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I have a doubt about the last part of the proof of Lemma 1.10 in Introduction to smooth manifolds by John Lee, so I will put just the last part of the proof.

Lemma 1.10 Every topological manifold has a countable basis of precompact coordinate balls.

"[...] If $V \subset U_i$ is one of these balls (a coordinate ball that is precompact in $U_i$), then the closure of $V$ in $U_i$ is compact, and because $M$ is Hausdorff, it is closed in $M$. It follows that the closure of $V$ in $M$ is the same as its closure in $U_i$, so $V$ is precompact in $M$ as well." $\square$

I know that we need to prove that $\overline{V}$ is compact in $M$, but by the Lemma 26.1 of Munkres's book Topology, it is clear that if $\overline{V}$ is compact in $U_i$, then $\overline{V}$ is compact in $M$, isn't it? Why show that $\overline{V}$ is closed in $M$ ensure that $\overline{V}$ is compact in $M$? Thanks in advance!

Lemma 26.1 (Munkres) Let $Y$ be a subspace of $X$, then $Y$ is compact if and only if every covering $Y$ by sets open in $X$ contains a finite subcollection covering $Y$.

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George
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  • I just came here to express the very same doubt! Bravo! It's not well-phrased in the book, but I think I see what was meant now – user20672 Jun 29 '20 at 11:45

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In general, if $U\subset X$ is a subset of some topological space $X$, and $A\subset U$ is a subset of $U$, then the closure of $A$ in $U$, and its closure in $X$, may be different. To be precise, if we call $\overline{A}^U$ the closure of $A$ in $U$, and $\overline{A}^X$ its closure in $X$, then $$\overline{A}^U=\overline{A}^X\cap U$$

I think the author's point is that $\overline{V}^{U_i}$, the closure of $V$ in $U_i$, being compact, has to be a closed subset of the Hausdorff space $M$. Thus it is closed in $M$, and we have, by definition of the closure of a subset as the intersection of all closed subsets containing said set $$\overline{V}^M\subset \overline{V}^{U_i}$$ Since one alwasy has $\overline{V}^{U_i}=\overline{V}^M\cap U_i\subset \overline{V}^M$, this proves that $$\overline{V}^M=\overline{V}^{U_i}$$ and hence $V$ is a relatively compact open ball in $M$.

Olivier Bégassat
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  • Oh, I think I understand now why $\overline{V}^M \subset \overline{V}^{U_i}$. Since $\overline{V}^{U_i}$ is the closure of $V$ in $U_i$, $\overline{V}^{U_i}$ is the intersection of all closed subsets containing $V \cap U_i$, but $V \subset U_i$, so $V \cap U_i = V$ and $\overline{V}^{U_i}$ is a closed set containing $V$, then follows that $\overline{V}^M \subset \overline{V}^{U_i}$ by the definition of closure of a subset, is it? – George May 01 '17 at 13:26
  • A priori, one should expect the inclusion $\overline{V}^{U_i}\subset\overline{V}^{M}$ to be strict : for instance if we were to take $U_i=(0,1)\subset\Bbb R=M$, and $V=U_i$, then $\overline{V}^{U_i}=(0,1)\subsetneq\overline{V}^{M}=[0,1]$. The compactness hypothesis imposed on $\overline{V}^{U_i}$ is crucial. – Olivier Bégassat May 01 '17 at 17:10
  • I understood, thanks a lot! – George May 01 '17 at 20:46
  • I just want to clarify one more time for myself what the author meant. $\overline{V}$ compact in $U_i$ means that $\overline{V}$ compact in $M$, and $M$ Hausdorff means that $\overline{V}$ must be closed in both $M$ and $U_i$. It's from this that we can argue the closures in $M$ and $U_i$ coincide, and $V$ precompact in $M$. Did I understand that right? – user20672 Jun 29 '20 at 11:51