This question was asked in my complex analysis quiz and I was unable to do it.
Can infinitely many points on the boundary $C$ of a domain be singular without $C$ being a natural boundary ?
I thnik it can be as on the boundary there are uncountable many points and if countably finite points are singular then it's not a problem as natural boundary in that case can be extended as there are points that are still regular.
But I want it to be checked. So I posted here.
Useful definitions: Suppose $f(z)$ is analytic in a domain $D$. A point $z_1$ is said to be a regular point of $f(z)$ if the function element $(f,D)$ can be analytically continued along some curve from a point in $D$ to the point $z_1$.
Any boundary point of $D$ that is not a regular point of $f(z)$ is said to be a singular point of $f(z)$.
Let $\Omega = \mathbb C \setminus [(-\infty,0]\cup \{1\}].$ Then $\partial \Omega = (-\infty,0]\cup \{1\}.$ Let $f(z)=\log z,$ the principal branch of the logarithm. Then $f$ is holomorphic on $\Omega$ and every $x\in (-\infty,0]$ is a singular point of $f.$ But $\partial \Omega$ is not a natural boundary for $f,$ as $f$ is holomorphic in a neighborhood of $1.$