# Statement about the implicit function theorem which I can't understand

Theorem: Let $F:X\subseteq \mathbb{R^n}\rightarrow \mathbb{R}$ be of class of $C^1$ and let $a$ be a point of the level set $S=\{x\in\mathbb{R^n}|F(x)=c\}$. If $F_{x_n}(a)\neq 0$ then there is a neighborhood $U$ of $(a_1,a_1\dots a_{n-1})\in\mathbb{R^{n-1}}$, a neighborhood $V$ for of $a_n\in\mathbb{R}$ and a function $f:U\subseteq\mathbb{R^{n-1}}\rightarrow V$ of class $C^1$ such that if $(x_1,x_2,...x_{n-1})\in U$ and $x_n \in V$ satisfy $F(x_1,x_2...x_n)=c$ ,then $x_n=f(x_1,x_2,...x_n)$.

I can't quite understand statement " If $F_{x_n}(a)\neq 0$ then there is a neighborhood $U$ of $(a_1,a_1\dots a_{n-1})\in\mathbb{R^{n-1}}$, a neighborhood $V$ for of $a_n\in\mathbb{R}$ and a function $f:U\subseteq\mathbb{R^{n-1}}\rightarrow V$ of class $C^1$ such that if $(x_1,x_2,...x_{n-1})\in U$ and $x_n \in V$ satisfy $F(x_1,x_2...x_n)=c$, then $x_n=f(x_1,x_2,...x_n)$."Why the statement is true and what idea it talks about?

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So, what exactly gives you trouble here? – fedja Jan 9 '12 at 12:16
As the topic,i dont quite understand the theorem why the statement is ture? – johnny Jan 9 '12 at 12:19

So for example if you plot the unit circle $S^1 \subseteq \mathbb{R}^2$, at any point on the circle which doesn't lie on the $x$-axis, you can find a neighbourhood on which the circle is the graph of a function. So for example we can view the portion of $S^1$ lying in the upper-half-plane as the graph of $\sqrt{1-x^2}$, and the portion in the lower-half-plane as the graph of $-\sqrt{1-x^2}$. Here's a nice illustration from Wikipedia:
In this illustration, the point $A$ has a neighbourhood which projects onto the $x$-axis; and we can recover the neighbourhood of $A$ on the graph by taking the set of points $(x,\sqrt{1-x^2})$ for $x$ in this projection onto the $x$-axis. On the other hand, there is no neighbourhood of the point $B$ which is locally the graph of a function, because it lies on the $x$-axis; this is the significance of the stipulation $F_{x_n}(a) \ne 0$ in your statement of the theorem.
The implicit function theorem takes this idea and places it in the arbitrary finite-dimensional case of functions $\mathbb{R}^n \to \mathbb{R}^m$ and subsets of $\mathbb{R}^{n+m}$, rather than just the case of functions $\mathbb{R} \to \mathbb{R}$ and subsets of $\mathbb{R}^2$. (In fact, your case is slightly less general: it considers functions $\mathbb{R}^{n-1} \to \mathbb{R}$ and subsets of $\mathbb{R}^n$.)