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It is known that any manifold of finite dimension $n$ has an finite atlas (with at most $n+1$ charts). My question is then the following : why don't we define an atlas of a manifold (in the finite dimensional case) as a finite set of charts that covers the manifold, for teaching purposes. Wouldn't that simplify the presentation and the theory (partition of unity, etc.) ?
Or in other words : do we encounter naturally manifolds with an infinite atlas (as part of their definition) ?

Jon-S
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    Requiring an atlas to be finite would be an additional restriction to check, for what purpose? Sometimes one would have to work around it -- e.g. if you define a manifold structure on a topological space by giving a separate chart for a neighborhood of each point and show that the charts fit together correctly, you wouldn't be able to just throw all of those local charts together in an atlas -- and unless the space is compact there might not be a finite subset of them that covers everything, so now you need to make new charts somehow ... – hmakholm left over Monica Apr 03 '16 at 11:27

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There's a relevant discussion on pp. 12-13 of Introduction to Smooth Manifolds by Lee:

"Our plan is to define a "smooth structure" on $M$ by giving a smooth atlas, and to define a function $f:M \to \mathbb R$ to be smooth if and only if $f \circ \phi^{-1}$ is smooth in the sense of ordinary calculus for each coordinate chart $(U,\phi)$ in the atlas. There is one minor technical problem with this approach: In general, there will be many possible choices of atlas that give the "same" smooth structure, in that they all determine the same collection of smooth functions on $M$. For example, consider the following pair of atlases on $\mathbb R^n$: \begin{align} \mathcal A_1 &= \{ (\mathbb R^n, \text{Id}_{\mathbb R^n}) \} \\ \mathcal A_2 &= \{(B_1(x),\text{Id}_{B_1(x)}) : x \in \mathbb R^n \}. \end{align} Although these are different smooth atlases, clearly a function $f:\mathbb R^n \to \mathbb R$ is smooth with respect to either atlas if and only if it is smooth in the sense of ordinary calculus.

"We could choose to define a smooth structure as an equivalence class of smooth atlases under an appropriate equivalence relation. However, it is more straightforward to make the following definition: A smooth atlas $\mathcal A$ on $M$ is maximal if it is not contained in any strictly larger smooth atlas. This just means that any chart that is smoothly compatible with every chart in $\mathcal A$ is already in $\mathcal A$. (Such a smooth atlas is also said to be complete.)"

littleO
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The oriented surface with infinite genus does not have a finite atlas.

Tsemo Aristide
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    It does, because every (finite-dimensional) manifold has a finite atlas. You just have to allow the coordinate domains to be disconnected. – Jack Lee Apr 03 '16 at 13:29
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    It depends of the definition that you are using, usually, the domain of a chart in an atlas is supposed to be connected. https://en.wikipedia.org/wiki/Atlas_%28topology%29#Formal_definition_of_atlas – Tsemo Aristide Apr 03 '16 at 14:53
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    What is an atlas with one chart for $S^2$? Remember, a chart has to be a homeomorphism onto its image. – Jack Lee Apr 03 '16 at 15:04