Following Lee's book on smooth manifolds.
I'm trying to understand the proof of the theorem (Every manifold is paracompact) and there are some topological claims that i don't understand. I've marked the parts that i don't get with 4 (Why's).
Definition: A $(M,T)$ is locally compact topological space if for every $x \in M$ and an open
neighborhood $(U_x \subset M) $, $(\exists K \subset M)$ compact : $U_x \subset K$.
Definition: Let $(M,T)$ a Hausdorff. We say the subset $U\subset M$ is precompact if $\overline{U}$ is compact.
Definition: Let $(M,T)$ is paracompact if every open cover of M admits an open, locally finite refinement
Definition: Let $(M,T)$ a Hausdorff and locally compact topological space. A sequence $(K_i)_{i=1}^{\infty}$ of compact subsets of $M $ is called an exhaustion of M by compact sets if:
$1)$ $M = \cup_i K_i$
$2)$ $K_i \subset Int(K_{i+1})$
Proposition 1: Let $(M,T)$ be a Hausdorff and locally compact topological space. Then $M$ has a basis of precompact open subsets.
Proposition 2: A second-countable, locally compact Hausdorff space admits an
exhaustion by compact sets.
Theorem: A $(M,T)$ Second countable, Hausdorff and locally compact topological space is Paracompact. In fact, given a an open cover $X$ of $M$,
and any basis $\mathbb{B}$ for the topology of $M$, there exists a countable, locally finite open
refinement of X consisting of elements of $\mathbb{B}$.
Proof: Given $M$; $X$, and $\mathbb{B}$ as in the hypothesis of the theorem.
Let $(K_j)_{j=1}^{\infty}$ be an
exhaustion of $M$ by compact sets (Proposition 2).
For every $j$, Let:
$V_j = K_{j+1}-Int(K_j)$
$W_j = Int(K_{j+2})-K_{j-1}$
$K_j = \emptyset$ is $j<1$
Then $V_j$ is compact (Why 1?)
and $W_j$ is open (Why 2?)
and $V_j \subset W_j$ (Why 3?)
For every $(x \in V_j)(\exists X_x \in \mathbb{X}).$
Because $\mathbb{B}$ is a basis, $(\exists B_x \in \mathbb{B})$ s.t. $x \in B_x \subset X_x \cap W_j$ (Why 4?)
Update: Continuing the proof:
$\{B_x \in \mathbb{B}\}_{x \in V_j}$ is an open cover of $V_j$. Then for compactness of $V_j$ there exists:
$B_j=\{\{B_{x_i} \in \mathbb{B}\}_{x \in V_j}\}_{i=1}^{n}$ finite cover of $V_j$. and
$\cdot$ $\cup_{j=1}^{\infty} B_j$ is a cover of $M$ (Why 5?)
$\cdot$ and a refinement of $X$ (Why 6?)
Best Answer
The set $V_j$ is compact since it is a closed subset of the compact set $K_{j + 1}$. The set $W_j$ is open since it is the intersection of two open sets $\text{Int}(K_{j + 2})$ and $M - K_{j - 1}$. The inclusion $V_j \subset W_j$ holds since $K_{j + 1} \subset \text{Int}(K_{j + 2})$ and $K_{j - 1} \subset \text{Int}(K_j)$. The set $X_x \cap W_j$ is open and contains $x$ so by the definition of a basis, there must exist $B_x \in \mathbb{B}$ such that $x \in B_x \subset X_x \cap W_j$.
The set $\cup_{j = 1}^{\infty} B_j$ is a cover of $M$ since the sets $V_j$ cover $M$ and each $B_j$ covers $V_j$. Each set $B_x$ is contained in some $X_x \in X$. Hence $\cup_{j = 1}^{\infty} B_j$ is a refinement of $X$.