# Are these conditions equivalent to the definition of regular coordinate ball?

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 $$r, $$\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 $$r, $$\varphi (B)=B_r(0), \quad\varphi (B')=B_{r'}(0).$$

Are they equivalent?

• 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. Oct 22, 2018 at 17:04
• @JackLee Thanks for your excellent book! Oct 22, 2018 at 19:29
• You’re welcome! Oct 22, 2018 at 19:36

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 $$B'' = \overline{B}$$.
$$\varphi^{-1}:B_{r'}(0)\to B'$$ $$\varphi^{-1}(B_{r}(0))=B$$ 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 $$\varphi (\bar B)=\bar B_{r}(0)$$.