As described here (and as I always thought was the most general definition of boundary), a possible definition of the boundary of a subset $S$ of a topological space $X$ is $\partial S = \overline S \backslash \mathrm{int}(S)$.

I am reading Allen Hatcher's book on algebraic topology, and they often refer to "the" boundary of a topological space, for instance by saying that $\partial D^2 = S^1$. How is this notion of boundary uniquely defined? Because if I take an injection of $D^2$ into $\mathbb R^2$, I'm going to get $S^1$, but if I take the injection of $D^2$ into $\mathbb R^3$, then $D^2$ becomes its own boundary. I know that the $\partial$ is an important operation in topology so I'm trying to figure it out, and I feel like it doesn't quite make sense here.

So can anyone describe precisely what is meant by $\partial X$ when $X$ is a topological space? Or at least explain what is meant by that in particular contexts where it is used.

  • $\begingroup$ This is not the topological notion of boundary as you observed. The context you're probably looking for is that of "topological manifolds with boundary". $\endgroup$ Sep 1 '13 at 7:05
  • $\begingroup$ @Anthony : I figured it wasn't the definition of the boundary of a topological space... but I guess I'll look up that topological manifold boundary thing! $\endgroup$ Sep 1 '13 at 7:09
  • $\begingroup$ As you have observed, $\overline{S}\setminus int(S)$ is not intrinsic to $S$, but also depends on $X$. If you view $D^2$ as a subset of itself, then its boundary becomes empty. Therefore in the context of manifolds a different definition is needed. +1 for a good question. $\endgroup$ Sep 1 '13 at 7:19
  • 1
    $\begingroup$ For geodesic metric spaces, there is also a notion of boundary obtained by considering equivalence classes of geodesic rays 'tending to infinity'. $\endgroup$
    – user68316
    Sep 1 '13 at 7:23
  • 1
    $\begingroup$ A more general definition would be to talk about the local homeomorphism type at a point in a topological space. For a manifold with boundary there are only two such points, interior points and boundary points. This has the advantage of making sense in arbitrary topological spaces. $\endgroup$ Sep 3 '13 at 18:30

The notion of boundary that you are looking for comes from the definition of topological manifolds with boundary. As opposed to a regular manifold $X$, a manifold with boundary has the property that each point in $X$ has an open neighborhood which is homeomorphic to an open set in the euclidean half space $\mathbb{R}_+^n=\{(x_1,\dots,x_n)\in\mathbb{R^n}:x_n\ge0\}$. Thus we then define $\partial X$ to be the points which when mapped to $\mathbb{R}_+^n$ have $x_n=0$.

This definition has the benefit that an embedding of $X$ into some other space does not change $\partial X$. Thus $\partial D^2=S^1$ irregardless of whether you view it as living in $\mathbb{R}^2$ or in $\mathbb{R}^3$.

  • 1
    $\begingroup$ When you say "all points in $X$", you mean "each point of $X$ has some neighborhood such that"? $\endgroup$ Sep 1 '13 at 9:57
  • $\begingroup$ @PatrickDaSilva thanks! I have fixed the error now $\endgroup$
    – E.O.
    Sep 1 '13 at 12:34
  • $\begingroup$ This seems to restrict to a very particular kind of space, i.e. those whose boundary is $(n-1)$-dimensional (i.e. in the sense that its boundary has the property that each point has an open neighborhood that maps homeomorphically to an open set in $\mathbb R^{n-1}_+$. I would think, for instance, that in this "manifold with boundary" context, if you see the letter $P$ as a topological space (where the hole in the $P$ is considered as its interior, i.e. $\pi_1(P) = 0$), then the tail of the $P$ would be a part of its boundary, which does not seem the case with your definition. $\endgroup$ Sep 1 '13 at 13:47
  • $\begingroup$ I'm asking because I see this chapter with cell-complexes in my book where they glue a bunch of stuff together and its possible to glue a line to a disk (to get the letter $P$ for instance). How does one go around that in the literature? Or is it even a problem? $\endgroup$ Sep 1 '13 at 13:48
  • $\begingroup$ @PatrickDaSilva you correct in that the boundary is necessarily $(n-1)$-dimensional. For your example though, $P$ is not a manifold. When you map the tail to $\mathbb{R}^2$, it will be a line. But lines are not open in $\mathbb{R}^2$ so $P$ is not a manifold. $\endgroup$
    – E.O.
    Sep 1 '13 at 22:20

The concept of boundary can be extended to the (regular) CW-complexes, as suggested here.

The boundary of a (regular) CW-complex $X$ is : $$\partial X := \overline{\bigcup_{n \ge 0}(\bigcup_{c \in \text{n-cells}} \partial c) / (\bigcup_{c \ne c'\in \text{n-cells}} (\partial c \cap \partial c'))}$$

Definition : The notation "$n$-cells" above, is the set of closed $n$-cells.

Example : Let $X$ be a topological space with the following simplicial complex structure :
enter image description here

All the sets :

  • $0$-cells $=\{ A,B,C,D,E,F \}$
  • $1$-cells $=\{ [A,B],[B,C],[C,D], [D,E],[E,A],[A,F]... \}$
  • $2$-cells $=\{ [A,B,F],[B,C,F],[C,D,F], [D,E,F],[E,A,F]\}$

Now :

  • $\partial A = \partial B = ... = \partial F = \emptyset$
  • $\partial [A,B] = \{A,B \}$ , $\partial [B,C] = \{B,C \}$ , ....
  • $\partial [A,B,F] = [A,B] \cup [B,F] \cup [A,F] $, $\partial [B,C,F] = [B,C] \cup [C,F] \cup [B,F] $, ...

So :

  • $(\bigcup_{c \in \text{0-cells}} \partial c) / (\bigcup_{c \ne c'\in \text{0-cells}} (\partial c \cap \partial c')) = \emptyset$
  • $(\bigcup_{c \in \text{1-cells}} \partial c) / (\bigcup_{c \ne c'\in \text{1-cells}} (\partial c \cap \partial c')) = \emptyset$
  • $(\bigcup_{c \in \text{2-cells}} \partial c) / (\bigcup_{c \ne c'\in \text{2-cells}} (\partial c \cap \partial c')) = (A,B) \cup (B,C) \cup (C,D) \cup (D,E) \cup (E,A)$

Conclusion : $\partial X = [A,B] \cup [B,C] \cup [C,D] \cup [D,E] \cup [E,A]$

Questions : Let $X$ be a topological space admitting a (regular) CW-complex structure :

  • Does $\partial X$ depend on the choice (regular) CW-complex structure ?
  • Can we extend this definition for all the topological spaces ?
  • $\begingroup$ Yeah... I actually hate CW complexes ; in Hatcher's book I feel like it is the most handwavy explanation I ever read. Maybe I am just not a CW complex kind of guy, but this thing is seriously vague to me. I don't understand most of it, nor do I understand your answer... sorry. (It's not your fault.) $\endgroup$ Sep 2 '13 at 2:39
  • $\begingroup$ @PatrickDaSilva : perhaps a good way to understand what does it mean, is by taking an example : look at various simplicial complexes structure (instead of CW complex) on a disk, and see that we obtain its boundary by this process. I hope I have helped you, else, never mind ! $\endgroup$ Sep 2 '13 at 6:37
  • $\begingroup$ Yeah no, didn't help. I think I just need to find various points of view until one of them rings a bell and then I'll figure it out. But thanks for trying, I appreciate it. $\endgroup$ Sep 2 '13 at 16:07
  • $\begingroup$ @PatrickDaSilva : I have edited an explicit example. Maybe it's more understandable now. $\endgroup$ Sep 2 '13 at 18:27
  • 1
    $\begingroup$ @PatrickDaSilva : I warn you that if the local dimension of the space is constant, then we can probably only deal with n-cells (with n the global dimension), as you write and as my example; but if the local dimension is not constant then we need to deal with r-cells with r≤n. I don't know if this definition depends on the choice, that why I ask, but I guess no (with an easy proof). $\endgroup$ Sep 3 '13 at 19:49

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.