# Is a single point boundaryless?

In the question at MathSE "Is a single point boundaryless?" I saw two contrary answers both make much sense to me. One is by wendy.krieger

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 $k>0$.

Another is by tomasz:

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.

So I am wondering if I can ask for some clarification here? Thank you.

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## migrated from mathoverflow.netJul 28 '13 at 6:56

This question came from our site for professional mathematicians.

The boundary of a point has pure dimension $-1$, and the only manifold of negative dimension is the empty manifold. I voted to closed as off-topic. –  Angelo Jul 28 '13 at 5:50
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## 2 Answers

think your confusion partly lies in not being very clear on what 'points' and boundaries of a space mean. For instance, if we forget about manifolds and such like, and just consider two linear equations over the vector space R2 that intersect once, then the point' is a vector in the vector space R2.

Similarly for other mathematical spaces points in the space have other meanings -- think for example what points' of interest might be in algebraic geometry or function spaces.

A topological manifold X with boundary (of dimension n) is a topological space where for each point in x∈X there exists a neighbourhood homeomorphic to Rn or the closed upper half space in Rn.

Here again we need to be careful -- Let D2 denote the open ball in R2. That is D2={x∈R2|x<1}. As a manifold D2 does not have boundary. As a subspace of R2 we see that D2 has boundary S1.

In short, if X is a topological n--manifold with boundary denoted ∂X, then ∂X is an (n−1)--manifold (with relative topology) such that the boundary of ∂X is empty -- ∂∂X=∅.

Thus to answer your question, you need to decide what sort of space you are in, what are the points and how they are determined, and also what you mean by a boundary. Hope this is helpful, I have tried to be as intuitive as possible.

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Thanks so much the generous user87998. I am really very grateful for your help and I learnt a lot from your wise answer. Thank you. –  WishingFish Jul 28 '13 at 19:23
So in my original question, I followed this definition of boundary: 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. Hence, can I think 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 19:36
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What you should remember is that a 'boundary' is not a condition of a point, but a relation between the point and some larger object.

So the same point, might be on the surface of a cube, but on the interior of one of its square faces. If you come to consider the 'boundary of the square', it's a different thing to the boundary of the cube.

A boundary can be a partition of points outside vs points inside, and supposes this is a non-trivial one (that is there are interior points etc). A point is a singleton, and therefore can not have a boundary in the space it occupies.

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