Sequence of closed sets (Milnor's proof)

Question

I've got a question about a descending sequence of closed sets. Milnor writes in his book "FROM THE DIFFERENTIABLE VIEWPOINT". In his proof of Sards Theorem he wrote: Let $f:U\rightarrow \mathbb{R}^p$ be a smooth map, with $U$ open in $\mathbb{R}^n$. Let $C_i$ denote the set of $x$ such that all partial derivatives of $f$ of order $\le i$ vanish at $x$. Thus we have a descending sequence of closed sets $C\supset C_1 \supset C_2...$. But I don't see why for example $C_1$ is closed in $C$ or more generally why $C_i$ is closed in $C_{i-1}$.

Answer 1

If $f$ is smooth, then all the partials are continuous. Now we have a general fact about topology:

If $G: A\rightarrow B$ is a continuous function, then $\{x: G(x)=y\}$ is closed, for every $y\in B$.

(OK, fine, this isn't strictly true: I need $B$ to have the (very basic) property that points are closed. But $\mathbb{R}^p$ certainly has this property.)

Now, each $C_i$ is an intersection of sets of the form "so-and-so partial is zero", and hence (by the above, and since each of the partials is continuous) an intersection of closed sets; and the intersection of closed sets is closed. So each $C_i$ is closed.