Construct Compact Exhaustion using Paracompactness

Question

Let $M$ be a topolgical $n$-manifold. I have to show that there exist a sequence $(K_i)_{i \in \mathbb{N}}$ of compact subspaces $K_i \subset M$ with properties

1. $K_i \subset K_{i+1} $ for all $i \in \mathbb{N}$

2. $\bigcup_i K_i = M$

I constructed it in following way: since $M$ top manifold it's second countable and therefore Lindelöf. So I can choose an open covering $\bigcup_{i \in \mathbb{N}} U_i$ with $U_i \cong B(0, \epsilon_i)$.

Here $B(0, \epsilon_i) \subset \mathbb{R}^n$ is an open $n$ ball around $0$ with radius $\epsilon_i$. I can choose such countable covering since $M$ is Lindelöf.

Then I define inductively

$K_1:= \overline{B(0, \epsilon_1- \frac{\epsilon_1}{2}})$

$K_{m+1} := \bigcup^{m+1} _{i=1} \overline{B(0, \epsilon_i- \frac{\epsilon_i}{n+2}})$

One can indeed easily prove that this consruction fulfils properties 1. and 2. so it solves the problem.

Now my question: One can also show that a topological manifold is paracompact. My construction above seems quite awkward and long to me.

Does there exist an abstract argument using paracompactness to show the existence of the sequence above?

I suppose that there is a way to use paracompactness to reach a "nicer" solution without such a construction. Does anybody see it?

Answer 1

Your construction is stronger than what's required; you have exhibited a sequence of open subsets $\Omega_1 \subset \Omega_2 \subset \dots$ such that each $K_n = \overline \Omega_n$ is compact with $K_n \subset \Omega_{n+1}$ is compact and $\bigcup_n \Omega_n = M.$ Instead, you could have for example taken a countable cover of open balls $B_i$ (each lying in a local chart) and setting $K_n = \bigcup_{i=1}^n \overline B_i.$

Paracompactness isn't too useful here I don't think, and in any case it's overkill for what's needed.

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Edit: In light of the comment below (originally the condition was only $K_i \subset K_{i+1}$ for all $i$), I don't think you can simplify the proof much if you work from first principles.

However if you take for granted the fact that partitions of unity exist subordinate to any open cover (which is one formulation of paracompactness), then you can simplify your proof.

The idea is to take a countable covering of $M$ by balls $B_i$ such that each $\overline B_i$ is compact. Then let $\{\psi_i\}$ be a subordinate partition of unity and define an exhaustion function by, $$ \varphi(x) = \sum_{i=1}^{\infty} i \psi_i(x). $$ Then one can verify that $\varphi$ is proper, that is $\varphi^{-1}(K)$ is compact for any $K \subset \mathbb R$ is compact. Indeed for any $K$ compact we have $K \subset [-N,N]$ for some $N,$ so then $\varphi^{-1}(K)$ is a closed subset of $\bigcup_{i=1}^{N+1} \overline B_i$ which is compact.

Then we can take $K_n = \varphi^{-1}([0,n]),$ noting that $K^{\circ}_n = \varphi^{-1}((-\infty,n)) = \varphi^{-1}([0,n)).$