Level set corresponding to the value $0$ for $f(x,y,z)=y^3-z^3$ is a smooth surface

multivariable-calculusvector analysis

I was attempting exercises in Vector Calculus by Peter Baxandall. enter image description here

Definition $3.8.11$: A set $S \subseteq \Bbb R^m$ is called a smooth surface if $S$ is a level set of a $C^1$ function $f: D \subseteq \Bbb R^m \rightarrow \Bbb R~$ s.t $~\text {grad}~ f(x) \ne 0~\forall~x \in S$

My Argument:

We can see that $S$ is the level set corresponding to the value $0$ of the $C^1$ function $f:D \subseteq \Bbb R^m \rightarrow \Bbb R~|~f(x,y,z)=y^3-z^3.~~~\therefore~\text{grad}f(x,y,z) = \Big(0,3y^2,-3z^2\Big)$. Clearly, $\text{grad}f(0,0,0)=0$ and $(0,0,0) \in S$. Thus, $S$ cannot be a smooth surface.
Why does $3.8.11$ not apply to the above function?

Could someone point out any error in the above argument? Thanks a lot!

Best Answer

Let's rewrite the definition slightly:

Definition $3.8.11$

A set $S\subset \Bbb{R}^m$ is called a smooth surface if there exists an open set $D\subset \Bbb{R}^m$ and a function $f:D\to \Bbb{R}$ such that $S = f^{-1}\left(\{0\}\right)$ and for every $x\in S$, $(\text{grad } f)(x) \neq 0$.

Note the "there exists" in the definition. Nowhere in the definition does it say anything about uniqueness. In other words, just because the given function $F$ which doesn't satisfy that condition, it does not mean $S$ is not a smooth surface. Just because $F$ does not satisfy the condition, it doesn't mean no other function can satisfy that condition.

For example, just because I can't run a $100$m dash in $9.58$ seconds, it doesn't mean nobody else can... Usain Bolt certainly can (and did) $\ddot{\smile}$.

For example, consider $\phi:\Bbb{R}^3\to \Bbb{R}$ given by $\phi(x,y,z) = y-z$. What is the level set $\phi^{-1}(\{0\})$? What is its gradient?

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