Topology I – Problem Set Two Fall 2011
... (b) A space X is Hausdorff if and only if each net converges to at most one point. (c) A function f : X → Y is continuous if and only if for every net g converging to a point x ∈ X, the net f ◦ g converges to f (x). 2. Let X be a T1 topological space. Prove that X is countably compact (as defined in ...
... (b) A space X is Hausdorff if and only if each net converges to at most one point. (c) A function f : X → Y is continuous if and only if for every net g converging to a point x ∈ X, the net f ◦ g converges to f (x). 2. Let X be a T1 topological space. Prove that X is countably compact (as defined in ...
Midterm Exam Solutions
... X/∼, which is connected by hypothesis, giving a contradiction. (4) Consider the set R with the finite-complement topology. (This is the topology in which the nonempty open sets are exactly the sets with finite complement.) Answer each of the following questions about this topological space, and give ...
... X/∼, which is connected by hypothesis, giving a contradiction. (4) Consider the set R with the finite-complement topology. (This is the topology in which the nonempty open sets are exactly the sets with finite complement.) Answer each of the following questions about this topological space, and give ...
G13MTS: Metric and Topological Spaces Question Sheet 5
... 9. Choose one of the sets A, B, C from Question 4 above. Prove that your choice is a connected subset of R2 . 10. Let E be a connected subset of a topological space X, and let F be a subset of X satisfying E ⊆ F ⊆ E. Prove that F is connected. (Hint: use the closed sets version of connectedness). 11 ...
... 9. Choose one of the sets A, B, C from Question 4 above. Prove that your choice is a connected subset of R2 . 10. Let E be a connected subset of a topological space X, and let F be a subset of X satisfying E ⊆ F ⊆ E. Prove that F is connected. (Hint: use the closed sets version of connectedness). 11 ...
Assignment 6
... c) Any two connected components are either equal or disjoint. The space is partitioned into its connected components. The space is connected if and only if it has only one connected component. d) The same statements as above with connected replaced by path connected. e) The closure of a connected s ...
... c) Any two connected components are either equal or disjoint. The space is partitioned into its connected components. The space is connected if and only if it has only one connected component. d) The same statements as above with connected replaced by path connected. e) The closure of a connected s ...
Tutorial Sheet 3, Topology 2011
... Note that the associated topology, by the definition of basis, is defined as follows: a set is open if it can be written as the union of basis elements. (Or if it is the empty set.) Note that each member of β is open by definition. The complement is (−∞, a) ∪ [b, ∞), which can be written as the unio ...
... Note that the associated topology, by the definition of basis, is defined as follows: a set is open if it can be written as the union of basis elements. (Or if it is the empty set.) Note that each member of β is open by definition. The complement is (−∞, a) ∪ [b, ∞), which can be written as the unio ...
Shape-From
... Intensity is a function of reflectance, and reflectance is a function of surface normals (p,q) and light source direction ...
... Intensity is a function of reflectance, and reflectance is a function of surface normals (p,q) and light source direction ...
Final - UCLA Department of Mathematics
... Problem 2. Let K, L be disjoint compact sets in a normal topological space X. Suppose that f : K → R and g : L → R are bounded and continuous functions on K, L respectively. Show that there exists a bounded continuous function h : X → R such that h(x) = f (x) for all x ∈ K and h(x) = g(x) for all x ...
... Problem 2. Let K, L be disjoint compact sets in a normal topological space X. Suppose that f : K → R and g : L → R are bounded and continuous functions on K, L respectively. Show that there exists a bounded continuous function h : X → R such that h(x) = f (x) for all x ∈ K and h(x) = g(x) for all x ...
Topology I – Problem Set Five Fall 2011
... is non-abelian, contains elements of infinite order, and that its center is trivial. 3. Let G , H, G0 , and H 0 be cyclic groups of orders m, n, m0 , and n0 respectively. If G ∗ H is isomorphic to G0 ∗ H 0 then m = m0 and n = n0 or else m = n0 and n = m0 . 4. Let M and N be n-dimensional manifolds, ...
... is non-abelian, contains elements of infinite order, and that its center is trivial. 3. Let G , H, G0 , and H 0 be cyclic groups of orders m, n, m0 , and n0 respectively. If G ∗ H is isomorphic to G0 ∗ H 0 then m = m0 and n = n0 or else m = n0 and n = m0 . 4. Let M and N be n-dimensional manifolds, ...
Solution - Stony Brook Mathematics
... 1. Determine the connected components of the space R` . Solution. Every connected component of R` consists of just one point, because no subset X ⊆ R` with at least two points can be connected. Indeed, if a, b ∈ X are distinct points, say with a < b, then U = X ∩ (−∞, b) and V = X ∩ [b, ∞) are disjo ...
... 1. Determine the connected components of the space R` . Solution. Every connected component of R` consists of just one point, because no subset X ⊆ R` with at least two points can be connected. Indeed, if a, b ∈ X are distinct points, say with a < b, then U = X ∩ (−∞, b) and V = X ∩ [b, ∞) are disjo ...
Existence of partitions of unity
... p ∈ Vβ ∩ (Wj+2 /Wj−1 ) for some β. Take a chart Up contained this open set and let f be a bump function which is identically 1 on an neighbourhood Np of p and whose support is within this chart. Now as p ranges over Wj+2 /Wj−1 , the Np cover Wj+1 /Wj so by compactness we can take a finite subcover. ...
... p ∈ Vβ ∩ (Wj+2 /Wj−1 ) for some β. Take a chart Up contained this open set and let f be a bump function which is identically 1 on an neighbourhood Np of p and whose support is within this chart. Now as p ranges over Wj+2 /Wj−1 , the Np cover Wj+1 /Wj so by compactness we can take a finite subcover. ...