0

When dealing with topological manifolds, the definition specifies that for every point in the manifold, there should exist at least one neighborhood around that point that is homeomorphic to an open subset of Euclidean space. However, does this requirement imply that all open neighborhoods around a given point need to be homeomorphic to Euclidean space, or can a point have multiple open neighborhoods with different topological properties?

Ajay Varghese
  • 3
  • Simply, whole space is also an open neighborhood of a point and it is generally not Euclidean. If you want nontrivial example : torus and annuli shaped neighborhood. – ChoMedit Oct 23 '23 at 10:22
  • @ChoMedit If that the case I can take Y curve with topology inherited from standard topology. Then take an open neighborhood of intersection point such that only one branch is in the open neighborhood. Then it would homeomorphic to Euclidean space in R1 – Ajay Varghese Oct 23 '23 at 10:29
  • The point is that not all neighrhood is homeomorphic to Euclidean space. I suggested counterexamples. In fact, if you choose a standard topology on Y-shaped space, then there is no open neighborhood containing only one branch and intersection point! – ChoMedit Oct 23 '23 at 10:49
  • @ChoMedit Are you saying that in torus, annuli shaped neighborhood is not homeomorphic to Euclidean space? – Ajay Varghese Oct 23 '23 at 12:15
  • 1
    @AjayVarghese The upper hemisphere in the sphere is an Euclidean neighbourhood of the north pole. The whole sphere is a non-Euclidean neighbourhood of the north pole. Hence : there exists an Euclidean neighbourhood of the north pole, but not all of its neighbourhoods are Euclidean. – Didier Oct 23 '23 at 13:40
  • @AjayVarghese Yes. That's what I tried to say. Annuli is not contractible and hence not homeomorphic to Euclidean space. – ChoMedit Oct 23 '23 at 14:58

0 Answers0