I'm studying Sard's Theorem in prof john lee's 'smooth manifolds'
The proof contains three step and there is a second step: F($C_k$ \ $C_{k+1}$) is measure zero for smooth map F and $C_k$ = {$x \in C$: for $1 \leq i \leq k$, all $i$-th partial derivatives of $F$ vanish at $x$}
To sum up the proof of second step,
Assume $C_{k+1} = \emptyset$
Let a $\in$ $C_k$ be arbitrary and let $y: U \rightarrow R$ denote some k th partial derivative of $F$ that has at least one nonvanishing first partial derivative at $a$.
Then $a$ is a regular point of the smooth map $y$, so there is a neighborhood $V_a$ of a consisting entirely of regular points of $y$.
Then $F(C_k \cap V_a)$ is measure zero. (Omitted the detail)
Since $U$ can be covered by countably many neighborhoods like $V_a$, it follows that $F$($C_k$ \ $C_{k+1}$) is contained in a countable union of sets of the form $F$($C_k \cap V_a$), and thus has measure zero.
My question is on the fifth phrase, "$U$ can be covered by countably many neighborhoods like $V_a$".
I understood all the other part but for that part.
Please tell me why $U$ can be covered by countably many neighborhoods like $V_a$
We know that $C$ is a closed subset of $U$, and for any $a \in C\subset U$, we have neighbourhood $V_a \subset U$. Therefore the collection $\{U \smallsetminus C\} \cup \{V_a : a \in C\}$ is an open cover for $U$. Since $\mathbb{R}^n$ is second countable (hence $U \subset \mathbb{R}^n$ is also second countable), and because every open cover for second countable space has a countable subcover, then we have countable cover say $\{U \smallsetminus C\} \cup \{V_{a_i} : a_i \in C\}$ for $U$.
By this, you may also proceed that since $$C \cap U = C \cap \Big(\bigcup_{i} V_{a_i} \cup (U \smallsetminus C)\Big) = \Big(C\cap \bigcup_{i} V_{a_i}\Big) \cup \big( C \cap (U \smallsetminus C)\big) = \bigcup_{i} (C \cap V_{a_i})$$ then $$ F(C \cap U) = F\Big( \bigcup_{i} (C \cap V_{a_i})\Big) = \bigcup_{i}F(C \cap V_{a_i}) $$ has measure zero, since $F(C \cap V_{a_i})$ is.