In Tu's book " An introduction to smooth manifold". There is a statemnent :
Topologist's Sine Curve is not a regular submanifold of $\Bbb R^2$.The reason is as following:
If $p$ is in the interval $I=\{(0,y)\in \mathbb{R^2} |−1<y<1\}$, then there is no adapted chart containing $p$,
since any sufficiently small neighborhood $U$ of $p$ in $\Bbb R^2$ intersects $S$ in infinitely many components.
How to derive a contradiction?
Best Answer
Let $p = (0,\eta) \in I$. Assume $(U,\phi) = (U,x^1,x^2)$ is an adapted chart rel. $S$ with $p \in U$. Then $q \in U \cap S$ iff $x^2(q) = 0$.
$\phi$ is a diffeomorphism between $U$ and an open subset $V \subset \mathbb R^2$. In particular, the Jacobian determinant of $\phi$ at $p$ must be nonzero. Now let us compute $\frac{\partial x^2}{\partial x}(p)$ and $\frac{\partial x^2}{\partial y}(p)$.
Since $x^2$ vanishes on $I \cap U$, we get $\frac{\partial x^2}{\partial y}(p) = 0$.
There exists a sequence $p_n = (h_n,\eta) \in S$ such that $h_n > 0$ and $h_n \to 0$. Thus $p_n \to p$ which implies $p_n \in U$ for $n \ge n_0$. Then $p + h_n(1,0) = (0,\eta) + (h_n,0) = (h_n,\eta) = p_n$, hence $$\frac{\partial x^2}{\partial x}(p) = \lim_{n \to \infty} \frac{x^2(p + h_n(1,0)) - x^2(p)}{h_n} = \lim_{n \to \infty} \frac{0 - 0}{h_n} = 0 .$$ This shows that the Jacobian determinant of $\phi$ at $p$ is zero, a contradiction.