Let $S$ be a surface in $\mathbb{R}^3$. Prove $\forall x\in S$ there is a neighborhood of $x$ diffeomorphic to $\mathbb{R}^2$.

I'm not sure how to prove this. I think it is close to the very definition of regular surface, therefore it would be trivially true: in a regular surface every point has a neighborhood diffeomorphic to an open set in $\mathbb{R}^2$, and aren't open sets in $\mathbb{R}^2$ diffeomorphic to $\mathbb{R}^2$?

But what happens if the surface is non-regular? For example $S=\{(x,y,z)\in \mathbb R^3: z^2=x^2+y^2\}$ is not a regular surfac., The proposition doesn't seem to apply to any neighborhood of $(0,0,0)$, around the vertex,


If $S \subset \mathbb{R}^3$ is a surface, then there exists charts $(U_{\alpha},f_{\alpha})$ such that $f_{\alpha}: U_{\alpha} \cap S \to f_{\alpha}(U_{\alpha} \cap S)$ is a diffeomorphism. We also require that for any other chart $(U_{\beta},f_{\beta})$ with $U_{\alpha} \cap U_{\beta}:=U_{\alpha \beta} \not = \emptyset$ the transition map;

$$f_{\alpha \beta} = f_{\beta}^{-1} \circ f_{\alpha}\Bigr|_{U_{\alpha \beta}}$$

is a diffeomorphism. Hence, by definition you have that each $x \in S$ has a neighborhood diffeomorphic to an open subset of $\mathbb{R}^2$. You can actually show more and say that each $x$ has a neighborhood which is diffeomorphic to an open ball in $\mathbb{R}^2$. It is enough to show that $\mathbb{R}^2$ is diffeomorphic to a ball of unit radius. Take the map;

$$x \mapsto \frac{x}{1+\|x\|}$$


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