Lectures on quasi-isometric rigidity
... Note: For every finitely-generated group G there exists a compact Riemannian manifold M (of every dimension ≥ 2) with an epimorphism π1 (M ) → G. Thus, we get another correspondence Groups −→ Metric Spaces: Riemann : G → X = a covering space of some M as above. Thus, we have a problem on our hands, ...
... Note: For every finitely-generated group G there exists a compact Riemannian manifold M (of every dimension ≥ 2) with an epimorphism π1 (M ) → G. Thus, we get another correspondence Groups −→ Metric Spaces: Riemann : G → X = a covering space of some M as above. Thus, we have a problem on our hands, ...
TOPOLOGY WEEK 2 Definition 0.1. A topological space (X, τ) is
... and X. The indiscrete topology is the topology on X only containing ∅ and X. Prove the following. (a) If X has the discrete topology, then every function from X to a topological space Y is continuous. (b) If X does not have the discrete topology, then there is a topological space Y and a function f ...
... and X. The indiscrete topology is the topology on X only containing ∅ and X. Prove the following. (a) If X has the discrete topology, then every function from X to a topological space Y is continuous. (b) If X does not have the discrete topology, then there is a topological space Y and a function f ...
fixed points and admissible sets
... {Si : i ∈ I}, and follows that ∩nk=1 Slk ⊆ ∩nk=1 B(xjk , rjk ). Since the subfamily {Si : i ∈ I} is a totally ordered, ∩nk=1 Slk is nonempty and from previous we can observe that ∩nk=1 B(xjk , rjk ) is a nonempty. From (1) follows that family {B(xj , rj ) : j ∈ J} has nonempty intersection and so S ...
... {Si : i ∈ I}, and follows that ∩nk=1 Slk ⊆ ∩nk=1 B(xjk , rjk ). Since the subfamily {Si : i ∈ I} is a totally ordered, ∩nk=1 Slk is nonempty and from previous we can observe that ∩nk=1 B(xjk , rjk ) is a nonempty. From (1) follows that family {B(xj , rj ) : j ∈ J} has nonempty intersection and so S ...
free topological groups with no small subgroups
... (i) Xz's metrizable; (ii) F (X) is metrizable ...
... (i) Xz's metrizable; (ii) F (X) is metrizable ...
1.2 Inductive Reasoning
... expect on the next telephone key. Look at a telephone to see whether your conjecture is correct. ...
... expect on the next telephone key. Look at a telephone to see whether your conjecture is correct. ...
3-manifold
In mathematics, a 3-manifold is a space that locally looks like Euclidean 3-dimensional space. Intuitively, a 3-manifold can be thought of as a possible shape of the universe. Just like a sphere looks like a plane to a small enough observer, all 3-manifolds look like our universe does to a small enough observer. This is made more precise in the definition below.