Talk:Boundary (topology)

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[edit] Boundary of a boundary

Note sure about this section. Consider a closed disk {(x,y) : x^2+y^2<=1}. Here the complement is {(x,y) : x^2+y^2>1} and the closure of the complement is {(x,y):x^2+y^2>=1}. The intersection of the two is {(x,y):x^2+y^2=1} the circle.

I think the fact that the boundary of a boundary of manifolds is the empty set has more to do with soothness and differential constraints rather than the fact that its not a topological space. --Salix alba (talk) 11:23, 28 January 2006 (UTC)

Ah I see what your getting at. Your considering the disk as the lying in R^3, in which case topoligical boundary will be the whole disk. However we should be careful with the notion of embedding. Manifold do not have to be embedded in a particular euclidean space, a manifold != the embedding of a manifold. Consider projective plane often though of as a set in R^4, but works fine as a topological 2-manifold defined by its charts but not necessarily embedded in R^4.

As manifold is now a non technical article maybe we need better definition of manifold with bounday here. --Salix alba (talk) 11:42, 28 January 2006 (UTC)

Heres a simple non rigrious proof of why the boundary of a boundary of a smooth manifold is the empty set. Note this does require the intrinsic view of a manifold, rather than an embedded manifold.

Let M be a smooth n-manifold with boundary and let N be its boundary. Consider a point p on the boundary of M, there is an open set containing p and an open set containing the origin and a chart f:A->B. Now f maps onto which is isomorphic to an open set in R^{n − 1}. The boundary of C is empty. QED. I don't get why the boundary of C is empty. Like, suppose B is the filled upper hemisphere of a sphere. Then C is a disc, which has a nonempty boundary. You only discover the empty boundary when you patch together all the neighborhoods, you cannot discover it locally. The easiest way to prove that the boundary of the boundary of a manifold is empty is to realize that the boundary of the boundary is given in the subspace topology. Every set has empty boundary in the subspace topology. -lethe ^talk 20:54, 28 January 2006 (UTC)

Filled upper hemisphere of sphere is not a smooth (i.e. differential) manifold with boundary the charts around the equator are not diffeomorphic to instead they are diffeomorphic to . --Salix alba (talk) 21:41, 28 January 2006 (UTC)

You're mistaken about that. The filled hemisphere is of course a smooth manifold with boundary. But more importantly to my argument, it is a neighborhood in [0,∞)×R². -lethe ^talk 22:53, 28 January 2006 (UTC)

Looking back at the revison history there was a much better treatment of manifold with boundary a while back, rather than the slighlty confused and possible incorect version we have now. --Salix alba (talk) 18:49, 28 January 2006 (UTC)

You're probably talking about this version. We decided that it rambled on too long, I have a bad habit of doing that sometimes. But why do you think the version we have now is incorrect? -lethe ^talk 20:56, 28 January 2006 (UTC)

I think tere is some stuff which could go back, at very least we need a link to [1] which is probably the best treatment. The link to boundary of a manifold just redirects back to manifold. Currently its saying they are different but we are not going to tell you how. Add the appropriate restrictions (differential manifold?) and topological def of boundary gives the same as boundary of manifold and also implies that the boundary of a manifold with boundary is empty. (sorry fuzzy on my terms, will think later) --Salix alba (talk) 21:41, 28 January 2006 (UTC)

The top half of a filled sphere is an interesting example, highlighting the difference between a topological and differentiable manifold. Viewed as a topological manifold it is a topological ball so all points on the boundary (including those on the equator) are homeomorphic to .

Now if we view it as a differentiable manifold we require that the charts need to be differentiable. Around the equator with its sharp edge differentiability fails for a chart onto open sets in . Hence we need we need charts which map onto open sets in . --Salix alba (talk) 21:50, 29 January 2006 (UTC)

[edit] Clearly Wrong

This section is clearly wrong. Indeed, one can characterize the nowhere dense sets as those sets whose closures are contained in a boundary set. Recall that a set is nowhere dense if the interior of its closure is empty. Since boundary sets are closed, their closure is trivially contained in a boundary set. So every boundary set is nowhere dense. This implies that the interior of a boundary set is empty, again because boundary sets are closed. So the topological boundary operator is in fact idempotent. —The preceding unsigned comment was added by 12.154.214.29 (talk • contribs) 21:49, 17 August 2006 (UTC)

I'm not quite sure what you're trying to say here: specifically, I'm not sure what you mean by "sets whose closures are contained in a boundary set". In any case, the boundary of a set can certainly be dense: for example, in the usual topology of the real line, the boundary of the set of rational numbers is the entire real line, which is obviously everywhere dense. —Ilmari Karonen (talk) 23:26, 18 August 2006 (UTC)

I think I became confused by two distinct but related notions of "boundary sets". The notion I had in mind is that a set $A$ is a boundary set if the closure of its complement cl$(A^c)$ is dense in the ambient space. It is used primarily in Descriptive Set Theory and other related branches of topology and analysis. My confusion explains my difficulties proving a theorem in the field, and is very irksome. —The preceding unsigned comment was added by 12.154.214.29 (talk • contribs) 23:56, 25 August 2006 (UTC2)