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In page 15 of Lee's book "Introduction to Smooth Manifolds", there's a paragraph as follows:

We say a set $B\subset M$ is a regular coordinate ball if there is a smooth coordinate ball $B'\supset \bar B$ and a smooth coordinate map $\varphi:B'\to \Bbb R^n$ such that for some positive real numbers $r<r'$, $\varphi (B)=B_r(0),\quad\varphi(\bar B)=\bar B_r(0),\quad$ and $\varphi (B')=B_{r'}(0).$

If we change the above definition as follows:

We say a set $B\subset M$ is a regular coordinate ball if there is a smooth coordinate ball $B'\supset B$ and a smooth coordinate map $\varphi:B'\to \Bbb R^n$ such that for some positive real numbers $r<r'$, $\varphi (B)=B_r(0), \quad\varphi (B')=B_{r'}(0).$

Are they equivalent?

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    $\begingroup$ Yes, they're equivalent, as two answers posted below have shown. Here's why I defined it the way I did: In my Introduction to Topological Manifolds, I had introduced the analogous concept (but without smoothness) -- see p. 103. But that definition came before the chapter on compactness, so I had to give a definition that didn't rely on compactness. For ISM, of course, I could have used compactness, but I decided it was worthwhile to make this definition look as closely analogous to the one in ITM as possible, to avoid confusion. A judgment call, and certainly open to argument. $\endgroup$
    – Jack Lee
    Oct 22, 2018 at 17:04
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    $\begingroup$ @JackLee Thanks for your excellent book! $\endgroup$ Oct 22, 2018 at 19:29
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    $\begingroup$ You’re welcome! $\endgroup$
    – Jack Lee
    Oct 22, 2018 at 19:36

2 Answers 2

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A regular coordinate ball in the sense of the first definition is obviously one in the sense of the second definition.

Conversely, let $B$ be a regular coordinate ball in the sense of the second definition. Let $B' \supset B$ and $\varphi : B' \to \mathbb{R}^n$ as described. We write $\varphi^{-1} : B_{r'}(0) \to M$ for the inverse of $\varphi$.

We have $\overline{B}_r(0) \subset B_{r'}(0)$ so that $B'' = \varphi^{-1}(\overline{B}_r(0)) \subset B'$ is the image of a compact set, thus is itself compact. Hence $B''$ is closed in $M$, and $B \subset B''$ implies $\overline{B} \subset B''$. The continuity of $\varphi^{-1}$ shows $B'' = \varphi^{-1}(\overline{B}_r(0)) \subset \overline{\varphi^{-1}(B_r(0))} = \overline{B}$, hence $B'' = \overline{B}$.

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$$\varphi^{-1}:B_{r'}(0)\to B'$$ $$\varphi^{-1}(B_{r}(0))=B$$ Thus $\varphi^{-1}(\bar B_{r}(0))$ is compact, because every compact subset of a Hausdorff space is closed, so $\varphi^{-1}(\bar B_{r}(0))=\bar B$, then we have $B'\supset \bar B$ and $\varphi (\bar B)=\bar B_{r}(0)$.

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