0
$\begingroup$

I am trying to understand Preimage orientation. So I got this question:

Definition. The boundary of $X$, consists of those points that belong to the image of the boundary of $\mathbf{H}^k$, the upper half-space $\mathbf{H}^k$ in $\mathbb{R}^k$, under some local parametrization.

So there's the problem - then a single point seems is the boundary, therefore is not boundaryless?

$\endgroup$
  • $\begingroup$ What is $H^k$? For the definition I'm used to seeing, it depends on the topology of the set. $\endgroup$ – Clayton Jul 26 '13 at 23:19
  • $\begingroup$ Hi @Clayton, it is Hyperbolic Space. $\endgroup$ – WishingFish Jul 26 '13 at 23:23
  • $\begingroup$ Based on the terminology being used, am I correct in guessing that $X$ is a manifold? (Or at least a Hausdorff space?) $\endgroup$ – Dan Jul 26 '13 at 23:35
  • 2
    $\begingroup$ $H^0$ is a one-point space. So the boundary of it is empty, as $H^0 = \mathbb R^0$. $\endgroup$ – Ryan Budney Jul 26 '13 at 23:43
  • 1
    $\begingroup$ You really should put your questions in context, otherwise they're just extremely confusing and hard to properly answer. $\endgroup$ – tomasz Jul 27 '13 at 0:52
1
$\begingroup$

A single point is a one-dimensional polytope, and is entirely of content, without boundary.

The problem comes when you try to divide space with an equal-sign (ie upper and lower half, or $x \gt a$ vs $x \lt a$), when dividing a point is meaningless. Therefore a point exists in an undividable space, and since a boundary is a division, a point cannot have a boundary.

The definition ought imply that $k>0$.

$\endgroup$
  • $\begingroup$ I think it's a matter of convention. If you have a manifold with boundary and take a submanifold, for well-behavedeness purposes you sometimes want the boundary of the submanifold to be its intersection with boundary of the larger manifold. In this case, if the submanifold is a singleton, the single point might or might not be in the boundary... $\endgroup$ – tomasz Jul 27 '13 at 12:19
  • $\begingroup$ If you take say, a face of a dodecahedron (which is part of the dodeca's boundary), then the face belongs to a 2d space, and not all parts of the pentagon belong to the boundary of the pentagon. In short, simply because a point was part of a surface in a higher dimension, it does not mean it has a surface when considered by itself. $\endgroup$ – wendy.krieger Jul 28 '13 at 1:53
  • $\begingroup$ Hi Wendy, thanks so much for your generous help. Can I take it in this way? A single point is $\mathbb{R}^0$. Since there is no upper half-space of $\mathbb{R}^0$, there is no boundary for a single point. $\endgroup$ – WishingFish Jul 28 '13 at 19:37
  • $\begingroup$ Hi @tomasz, I have been thinking over your comment. Do you mind writing more of your thoughts, like when the singleton is the boundary, and when is not? Or even better, put as an answer? Thank yoU~ $\endgroup$ – WishingFish Jul 28 '13 at 19:45
  • $\begingroup$ @WishingFish: Actually, I'm not sure of what I've written before. Boundary should be a submanifold of codimension $1$, and there's no such thing for points. I suppose there might be cases where a convention where a point is its own boundary might be more elegant, but I don't think this is interesting enough to try and find them. Manifolds of dimension zero are of little interest on their own, anyway. $\endgroup$ – tomasz Jul 28 '13 at 19:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.