Is a single point boundaryless?

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?

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

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 $k>0$.
• 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. – WishingFish Jul 28 '13 at 19:37
• @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. – tomasz Jul 28 '13 at 19:51