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Let ($M,A$) be a smooth manifold. What I want to prove/disprove is that every open set $U \subset M$ can be the support of a chart of the smooth manifold.

If there exists a chart ($W,\psi $) such that $U \subset W$, then it is easy to show that ($U, \psi_{|U} $) is a chart of the smooth manifold. However if we do not know of such a $W$, then what do we do?

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    $\begingroup$ $U=M$ is an open set, and $M$ certainly cannot be the domain of a chart if $M$ isn't homeomorphic to $\mathbb{R}^n$ (say it isn't simply connected), and examples of this abound. It is very easy to observe the same phenomena in strict subsets of $M$. If the answer to your question were yes, smooth manifolds would be rather uninteresting. $\endgroup$
    – jawheele
    Feb 16, 2020 at 6:16
  • $\begingroup$ @jawheele Oh okay. I didn’t know about the notion of simply connected sets. Just out of curiosity, why do you say the last statement (smooth manifolds would be uninteresting)? $\endgroup$
    – ModCon
    Feb 16, 2020 at 8:35
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    $\begingroup$ Because if the answer were yes, then every smooth manifold would be equivalent (diffeomorphic) to $\mathbb{R}^n$, and that defeats the purpose of the definition (to generalize beyond $\mathbb{R}^n$). On the note of simply connected, think of the plane without the origin, $M = \mathbb{R}^2 \backslash \{0\}$. A working counterexample that is simply connected would be the sphere, $M=S^2$, which cannot be homeomorphic to $\mathbb{R}^2$ because it's compact. $\endgroup$
    – jawheele
    Feb 16, 2020 at 16:36
  • $\begingroup$ @jawheele You shoukd give an official answer. $\endgroup$
    – Paul Frost
    Feb 18, 2020 at 10:17
  • $\begingroup$ Wait, how do you show that $(U, \psi|_U)$ is a chart of the smooth manifold? The proof of that would be very helpful for what I'm currently working on $\endgroup$
    – jsmith
    Oct 26, 2022 at 3:38

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The answer to the title question is, in general, no. My comment paints the picture a bit more simplistic than it is by assuming a definition of coordinate chart requiring its image to be $\mathbb{R}^n$ rather than just an open subset of $\mathbb{R}^n$, but this by no means universal.

To answer completely, we must understand what a smooth manifold is. The first definition one sees is often a pair $(M,A)$, where $M$ is a (Hausdorff, second countable) topological space and $A=\{U_\alpha,\phi_\alpha\}_{\alpha \in I}$ ($I$ some indexing set) is a smooth atlas, i.e. $M = \cup_{\alpha \in I} U_\alpha$ with each $U_\alpha$ open, each $\phi_\alpha : U_\alpha \to \mathbb{R}^n$ is a homeomorphism onto an open subset of $\mathbb{R}^n$, and $\phi_\alpha \circ \phi_{\beta}^{-1} : \phi_\beta(U_\alpha \cap U_\beta) \subset \mathbb{R}^n \to \mathbb{R}^n$ is smooth when its domain is nonempty. In this picture, a coordinate chart is simply defined as an element of $A$, so that the open cover $\{U_\alpha\}$ comprises all possible domains of coordinate charts. Hence, here it certainly need not be the case that every open set is the domain of a coordinate chart, as all that is required of $\{U_\alpha\}$ is that it cover $M$. Even in the simplest case of $M=\mathbb{R}^n$, we may choose $I=\mathbb{Z}^n$ and for $\mathbf{k} = (k_1,k_2,...,k_n) \in I$ take $U_{\mathbf{k}} = B_{\mathbf{k}}(2 \sqrt{n})$, the ball of radius $2 \sqrt{n}$ centered at $\mathbf{k}$ (viewed as an element of $\mathbb{R}^n$), and $\phi_{\mathbf{k}}$ the identity map. This gives a perfectly valid atlas for $M=\mathbb{R}^n$ making it a smooth manifold, but the coordinate chart domains $\{U_{\mathbf{k}}\}$ are only the balls of radius $2 \sqrt{n}$ centered at points with integral coordinates!

However, there is a better picture that does not distinguish between atlases that make $M$ into the "same" smooth manifold. We say that two smooth atlases $A_1, A_2$ on the same topological space $M$ are compatible if $A_1 \cup A_2$ is a smooth atlas, so that all the transition maps between coordinate charts from either atlas are smooth. Starting with a single atlas $A$, then, we may instead consider the unique maximal atlas $\mathcal{A}$ containing $A$ given by taking the union of all atlases compatible with $A$-- $\mathcal{A}$ then contains every possible coordinate chart that could be added to $A$. Since compatible atlases essentially give the same structure to $M$, a smooth manifold is often defined as a pair $(M, \mathcal{A})$ where $\mathcal{A}$ is a maximal atlas (often called a smooth structure). We're now led to a mildly less trivial problem: given a smooth manifold $(M,\mathcal{A})$, does every open set $U \subset M$ appear as the domain of some coordinate chart? Maximal atlases are absurdly large, so it's easy to think the answer must be yes, but it's not too hard to see that, in fact, the answer is yes if and only if $M$ is diffeomorphic (with respect to the smooth structure $\mathcal{A}$!) to an open subset of $\mathbb{R}^n$, as this latter statement is simply saying that $U=M$ is the domain of some coordinate chart in $\mathcal{A}$ (by maximality), and you observed in the question statement that restricting this to any open $U \subset M$ gives a compatible coordinate chart, which then must be in $\mathcal{A}$ (again by maximality).

To arrive at my answer of no to the general case, then, we need only observe that there exist $n$-dimensional smooth manifolds that are not diffeomorphic to any open subset of $\mathbb{R}^n$. A rather broad and simple class of examples is that of nonempty compact manifolds (say the spheres $S^n$ or tori $(S^1)^{\times n}$), as the compact subsets of $\mathbb{R}^n$ are closed and bounded (and therefore not open, as $\mathbb{R}^n$ and $\emptyset$ are the only clopen subsets of $\mathbb{R}^n$ by connectedness). However, there are many more examples: no non-orientable $n$-manifold $M$ is diffeomorphic to an open subset of $\mathbb{R}^n$, as the definition of non-orientability requires that $M$ cannot be covered by a single coordinate chart (this single coordinate chart would necessarily comprise an oriented atlas). A somewhat more surprising example is that, for $n=4$ (and only $n=4$), there exist maximal atlases on the topological space $\mathbb{R}^4$ that make it a smooth manifold that is not diffeomorphic to any open subset of the standard $\mathbb{R}^4$. Smooth functions on these so-called exotic $\mathbb{R}^4$ are not smooth according to the standard definition of differentiability!

This unique property of $n=4$ is especially interesting considering that classical general relativity models the universe as a smooth $4$-manifold, leading to the question of what it might look like or mean if the universe had an exotic smooth structure.

Finally, a word on why the answer had to be no if smooth manifolds are to be a useful concept: the motivation of the definition of a smooth manifold is that we'd like to capture important local properties of $\mathbb{R}^n$ that allow us to do calculus, but in the most general setting that it makes sense. If we found that, even in this most general setting, every smooth $n$-manifold was diffeomorphic to an open subset of $\mathbb{R}^n$, then we'd have found that the setting wasn't actually any more general than doing calculus in $\mathbb{R}^n$ to begin with. So, some mathematician would have proved this once, and no one would ever bother to work with the definition of a smooth manifold again since one might as well just say "let $M$ be an open subset of $\mathbb{R}^n$".

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