Prescribed finite open cover of compact set Let $A \subseteq \mathbb{R}^n$ be an open set, and $h: A \rightarrow \mathbb{R}$ a continuous function.
Suppose $K\subseteq A$ is compact and that for every $p \in K \cap h^{-1}(\{0\})$ I have an open neighbourhood $N_p$ of $p$.

I need to find a finite open cover $\mathcal{C}$ of $K$ such that:

*

*it includes a finite cover $\mathcal{N}$ of $K \cap h^{-1}(\{0\})$ by the open sets $N_p$ defined above

*no set of $\mathcal{C}$ other than those in $\mathcal{N}$ intersect $K \cap h^{-1}(\{0\})$

My try: $h^{-1}(\{0\})$ is closed relatively to $A$, then $K \cap h^{-1}(\{0\})$ is closed relatively to $A$ and also bounded, so it is compact relatively to $A$ (?).
Since $N_p$ is an open (also relatively to $A$) cover of $K \cap h^{-1}(\{0\})$, by relative compactness I can find a finite subcover $\mathcal{N} = \{N_1,...,N_m\}$.
Here I'm stuck in finding finitely many open sets to complete $\mathcal{N}$ to an open cover of $K$ with the second property (and I think the passage with the question mark is wrong).
My other try: if I had open sets that don't intersect $K \cap h^{-1}(\{0\})$ that complete $\{N_p\}_{p \in K \cap h^{-1}(\{0\})}$ to an open cover of $K$ then the compactness of $K$ would yield an open cover of $K$ with the desired properties.
How can I proceed? Is the construction even possible?
 A: Yes this is possible.
Let $\mathcal{N} := \{N_p : p \in K \cap h^{-1}(\{0\})\}$.
The set $\mathcal{N}$ consists of open sets in $A$, so $\cup \mathcal{N}$ is open in $A$ as well and thus $K \cap \cup \mathcal{N}$ is open in $K$. Hence $K \setminus K \cap \cup \mathcal{N} = K \setminus \cup \mathcal{N}$ is closed in $K$ and a subset of $K$ and since $K$ is compact, it is compact as well. 
Now consider an arbitrary cover $\{U_i\}_{i \in I}$ of $K \setminus \cup \mathcal{N}$. Since it is compact, we get a finite subcover $\{U_i\}_{i = 1}^m$. 
Since $h^{-1}(\{0\})$ is closed in $A$, $K \cap h^{-1}(\{0\})$ is closed in $K$ and thus compact, since $K$ is compact. Since $\mathcal{N}$ is an open cover of $K \cap h^{-1}(\{0\})$ we get a finite subcover. Denote this $\mathcal{N}'$.
Now we claim that the set $\mathcal{N}' \cup \{U_i\}_{i = 1}^m$ is finite consists of open sets, covers $K$ and the only sets intersecting $K \cap h^{-1}(\{0\})$ are those in $\mathcal{N}$: 


*

*It is the union of finite sets

*All the sets are open by construction 

*If $p \in K \cap h^{-1}(\{0\})$ then $p \in \cup \mathcal{N}$, and if $p \not\in K \cap h^{-1}(\{0\})$, then by construction $p \in U_i$ for some $i \in \{1, \ldots, m\}$

*All the $U_i$ are contained in $K \setminus \cup \mathcal{N}$.



To adress your attempt, both are a good start. Though, in the second one, you are just guaranteed some open cover. It may not satisfy any of your two conditions.
