It is a three variables version of IFT, and the proof I found on a website given as follows:
WLOG, the set S can be taken to be a parallelpiped, a box,centered on P because within the set S there will be at least one such box. Within any such box there will be another box containing P such that ∂F/∂z has the the same sign as at P. Let the box be given as a triple (a,b,c) such that
|x-x0| ≤ a |y-y0| ≤ b |z-z0| ≤ c
WLOG, the sign of ∂F/∂z at P can be taken to be positive.
This means that
F(x0,y0,z0+c) > 0 F(x0,y0,z0-c) < 0
Now consider any point (x1,y1) such that
|x1-x0| ≤ a |y1-y0| ≤ b
Because F(x0,y0,z0+c) > 0 and F(x,y,z) is continuous, F(x1,y1,z0+c) > 0. Likewise F(x1,y1,z0-c) < 0. With x and y held fixed at x1 and y1, G(z)=F(x1,y1,z) is a function such that G(z0+c) > 0 and G(z0-c) < 0. Therefore there is some z between z0-c and z0+c such that G(z)=0; i.e., F(x1,y1,z)=0. Moreover this value of z is unique. Since this holds for any (x,y) such |x-x0| ≤ a |y-y0| ≤ b
over this domain there exists a function z=f(x,y) such that F(x,y,z)=0.
I don't understand why by continuity of F(x,y,z) F(x1,y1,z0+c) has the same signs as F(x0,y0,z0+c).
It confuses me a lot. Hope anyone could help!!
Big thanks!
There are some hypotheses which you should add (or didn't tell about?). $F$ is obviously assumed continuous but also $\partial F/\partial z$ should be assumed continuous. Then there is a story concerning 'choices'. Given the first parallelepiped you may shrink $a$ and $b$ so as to satisfy the condition you mention and keeping the sign of $\partial F/\partial z$. From that the argument works. Sort of nice as it does not assume $F$ to be $C^1$ in all variables.