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- Thread starter Laur
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In what sense do you mean equals? Diffeomorphic?

I would think do. The disjoint union of two manifold is a manifold, right? But any neighborhood has only one connected component but the whole manifold has two.

ETA: Assuming both pieces are of the same dimension.

I would think do. The disjoint union of two manifold is a manifold, right? But any neighborhood has only one connected component but the whole manifold has two.

ETA: Assuming both pieces are of the same dimension.

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By "equals", i mean that the neighbourhood is dense in the manifold.

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quasar987

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A sub question would be: if a manifold does have a handle body decomposition is it the union of an open cell and its boundary? This should not be too hard.

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Yes!

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.

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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.

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Thinking a bit more about my idea for a proof, the following statement can be proven

- if S is a nonempty collection of subsets of the manifold which are each homeomorphic to R^{n}, and totally ordered (by inclusion), then the union of S is also homeomorphic to R^{n}.

This can be done by applying the generalized http://en.wikipedia.org/wiki/Jordan%E2%80%93Sch%C3%B6nflies_theorem" [Broken] to 'glue' together the different homeomorphisms. It follows that any open subset homeomorphic to R^{n} is contained in a maximal such subset.

The question of whether such a maximal euclidean neighbourhood has closure equal to the entire manifold still remains.

- if S is a nonempty collection of subsets of the manifold which are each homeomorphic to R

This can be done by applying the generalized http://en.wikipedia.org/wiki/Jordan%E2%80%93Sch%C3%B6nflies_theorem" [Broken] to 'glue' together the different homeomorphisms. It follows that any open subset homeomorphic to R

The question of whether such a maximal euclidean neighbourhood has closure equal to the entire manifold still remains.

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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.

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Well, the unit open ball in R

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.

Your example shows that the boundary of such a neighborhood can be a figure of eight, but it isn't maximal.

Showing that a Euclidean neighborhood whose closure isn't the entire manifold can be enlarged sounds difficult. This is because the boundary of the neighbourhood can be very complicated, and fractal-like.

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