Sorry for the vague title, but the question is fairly long:
Let $\{A_\alpha\}$ be a ndb-finite closed cover of $X$. Consider $x_0\in X$, and let $A_{\lambda_i}$ be (all) the $A_\alpha$ that contain $x_0$. Let $V_{\lambda_i}\subset A_{\lambda_i}$ be nbds of $x_0$ in $A_{\lambda_i}$. Prove: $\bigcup V_{\lambda_i}$ contains a nbd of $x_0$ in $X$.
Here is my attempt:
By nbd-finiteness of $A_\alpha$ there is nbd $V^0$ of $x_0$ in $X$ s.t. $A_\alpha\cap V^0\not=\emptyset$ for only finitely many $\alpha$; call these $\alpha_i$. Then we iterate over the $\alpha_i$:
- If $\alpha_i=\lambda_j$ for some $j$ then we do nothing,
- Otherwise we set $V^{i+1}=V^i\cap A_{\alpha_i}^{-1}$
Call the result $V$. This process ends since there are only finitely many $\alpha_i$. Again from the finiteness and the fact that the $A_\alpha$ are closed it follows that $V$ is open. And finally $x_0\in V$ since $x_0$ is only contained in the $A_{\lambda_i}$ and so is never removed from any $V^i$.
Now all we do is say $V_{\lambda_i}=A_{\lambda_i}\cap U_{\lambda_i}$ for open $U_{\lambda_i}$ in $X$. Now it's not hard to see that $$\left(\bigcap U_{\lambda_i}\right)\bigcap V\subseteq \bigcup V_{\lambda_i}$$ And that it's open and contains $x$.
I have two questions:
- Is this correct?
- Is there a shorter, more direct proof? For some reason I never really like 'algorithmic' proofs like this. It also took me quite a while to come up with this, and I'm wondering if I'm just overlooking a more simple way of proving this.
Ad 1: Yes, it is correct - assuming $A_{\alpha_i}^{-1}$ is used as a notation for the complement of $A_{\alpha_i}$.
Ad 2: I'm not sure whether you find the following more direct, it's not essentially different from your proof, but avoids the explicit stepwise construction of $V$:
The crux of the matter is that the union of a locally finite [neighbourhood-finite, but locally finite is the more common terminology] family of closed sets is closed. If you know that, you can directly see that
$$U_0 = X\setminus \bigcup \{ A_\alpha : x_0 \notin A_\alpha\}$$
is open. Furthermore, for each $\lambda_i$, we can write $V_{\lambda_i} = A_{\lambda_i} \cap U_{\lambda_i}$ with a neighbourhood $U_{\lambda_i}$ of $x_0$. Then
$$U = U_0 \cap \bigcap U_{\lambda_i}$$
is a neighbourhood of $x_0$ (since it is a finite intersection of neighbourhoods), and we have
$$U = \bigcup (U\cap A_{\lambda_i}) \subset \bigcup (U_{\lambda_i} \cap A_{\lambda_i}) = \bigcup V_{\lambda_1}.$$
If you don't know that the union of a locally finite family of closed sets is closed, the proof isn't hard. It's essentially the same thing again. We prove the equivalent formulation that
$$\overline{\bigcup B_\alpha} = \bigcup \overline{B_\alpha}\tag{$\ast$}$$
for every locally finite family $\{ B_\alpha \}$. By the monotonicity of the closure, we have
$$\overline{B_\beta} \subset \overline{\bigcup B_\alpha}$$
for all $\beta$, and hence
$$\bigcup \overline{B_\alpha} \subset \overline{\bigcup B_\alpha}.$$
That inclusion holds for all families, whether locally fintie or not. Conversely, let
$$x \notin \bigcup \overline{B_\alpha}.$$
By the local finiteness of the family, there is an open neighbourhood $U_0$ of $x$ such that $U_0$ intersects only finitely many $B_\alpha$, say $B_{\alpha_i}$ for $1 \leqslant i \leqslant k$. Since $x \notin \overline{B_{\alpha_i}}$, there is a neighbourhood $U_{\alpha_i}$ of $x$ with $U_{\alpha_i} \cap B_{\alpha_i} = \varnothing$, for all $1 \leqslant i \leqslant k$. Then $U = U_0 \cap \bigcap\limits_{i = 1}^k U_{\alpha_i}$ is a neighbourhood of $x$ that doesn't intersect any $B_\alpha$, hence
$$U \cap \bigcup B_\alpha = \varnothing,$$
and $x \notin \overline{\bigcup B_\alpha}$, which shows that
$$\overline{\bigcup B_\alpha} \subset \bigcup \overline{B_\alpha}$$
for locally finite $\{ B_\alpha \}$.