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How to prove that the set

$S =\left\{(x,y,z)\in \mathbb{R}^3 \mid z=0, \ \ x^2+y^2\leq1 \right\}$

is not a regular surface ?

Intuitively I can see two possible reasons why $S$ is not regular: One is that for any open set $V$ of $S$ (with subspace topology), $V$ is not homeomorphic to an open set of $\Bbb R^2$ (but I don't know how to prove that).

Another is that there is no tangent plane at any point in the boundary.

But I have not a clue of how to go further.

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Yes, any neighborhood of a point on the unit circle cannot, by the Inverse Function Theorem, be diffeomorphic to an open set in $\Bbb R^2$. Without derivatives, you need a theorem called Invariance of Domain.

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    $\begingroup$ @ThomasShelby: Suppose $U\subset\Bbb R^2$ is an open set (say containing $0$), $V\subset S$ a neighborhood of a boundary point $p$, and $f\colon U\to V$ is a diffeomorphism with $f(0)=p$. Viewing $f$ as a map from $U$ to $\Bbb R^2$, the inverse function theorem tells us $f$ is an open map. This means that $p$ must be an interior point of $V$. $\endgroup$ Jan 26 '20 at 17:29

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