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Does every (continous and second countable) topological manifold have an Euclidean neighbourhood around each of its points whose closure equals the whole manifold?
Yes!Does every (continous and second countable) topological manifold have an Euclidean neighbourhood around each of its points whose closure equals the whole manifold?
I like that ideaYes!
Consider the set S of all neighbourhoods of a point P which are homeomorphic to the open ball. Try showing that this set has a maximal element (for this you just need to show that the union of a chain in S is again in S). Then that the closure of such a maximal element is the whole manifold.
I haven't worked through the details, but it seems like this argument should work.
Well, the unit open ball in R^{n} is the nested union of the open balls {x:|x|<a} for a < 1. It follows that every Euclidean neighborhood is a nested union of such neighborhoods.Interesting. One can have a nested chain of open disks in a manifold whose union does not have the whole manifold as its boundary.
Take the height function on a torus standing on end on the the plane. From the bottom, f(x) < a, for a small enough is a a nested sequence of open disks that has a figure eight as its boundary and leaves out all of the torus above the first saddle point of f.