Sard's Theorem: Suppose $M$ and $N$ are smooth manifolds with or without boundary and $F:M\to N$ is a smooth map. Then the set of critical values of $F$ has measure zero in $N$.
In what appears to be a standard proof of Sard's Theorem for Manifolds, one can reduce using charts to the case of $F:U\subset \mathbb{R}^m\to \mathbb{R}^n.$ This is clear enough. Eventually, we come to a statement where we define a decreasing sequence of subsets $$ C\supset C_1\supset C_2\supset \cdots$$ where $C$ is the set of critical points of $F$, our smooth function, and $$C_k=\{x\in C: \text{for}\: 1\le i\le k, \:\text{all}\:i^{th}\: \text{partial derivatives of}\:F\:\text{vanish at}\:x\} .$$ Now, I am clearly misunderstanding something here, because there is a line in the proof which states:
Since $C_{k+1}$ is closed in $U$, we can discard it and assume that at every point of $C_k$ there is some $(k+1)^{st}$ partial derivative of $F$ that does not vanish.
So, we reduce to the case of $C_k\setminus C_{k+1}$. Per definition, this leads me to believe that if $x\in C_k\setminus C_{k+1}$, there must exist functions with vanishing $k^{th}$ order partials, but non-vanishing $(k+1)^{st}$ order partials, contrary to my presupposed notion. Could anyone illuminate me as to my error?