# Why are manifolds defined with open sets?

First a small disclaimer that I have been introduced to manifolds but I am not extremely comfortable with them in the general case yet, however I am taking a course on curves and surfaces (which is almost over) and thus am quite familiar with them (i.e. I know what manifolds are, but I am more used to working with curves and surfaces in $\mathbb R^3$ and $\mathbb R^2$).

My question is why is must we restrict ourselves to defining manifolds as homeomorphic to open subsets of $\mathbb R^n$. Allowing us to use closed sets would allow us, for example to cover $S^2$ with a single surface patch, as opposed to two, which seems like an attractive property.

I'm going to go on to study more advanced differential geometry very soon and I would love to have a good motivation for this property. Thanks in advance!

• Basically, to work with calculus you need domains to be open sets. :) – Ted Shifrin Mar 30 '17 at 17:01
• I think you can find your answer in Why is a topology made up of 'open' sets?. – MohammadSh Mar 30 '17 at 17:03
• @nyquist_plot: A small correction: manifolds are not homeomorphic to open subsets of $\Bbb R^n$, but locally-homeomorphic. – Alex M. Mar 30 '17 at 17:05
• @nyquist_plot The "surface patch" you are proposing for $S^2$ is the identity? Because the sphere is not homeomorphic to a subset of $\mathbb{R}^2$. And among many problems, note that if the definition were to consider closed subsets, not even dimension of connected manifolds would be well-defined. Besides, differential topology would stop short without the inverse function theorem, the fact that an atlas provides an open cover etc etc etc. – Aloizio Macedo Mar 30 '17 at 17:12
• Possibly a good property of locally homeomorphic to open sets in $\Bbb R^n$ is that manifolds will be locally path-connected, which would be good for doing calculus because we want to study how the value of function changes along a curve (i.e. differentiation). This property cannot be captured by closed sets because there are badly behaving closed sets, e.g. Cantor set. – edm Mar 30 '17 at 17:30

• In a Hausdorff space, points are closed; calling a Hausdorff space a "manifold" if for every point $p$, there exists a closed set $V$ containing $p$ and a homeomorphism from $V$ to some closed set in a Cartesian space is no condition at all. (Take $V = \{p\}$.)