2
$\begingroup$

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 $Y$.

$\endgroup$
5
$\begingroup$

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 $M$.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – George May 1 '17 at 13:26
  • $\begingroup$ 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. $\endgroup$ – Olivier Bégassat May 1 '17 at 17:10
  • $\begingroup$ I understood, thanks a lot! $\endgroup$ – George May 1 '17 at 20:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.