In 4 space these balls are puffed up into the fourth dimension but one can deflate them so that they reside in 3 space. In the case of the 3 sphere its equator is a regular two dimensional sphere and the two pieces are ordinary solid balls. So all of the boundary points disappear and the sphere has no boundary. When one glues the two large closed balls back together to get the sphere again, these two sets around the equator are pasted along their boundaries to make a small open ball. In the process a small, open ball around a point on the equator is split into two sets that look like small open subsets of Euclidean half space. The answer is no as proves in post #16.Ī Picture: If one slices a n dimensional sphere along its equator, it falls apart into two n dimensional closed balls whose boundaries are the shared equatorial n-1 sphere. This is easily seen from the definition of a manifold with boundary.Ī Caveat: One might ask whether an open neighborhood of a boundary point on a manifold is itself homeomorphic to an open subset of Euclidean space in which case the definition of boundary would be redundant. This manifold does not have a boundary as says in post #5. The manifold-boundary of a manifold is another manifold since each point on the boundary is surrounded by a open subset of the boundary of the Euclidean half space. Using that proof one can start with a parameterization around one point and others can be obtained by following the parameterization by a rotation of the sphere. So a sphere has no boundary as a manifold. So its topological boundary as a subset of Euclidean space and its "manifold boundary" are the same set.Ī sphere is everywhere locally homeomorphic to an open subset of Euclidean space. The Euclidean half space is itself a manifold with boundary. Its boundary is those points that are mapped to the boundary of the Euclidean half space i.e. The topological boundary of a half space in Euclidean space, for instance the points whose x coordinate is non-negative, is those points whose x coordinate is equal to zero.Ī manifold with boundary is defined as locally homeomorphic to a open set of a Euclidean half space of fixed dimension. So it is its own topological boundary as a subset of Euclidean space. The topological boundary of a subset of a topological space is those points which are in its closure that are not in its interior.Īs a subset of Euclidean space a sphere is closed and has no interior. No topological space by itself is a topological boundary since every point in it is an interior point. But it doesn't use embeddings or special cases either. Post #6 has an answer which doesn't use the definition. Even worse, you need the latter to define the former, but they are not the same.
![no boundary no boundary](http://2.bp.blogspot.com/-dokPk3UZXJY/TZxY40egiyI/AAAAAAAAEIs/m4aiwQtqWbg/s1600/sr-71_blackbird.jpg)
![no boundary no boundary](https://i.ytimg.com/vi/uPf3Uu-mVdU/maxresdefault.jpg)
A boundary of a manifold has a certain definition, the boundary of a subset of #(\mathbb^n,\|,\|_p)# has another. This only creates misunderstandings, confusion and teach the wrong facts. Why should you? If you talk about manifolds and boundaries in the same context, then use the correct definition and do not mix two different contexts.
![no boundary no boundary](https://i5.walmartimages.com/asr/5310260a-b564-4c9f-af13-4ee0281ade4c_1.34e1d46138b9f6877a7d87c1ecccc286.jpeg)
If you want to see it like this, never ever use the word manifold. But even as a ball it sends the completely wrong signal to define the topology in the surrounding Euclidean space and speak of boundaries like subsets of that space. The spheres can be defined multiple ways and most of them don't look like a ball.
![no boundary no boundary](http://i.ytimg.com/vi/ppUE0JKLhEc/maxresdefault.jpg)
That would allow it to inherit a topology and a metric (although not the path-length metric along the surface). I believe that a 3-sphere is defined as embeddable in the 4-dimensional Euclidean space.