I think I can see why it's Hausdorff, take two points on the sphere of distance from each other, then 'draw' two open balls around both these points of radius less than , so it should be Hausdorff.
I'm not sure about proving 2nd countability, draw open balls of radius 1 around all the points with rational coordinates?
And I've no idea about proving it's locally Euclidean.
And finally, just to clarify I'm not doing this wrong, then open sets of are the same open sets used in ? (i.e. the open sets as given by the distance metric?)
Feb 3rd 2013, 02:13 AM
Re: Spheres as manifolds
We can view Sn as a (closed) subset of Rn so it automatically inherits hausdorfness and second countabiltiy. To show it is a manifold, you need to cover it by charts. For example, lets take S1. The first chart you would take would be the open upper hemisphere, ie to the top half of the circle minus the points (0, 1) and (0-1). We need to do this to ensure this set is open in the subspace topology. Now as your coordinate function, just take the map f (x,y) = x. This just projects those points down onto the real line, showing this open set is homeomorphic to R. Then you just cover the rest of the circle using similar ideas ( hint, you will need 3 more patches in total). Then you show that the transition functions are smooth, which isnt hard to check. To extend this to S^n isnt too