(Differentiable Jordan Curve Theorem) Let $\alpha:[0;1]\to\mathbb{R}^2$ be a plane, regular, closed, simple curve. Then $\mathbb{R}^2\setminus\alpha([0;1])$ has exactly two connected components.
If I denote $C_1,C_2$ is connected components of $\mathbb{R}^2\setminus\alpha([0;1])$ then why $\partial C_1=\partial C_2=\mathrm{Im}(\alpha)$??
Help me, please!
In $\mathbb{R^2}$, the boundary of $C_1$ is defined to be $$\partial C_1 = Cl(C_1) \cap Cl(X-C_1)=Cl(C_1)\cap Cl(C_2\cup Im(\alpha)).$$ Because $C_1$ and $C_2$ are disjoint and $Im(\alpha)\subset Cl(C_1)$, this is equal to $Im(\alpha )$. You can do the same with $C_2$, so that we proved what we wanted.