# A maybe naive question about the definition of manifolds and the common analogy with real-world atlases

Maybe this question is a little bit naive, but there might be a good explanation. I am reading about manifolds, manifolds of dimension $d$ are defined, for example to be subsets $M \subseteq \mathbb R^n$ such that for every $p \in M$ there exists an open subset $U \subseteq \mathbb R^n$ and an open subset $V \subseteq \mathbb R^d$ such that $U \cap M$ and $V$ are homeomorphic. Then notions as charts (or coordinates) and atlases are introduced.

Often an analogy is brought up to atlases as how we know them from geography classes at school, i.e. books which chart the earth. But in these, the patches that are charted, for example a map of some portion of asia, are rectangles, and rectangles more closely resemble closed sets (or compact sets), i.e. they have a boundary and contain every limit point. So taking this analogy verbal a more suitable definition might be to define a manifold of dimension $d$ to be a subset $M \subseteq \mathbb R^n$ such that every point of $M$ is homeomorphic to a rectangle $[a_1, b_1]\times \ldots \times [a_d, b_d]$ of $\mathbb R^d$. But why is this definition not used? To summarize:

Why we require every point of a $d$-dimensional manifold to have a neighbourhood homeomorphic to an open subset of $\mathbb R^d$ and not (which is more in the spirit of real-world atlases) to require a subset around every point that is homeomorphic to a $d$-fold rectangle of $\mathbb R^d$?

• Your question is why manifold is locally ${\bf R}^n$ ? Or why we use open ball in ${\bf R}^n$ instead of closed ball ? – HK Lee Mar 19 '16 at 11:40
• It is locally $\mathbb R^d$ by definition; my question is more the second part of your comment, why not used closed balls (or closed rectangles as said in the question)? – StefanH Mar 19 '16 at 11:47
• I do not imagine it makes a whole lot of difference whether you use open neighborhoods or closed neighborhoods that contain an open neighborhood, but closed neighborhoods that do not contain an open neighborhood, such as the one containing no additional points lead to non-$\mathbf{R}^n$-like manifolds. – hkBst Mar 19 '16 at 12:31

I suspect that the reason is that when you take two open rectangles in $\mathbb R^n$ and their intersection is nonempty, the intersection contains an open rectangle of dimension $n$. For closed rectangles, consider something like the square $[0, 1] \times [0, 1]$ and the square $[1, 2] \times [0, 1]$. The intersection of these is a close rectangle (namely $\{1\} \times [0, 1]$, but that rectangle is 1-dimensional instead of two-dimensional, so saying things about derivatives in the intersection matching nicely with derivatives on either side gets messy.