## nonvanishing boundary of boundary

For any CW complex, the boundary of the boundary is the empty set. Outside of CW complexes, are there sets for which the boundary of the boundary is nonempty? Examples in one or two dimensions preferred.

EDIT: Based on the comments below I'm trying to make the question more precise:

• The notion of boundary requires a subspace S < X. It is defined by $\partial S = \overline{S} \cap \overline{X \backslash S}$ (taken from Munkres or Wikipedia)
• X should be Hausdorff and have countable basis. The Cantor set, being homeomorphic to {0,1}N is not countable basis.

It could be Hausdorff + countable basis + subspace of Rn are enough to ensure ∂2=0. Originally, I was hoping to find spaces "close" to being CW complexes for which ∂2 ≠ 0.

EDIT: The square of the boundary map should be defined by $\partial^2 S = \overline{\partial S} \cap \overline{S \backslash \partial S}$ - this is the boundary of $\partial S$ relative to S. I am looking for topological subspaces $S \subset X = \mathbb{R}^n$ for which $\partial^2 S \not = \varnothing$. Maybe Hausdorff and countable basis is enough to ensure this can't happen. Certainly this rules out the disk in R2. Perhaps such a counterexample could arise as "limit" of CW complexes (e.g. in the Hausdorff metric).

EDIT:There is still a defect in the above definitions. What definition do you suggest so that the boundary of the boundary of the disk is empty? Can this definition be extended to arbitrary sets? Will the boundary of the boundary still be empty for all sets?

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@Regenbogen, a one point set might be simpler in that case. I was therefore assuming that something else is intended, but I'm not sure what exactly. – Jonas Meyer May 13 2010 at 0:32
Yes, every nonempty nondense subspace will do. But if we're talking about manifolds with boundary, the answer is no. @John Mangual, will you please clarify what type of objects you're interested in and what their boundary means? – Jonas Meyer May 13 2010 at 0:44
John Mangual, you asked the original question, so I was wondering what definition you are using. Boundary for CW complexes and manifolds is very different from topological boundary of subspaces of a topological space. E.g., consider a closed disk in the plane. As a manifold (or subset of R^2), its boundary is a circle. As a manifold, the boundary of the circle is empty, whereas as a subset of the plane the boundary of the circle is the circle. See for example en.wikipedia.org/wiki/Boundary_%28topology%29, en.wikipedia.org/wiki/…. – Jonas Meyer May 13 2010 at 1:09
Your more precise question is still not what you want; Jonas' counterexample of a closed disk in R^2 still works. – Qiaochu Yuan May 13 2010 at 13:33
Am I the only one who thinks that this whole enterprise was founded on the fact that the same word was used for two different concepts? – Yemon Choi May 14 2010 at 8:17