f ); (2) f : (X, f ∗ OY ) → (Y, OY ) is continuous; (3) f ∗ OY ...
Compact groups and products of the unit interval
... [0,1]W(G) as a quotient space. 1-1. Definitions and Notation (a) A finite subset {xltx2,..., xk) of a group G is said to be independent if it does not contain the identity, 1, of G and x">x"2... x£* = 1, for integers nx,..., nk, implies that x"1 = x"' = ... = x£* = 1. An infinite subset X of a group ...
... [0,1]W(G) as a quotient space. 1-1. Definitions and Notation (a) A finite subset {xltx2,..., xk) of a group G is said to be independent if it does not contain the identity, 1, of G and x">x"2... x£* = 1, for integers nx,..., nk, implies that x"1 = x"' = ... = x£* = 1. An infinite subset X of a group ...
Categorically proper homomorphisms of topological groups
... In fact, we not only extend but slightly generalize the known object-level results since, unlike the authors of [12, 19] and of most papers on topological groups, we do not assume a priori that the topologies of our objects must be Hausdorff. The reason for this, however, is not the aim for generali ...
... In fact, we not only extend but slightly generalize the known object-level results since, unlike the authors of [12, 19] and of most papers on topological groups, we do not assume a priori that the topologies of our objects must be Hausdorff. The reason for this, however, is not the aim for generali ...
IOSR Journal of Mathematics (IOSR-JM)
... Proof: Let x and y be any two distinct points of X. Then f x and f y are different points of Y because f is injective. Since Y is Hausdorff, there exist disjoint open sets U and V in Y containing f x and f y respectively. Since f is continuous and U ∩ V = ϕ, f −1 U and f −1 V are disjoint open sets ...
... Proof: Let x and y be any two distinct points of X. Then f x and f y are different points of Y because f is injective. Since Y is Hausdorff, there exist disjoint open sets U and V in Y containing f x and f y respectively. Since f is continuous and U ∩ V = ϕ, f −1 U and f −1 V are disjoint open sets ...
More on Semi-Urysohn Spaces
... R and S containing x and y respectively. Note that S and X \ R are the required sets from the definition of U-open sets. Thus X \ {x} is U-open or equivalently {x} is U-closed. (4) ⇒ (2) Let x ≠ y . By (4), X \ {x} is U-open and hence there exists a regular closed set V and a regular open set U such ...
... R and S containing x and y respectively. Note that S and X \ R are the required sets from the definition of U-open sets. Thus X \ {x} is U-open or equivalently {x} is U-closed. (4) ⇒ (2) Let x ≠ y . By (4), X \ {x} is U-open and hence there exists a regular closed set V and a regular open set U such ...
TOPOLOGY 2 - HOMEWORK 1 (1) Prove the following result
... f ∗ is bijective (ie, one-to-one and onto) then f ∗ is a homeomorphism. (Hint: For the second part, prove and then use the following fact: any continuous map from a compact space X to a Hausdorff space Y takes closed subsets of X to closed subsets of Y .) (2) Prove that S 1 × S 1 is homeomorphic to ...
... f ∗ is bijective (ie, one-to-one and onto) then f ∗ is a homeomorphism. (Hint: For the second part, prove and then use the following fact: any continuous map from a compact space X to a Hausdorff space Y takes closed subsets of X to closed subsets of Y .) (2) Prove that S 1 × S 1 is homeomorphic to ...
Math 541 Lecture #1 I.1: Topological Spaces
... If X has the trivial topology, then the closure of any point x ∈ X is all of X, and any real-valued continuous function defined on X is constant. If X has the discrete topology, then the closure of any point x ∈ X is itself, and any real-valued function defined on X is continuous. A function from on ...
... If X has the trivial topology, then the closure of any point x ∈ X is all of X, and any real-valued continuous function defined on X is constant. If X has the discrete topology, then the closure of any point x ∈ X is itself, and any real-valued function defined on X is continuous. A function from on ...
Topology Semester II, 2015–16
... Question 1. Show that if a topological space X has a countable basis {Bn }, then every basis C of X contains a countable basis for X. Answer. Let B := {Bn | n ∈ N} be a basis for the topological space X. Since C is a basis of X, for every m ∈ N there is an element C ∈ C such that C ⊂ Bm . Similarly, ...
... Question 1. Show that if a topological space X has a countable basis {Bn }, then every basis C of X contains a countable basis for X. Answer. Let B := {Bn | n ∈ N} be a basis for the topological space X. Since C is a basis of X, for every m ∈ N there is an element C ∈ C such that C ⊂ Bm . Similarly, ...
Math 6210 — Fall 2012 Assignment #3 1 Compact spaces and
... i (∞i ) cover X if and only if i∈I Xi = ∅. Exercise 14. Let F be a collection of subsets of a set X. Show that F is an ultrafilter if and only if it satisfies the following beautiful condition:5 If X is the disjoint union of subsets A1 , . . . , An then there is exactly one i such that Ai ∈ F . 1 Co ...
... i (∞i ) cover X if and only if i∈I Xi = ∅. Exercise 14. Let F be a collection of subsets of a set X. Show that F is an ultrafilter if and only if it satisfies the following beautiful condition:5 If X is the disjoint union of subsets A1 , . . . , An then there is exactly one i such that Ai ∈ F . 1 Co ...
Manifolds
... spaces (the other is function spaces). Much of your work in subsequent topology courses (in particular 22M:133, 201, 200, 203) focuses on manifolds. (In Analysis and Differential Equations course you will study a lot of function spaces). The solution sets of equations are almost always manifolds. Fo ...
... spaces (the other is function spaces). Much of your work in subsequent topology courses (in particular 22M:133, 201, 200, 203) focuses on manifolds. (In Analysis and Differential Equations course you will study a lot of function spaces). The solution sets of equations are almost always manifolds. Fo ...
Weak-continuity and closed graphs
... (P) For each (x, y) ф G(f), there exist open sets U c X and V c Y containing x and y, respectively, such that f(U) n Intľ(Clľ(V)) = 0. Proof. Let (x, y) ф G(f)9 then y Ф /(x). Since Уis Hausdorff, there exist disjoint open sets Vand JVcontaining y and/(x), respectively. Thus, we have Int^Clj^V)) n n ...
... (P) For each (x, y) ф G(f), there exist open sets U c X and V c Y containing x and y, respectively, such that f(U) n Intľ(Clľ(V)) = 0. Proof. Let (x, y) ф G(f)9 then y Ф /(x). Since Уis Hausdorff, there exist disjoint open sets Vand JVcontaining y and/(x), respectively. Thus, we have Int^Clj^V)) n n ...
The Hausdorff topology as a moduli space
... could certainly have been obtained by Hausdorff himself at that time. What stopped that from happening has to do with the history of mathematics. Before Grothendieck’s influence, it would not have been common practice to look for a “modular interpretation” of such constructions. In fact, one could i ...
... could certainly have been obtained by Hausdorff himself at that time. What stopped that from happening has to do with the history of mathematics. Before Grothendieck’s influence, it would not have been common practice to look for a “modular interpretation” of such constructions. In fact, one could i ...
Mid-Term Exam - Stony Brook Mathematics
... V 1 ×V 2 is closed in X ×Y , since its complement [(X −V 1 )×Y ]∪[X ×(Y −V 2 )] is open. Hence (x, y) ∈ V1 × V2 ⊂ V1 × V2 ⊂ V 1 × V 2 ⊂ W1 × W2 ⊂ W and so U = V1 × V2 fulfills our requirement, and X × Y is regular, as claimed. Remark. In fact, V1 × V2 = V 1 × V 2 , but you do not need this here. 9. ...
... V 1 ×V 2 is closed in X ×Y , since its complement [(X −V 1 )×Y ]∪[X ×(Y −V 2 )] is open. Hence (x, y) ∈ V1 × V2 ⊂ V1 × V2 ⊂ V 1 × V 2 ⊂ W1 × W2 ⊂ W and so U = V1 × V2 fulfills our requirement, and X × Y is regular, as claimed. Remark. In fact, V1 × V2 = V 1 × V 2 , but you do not need this here. 9. ...
Algebra II — exercise sheet 9
... isomorphism of varieties. Extend this to an example of a morphism of projective varieties with the same properties. Solution: The map is given by polynomials, so it is regular. It is bijective, because for any point (x, y) ∈ X, we have y 2 = x3 and (with K algebraically closed), there are two square ...
... isomorphism of varieties. Extend this to an example of a morphism of projective varieties with the same properties. Solution: The map is given by polynomials, so it is regular. It is bijective, because for any point (x, y) ∈ X, we have y 2 = x3 and (with K algebraically closed), there are two square ...
Topology, MM8002/SF2721, Spring 2017. Exercise set 3 Exercise 1
... • Show that if X and Y are Hausdorff, so is X × Y . • Show that if X and Y are first countable, so is X × Y . • Show that if X and Y are second countable, so is X × Y . Which of these statements hold for all finite products? Which for arbitrary products? Exercise 7. Consider an arbitrary product of ...
... • Show that if X and Y are Hausdorff, so is X × Y . • Show that if X and Y are first countable, so is X × Y . • Show that if X and Y are second countable, so is X × Y . Which of these statements hold for all finite products? Which for arbitrary products? Exercise 7. Consider an arbitrary product of ...
MATH0055 2. 1. (a) What is a topological space? (b) What is the
... (b) What does it mean to say that a topological space is connected ? Let X, Y be connected topological spaces and A ( X, B ( Y proper subsets. Prove that X × Y \ A × B is connected. [You may assume without proof that X is homeomorphic to any X × {y}, y ∈ Y , and similarly for Y .] ...
... (b) What does it mean to say that a topological space is connected ? Let X, Y be connected topological spaces and A ( X, B ( Y proper subsets. Prove that X × Y \ A × B is connected. [You may assume without proof that X is homeomorphic to any X × {y}, y ∈ Y , and similarly for Y .] ...
Relations on topological spaces
... Theorem I. If X is a compact Hausdorff space and if P is a non-void closed transitive relation on X then there is at least one P-minimal element. The p r o o f of this, as one would expect, is not difficult and we indicate the train of the reasoning. A P-chain (for any relation P on any space X) is ...
... Theorem I. If X is a compact Hausdorff space and if P is a non-void closed transitive relation on X then there is at least one P-minimal element. The p r o o f of this, as one would expect, is not difficult and we indicate the train of the reasoning. A P-chain (for any relation P on any space X) is ...
Part II - Cornell Math
... (x, y). We can show that U ∩ G f = ∅ which proves that G f is closed: let (a, b) ∈ U. Then f (a) ∈ U f (x) . Since b ∈ Uy , b , f (a). (b) Let X be any topological space that is not Hausdorff and let f : X → X be the identity map on X. Then by HW3 question 1, we know that the diagonal ∆ = G f is not ...
... (x, y). We can show that U ∩ G f = ∅ which proves that G f is closed: let (a, b) ∈ U. Then f (a) ∈ U f (x) . Since b ∈ Uy , b , f (a). (b) Let X be any topological space that is not Hausdorff and let f : X → X be the identity map on X. Then by HW3 question 1, we know that the diagonal ∆ = G f is not ...
Felix Hausdorff
Felix Hausdorff (November 8, 1868 – January 26, 1942) was a German mathematician who is considered to be one of the founders of modern topology and who contributed significantly to set theory, descriptive set theory, measure theory, function theory, and functional analysis.