MHB Implicit function Theorem for R^3

[kalvin]

Let $U \subset \Bbb{R}^3$ be open and let $f : U → \Bbb{R}$ be a $C^1$
function. Let$ (a, b, c) \in U$
and suppose that$ f(a, b, c) = 0$ and $D_3f(a, b, c) \ne 0.$ Show there is an open ball$ V \subset \Bbb{R}^2$ containing $(a, b)$ and a $C^1$
function $\phi : V → \Bbb{R}$ such that $\phi(a, b) = c$ and
$f(x, y, \phi(x, y)) = 0$ for all$ (x, y) \in V$ . We call $\phi$ an implicit function determined by $f$ at $(a, b)$.

I know who how to do the proof in R^2 bout proving it for R^3 seems to be a lot trickier, any help

 

[Fallen Angel]

Hi,

Consider $\Bbb{R}^{3}$ as $\Bbb{R}^{2}\times \Bbb{R}$, this is , take $(a,b,c)=(z,c)$, where $z=(a,c)$ and now the proof is identical to the one you know.

 

[kalvin]

Hi,

Consider $\Bbb{R}^{3}$ as $\Bbb{R}^{2}\times \Bbb{R}$, this is , take $(a,b,c)=(z,c)$, where $z=(a,c)$ and now the proof is identical to the one you know.

Hi, I was just wondering if it was typo for z. Should z = (a, b) ? and not (a, c)? If it is not a typo do you think you could explain?

Thank you

 

[Fallen Angel]

Sorry, it was a typo, it should read $z=(a,b)$