# Carroll's Spacetime and Geometry - Notion of open subset of a manifold

In Sean Carroll's Spacetime and Geometry, an introductory section on manifolds contains the following:

A chart or coordinate system consists of a subset $$U$$ of a set $$M$$ along with a one-to-one map $$\phi : U \to \mathbf R^n$$, such that the image $$\phi(U)$$ is open in $$\mathbf R^n$$, [...]. (Any map is onto its image, so the map $$\phi : U \to \phi(U)$$ is invertible if it is one-to-one.) We then can say that $$U$$ is an open set in $$M$$. A $$C^\infty$$ atlas is an indexed collection of charts $$\{(U_\alpha, \phi_\alpha)\}$$ that satisfies two conditions:

1. The union of the $$U_\alpha$$ is equal to $$M$$; that is, the $$U_\alpha$$ cover $$M$$.
2. The charts are smoothly sewn together. More precisely, if two charts overlap, $$U_\alpha \cap U_\beta \ne \emptyset$$, then the map $$(\phi_\alpha \circ \phi_\beta^{-1})$$ takes points in $$\phi_\beta(U_\alpha \cup U_\beta) \subset \mathbf R^n$$ onto an open set $$\phi_\alpha(U_\alpha \cup U_\beta) \subset \mathbf R^n$$, and all of these maps must be $$C^\infty$$ where they are defined. [...]

[...] a $$C^\infty$$ $$n$$-dimensional manifold is simply a set $$M$$ along with a maximal atlas, one that contains every possible compatible chart.

On a first reading, the statement

We then can say that $$U$$ is an open set in $$M$$

seems to say that $$\phi$$ being an invertible function from $$U$$ to the open subset $$\phi(U) \subset \mathbf R^n$$ implies that $$U$$ is an open subset of $$M$$. This would certainly be the case if $$M$$ was a metric space and $$\phi$$ was continuous (Rudin, Theorem 4.8). But, looking closer, it was never established that $$\phi$$ is continuous, nor that $$M$$ is a metric space.

Now, I am not very familiar with manifolds, but could it be that the author actually means that the notion of an open subset of $$M$$ is defined through this construction? I suppose we could use $$\phi$$ to define a distance function $$d: U \to \mathbf R$$, $$d(p, q) = |\phi(p) - \phi(q)|,$$ but this would only make $$U$$, and not $$M$$, into a metric space. But maybe the idea is something similar to this?

Looking at Wikipedia, it seems like maybe the correct idea is instead to use the sets $$U_\alpha$$ to define a topology $$\tau$$ on $$M$$, and then any member of $$\tau$$ is open by definition, which seems quite different from the notion of openness for metric spaces. In that case it seems Carroll calls $$U$$ open a bit prematurely, but maybe this is still what he means?

So, what notion of openness is Carroll actually referring to?

• Your penultimate paragraph is exactly correct. Nov 5, 2021 at 3:09
• In physics you generally assume continuity and some degree of differentiability, otherwise you cannot compute. This simply means the $\phi$'s become at the very least homeomorphisms. Then the topology of $\mathbb{R}$ carries over. By the way the metric notion of openness is just a specific case of the topological one and are completely compatible. Once you have a metric, thus a distance, balls will be open sets in the topological sense Nov 5, 2021 at 9:38

In the other approach, the manifold has no a priori topology. This can be motivated by the fact that in the first approach, if $$U\subseteq M$$ is a set which fits into the domain of a single chart, then $$U$$ is open if and only if its image in $$\mathbb R^n$$ is open. Open sets are also open under arbitrary unions, so all open sets (i.e. the entire topology) is generated by open sets of $$\mathbb R^n$$ through chart maps.
Thus in the second approach, manifolds are defined to be just sets a priori, which has a collection $$\{(U_\alpha,\varphi_\alpha)\}_\alpha$$ of charts, where each map $$\varphi_\alpha:U_\alpha\rightarrow \varphi_\alpha(U_\alpha)$$ is required to be 1) a bijection, 2) to have an open image, i.e. $$\varphi_\alpha(U_\alpha)\subseteq\mathbb R^n$$ is open in $$\mathbb R^n$$ and 3) have $$C^r$$ transition functions between each other.
The topology on $$M$$ is then defined by declaring that open sets are generated by images of open sets of $$\mathbb R^n$$ through the chart maps. Of course if one wants to add the Hausdorff and second countability axioms, then one still needs to declare that only those atlases are allowed for which the induced topology satisfies these properties.