I am writing a short report on de Rham cohomology, and I'm approaching it from a geometric perspective, much like (and with reference to) this article (written by a MSX member) http://www3.nd.edu/~lnicolae/shape.pdf
In it it's stated that k-chains have boundaries which are k-1 chains. Here is how I get confused:
I am using the definition $\partial C = \overline{C}\setminus\mathrm{Interior(C)}$. If I have then 1-chain $I=(0,1)\subset \mathbb{R}$ then $\partial I = \overline{I}\setminus\mathrm{Interior(I)} = [0,1]\setminus (0,1) = \{0,1\} = \{0\}\cup\{1\}$, which is a union of 0-chains as I should get.
But what about $I\subset \mathbb{R}^2$? Now I'm sure that $\overline{I}=[0,1]$ as before but this time $I$ has no interior, because now any open ball centred on the line will of course contain points off the line.. so $\partial I = [0,1]$? What am I misunderstanding here?
You might want to think of k-chains as smooth maps (possibly embeddings) $\phi: S_k\rightarrow M$ (or $\mathbb{R}^n$)where $S_k$ is a standard $k$-dimensional simplex, instead of subsets in $M$ or $\mathbb{R}^n$. The boundary is then the restriction of the map to the boundary. If you need to think about sets in $\mathbb{R}^n$ consider the image of these maps. ($S_1$ would correspond to your interval $I$ and it's boundary consists of two points. The boundary operator reduces dimension).