Lengths of simple loops on surfaces with hyperbolic metrics Geometry & Topology G
... systems. Our main result, theorem 1.1, establishes Lipschitz type estimates for the length pairing expressed in terms of these coordinates. As a consequence, we give a new proof of a result of Thurston–Bonahon ([13], see [2, proposition 4.5] for a proof) that the length pairing extends to a continuo ...
... systems. Our main result, theorem 1.1, establishes Lipschitz type estimates for the length pairing expressed in terms of these coordinates. As a consequence, we give a new proof of a result of Thurston–Bonahon ([13], see [2, proposition 4.5] for a proof) that the length pairing extends to a continuo ...
Connected and hyperconnected generalized topological spaces 1
... Definition 2.3 3Let gX and gY be generalized topologies on X and Y , respectively. Then a function f : X → Y is said to be (α, gY )-continuous if for each g−open set U in Y , f −1 (U ) is g − α-open in X, (σ, gY )-continuous if for each g−open set U in Y , f −1 (U ) is g−semi-open in X (π, gY )-cont ...
... Definition 2.3 3Let gX and gY be generalized topologies on X and Y , respectively. Then a function f : X → Y is said to be (α, gY )-continuous if for each g−open set U in Y , f −1 (U ) is g − α-open in X, (σ, gY )-continuous if for each g−open set U in Y , f −1 (U ) is g−semi-open in X (π, gY )-cont ...
Lecture Notes - Mathematics
... Given a collection of polynomials defining an affine algebraic variety, it is not at all obvious, a priori, whether or not the variety is irreducible, or how many components it has.2 For hypersurfaces, it is easy to describe the irreducible components, as well as choose a “nice” defining equation: E ...
... Given a collection of polynomials defining an affine algebraic variety, it is not at all obvious, a priori, whether or not the variety is irreducible, or how many components it has.2 For hypersurfaces, it is easy to describe the irreducible components, as well as choose a “nice” defining equation: E ...
Chapter 9 The Topology of Metric Spaces
... Exercise 19 Prove that in R2 there exists a countable family B of open balls which form a basis for all the open sets in the topology of the space: every open set O in the topology of R2 can be written as the (necessarily countable) union of sets from B. Hint: prove that the set of all balls whose c ...
... Exercise 19 Prove that in R2 there exists a countable family B of open balls which form a basis for all the open sets in the topology of the space: every open set O in the topology of R2 can be written as the (necessarily countable) union of sets from B. Hint: prove that the set of all balls whose c ...
Math 54 - Lecture 18: Countability Axioms
... Definition A space X is second-countable if it has a countable basis for its topology, say U1 , . . .. That is, given any open set U and point x ∈ U , there is Un ⊂ U with x ∈ Un . Note: any second-countable space is first-countable, as we can take for a basis at x ∈ X the sequence of all Un which c ...
... Definition A space X is second-countable if it has a countable basis for its topology, say U1 , . . .. That is, given any open set U and point x ∈ U , there is Un ⊂ U with x ∈ Un . Note: any second-countable space is first-countable, as we can take for a basis at x ∈ X the sequence of all Un which c ...
Exterior Angle Theorem from 4.1 1 A B C
... Polygon Exterior Angles Theorem The sum of the measures of the exterior angles of a convex polygon, one angle at each vertex is 360. ...
... Polygon Exterior Angles Theorem The sum of the measures of the exterior angles of a convex polygon, one angle at each vertex is 360. ...
Polygon
... you have connected all vertices (lines can’t overlap) 2nd: The quadrilateral has 4 sides and is made up of 2 triangles 3rd: A triangle has 180 4th: Therefore the sum of the interior angles is (2 triangles x 180) for a sum of 360 Diagonal ...
... you have connected all vertices (lines can’t overlap) 2nd: The quadrilateral has 4 sides and is made up of 2 triangles 3rd: A triangle has 180 4th: Therefore the sum of the interior angles is (2 triangles x 180) for a sum of 360 Diagonal ...
Normality on Topological Groups - Matemáticas UCM
... Corollary 4.2? This can be formulated as: Question 4.4. (a) Let (G, τ ) be a nonnormal topological group such that G endowed with the Bohr topology is neither normal. Are all topologies on G which admit the same dual group G∧ also nonnormal? (b) The same question of (a) changing nonnormal, by normal ...
... Corollary 4.2? This can be formulated as: Question 4.4. (a) Let (G, τ ) be a nonnormal topological group such that G endowed with the Bohr topology is neither normal. Are all topologies on G which admit the same dual group G∧ also nonnormal? (b) The same question of (a) changing nonnormal, by normal ...
Bounded negativity of Shimura curves
... 1. Shimura curves on Shimura surfaces not isogeneous to a product An Shimura surface not isogeneous to a product is a connected algebraic surface that can be written as a quotient X = Γ\G/K, where G = GQ (R) is the set of R-valued points in a connected semisimple Q-algebraic group GQ , where K ⊂ G i ...
... 1. Shimura curves on Shimura surfaces not isogeneous to a product An Shimura surface not isogeneous to a product is a connected algebraic surface that can be written as a quotient X = Γ\G/K, where G = GQ (R) is the set of R-valued points in a connected semisimple Q-algebraic group GQ , where K ⊂ G i ...
Hyperbolic geometry 2 1
... It’s convenient to regard the points of the Cartesian plane as complex numbers. Thus H2 consists of all complex numbers with positive imaginary part. We’ve already seen one example of an isometry, namely inversion in a geodesic. If the geodesic is taken to be the y -axis, the effect of this isometry ...
... It’s convenient to regard the points of the Cartesian plane as complex numbers. Thus H2 consists of all complex numbers with positive imaginary part. We’ve already seen one example of an isometry, namely inversion in a geodesic. If the geodesic is taken to be the y -axis, the effect of this isometry ...
The Topological Version of Fodor`s Theorem
... By the local compactness of X then there is a bounded open set U in X with C1 ∩ C2 ⊂ U . However, then C1 \ U and C2 \ U would be two disjoint members of C(X), contradicting that X is good. Definition 2. Let X be a locally compact, noncompact T2 space. We say that a set S ⊂ X is stationary in X if i ...
... By the local compactness of X then there is a bounded open set U in X with C1 ∩ C2 ⊂ U . However, then C1 \ U and C2 \ U would be two disjoint members of C(X), contradicting that X is good. Definition 2. Let X be a locally compact, noncompact T2 space. We say that a set S ⊂ X is stationary in X if i ...