A closed subset of a manifold is precompact (Lemma 1.10 from Lee)

Question

I can't understand the end of the proof of the following lemma from "Introduction to smooth manifolds" by Lee

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

In order to prove it he takes a manifold $M$ which is covered by a countable subcover $\{(U_i,\phi_i)\}$.

A the end of the proof he concludes that, given a precompact ball $V \subset U_i $, $\bar{V}$ is compact in $U_i$ and $\bar{V}$ is closed in $M$. And this is clear for me.

I don't understand the sentence that concludes the proof: " (given that $\bar{V}$ 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."

What does it mean "is the same as"? How can I conclude that $V$ is precompact?

Can anyone help me?

Does this answer your question? Necessity of closedness in going from precompactness in chart to precompactness in the manifold — laura_b, Apr 20 '20 at 18:51

score 0 · Accepted Answer · answered Mar 20 '20 at 10:46

0

It means the following:

Give $U_i$ the subspace topology. Then as $V \subseteq U_i$, we can take the closure of $V$ inside $U_i$ with the subspace topology: $cl_{U_i}(V)$. But also $V \subseteq M$, so we can take the closure of $V$ in $M$, which is $cl_M(V)$. The author claims that

$$cl_{U_i}(V) = cl_{M}(V)$$

answered Mar 20 '20 at 10:46

J. De Ro

21,438

but why can he claim that? – Phi_24 Mar 20 '20 at 14:26
2

$\overline{V}$ is compact in $U_i$ thus also compact in $M$. A compact set is closed in a Hausdorff space. Does that answer your question? – J. De Ro Mar 20 '20 at 14:30

A closed subset of a manifold is precompact (Lemma 1.10 from Lee)

1 Answers1