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Let S be a smooth surface and p a point in S with positive Gaussian curvature. How can i prove that p has a neighbourhood U such that $U \setminus \{p\} $ lies entirely on one side of the plane $p+T_p (S) $?

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I'll give the idea briefly. Applying an isometry of $\Bbb R^3$ if necessary, you can assume that $p = 0$ and that $T_pS$ is the plane $z=0$. Then, near $p$, $S$ is the graph of the second fundamental form. In other words, the graph of $$(x_1,x_2, \sum h_{ij}x_ix_j + {\rm error})$$Near $p$, the sign of $\sum h_{ij}x_{i}x_j + {\rm error}$ is the same of the quadratic expression $\sum h_{ij}x_ix_j$. If $K(p) > 0$, $(h_{ij})$ is definite, and if $K(p) < 0$, $(h_{ij})$ is indefinite. Done.

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