How to prove that set of regular points of an analytic function is open This question is from Pg 457 of Ponnusamy and Silvermann's Complex analyis book.

How to prove that set of regular points of an analytic function is open ?

$z_1$ is called regular point wrt analytic function f(z) if for $z_1$ there exists a curve such that  function element (f,D) can be analytically from the point in D to point $z_1$.
If $z_1$ is assumed to be analytic then it doesn't means that i can draw an open 2 -dimensional  ball around $z_1$ so that each point inside the ball is also regular point.
I think I need to assume atleast 1 sigular point in that open ball but I am unable to find any contradiction.Can you please tell which result should i use?
In case it is helpful, I have proved that the set of singular points of an analytic function form a closed set.
Singular points are those points on the boundary of Domain of function element which are not  regular points.
 A: The claim follows from the fact that a domain in the book is defined as an open (non-empty) connected subset of the complex plane.
Using the notation in the book: Let $(f,D)$ be a
function element (i.e. a domain $D$ and an analytic function $f$ on $D$). A point $z_1$ is a regular point if there is a chain of function elements $(f_k,D_k)$, $k=0,...,n$ and a continuous curve $\gamma:[0,1]\mapsto {\Bbb C}$ with the following properties:
(1) For $0\leq k<n$ we have  $D_k\cap D_{k+1} \neq \emptyset$ and ${f_k}_{|D_k\cap D_{k+1}} = {f_{k+1}}_{|D_k\cap D_{k+1}}$.
(2) $z_0=\gamma(0)\in D\cap D_0$, $\gamma(1)=z_1\in D_n$ and $\gamma([0,1])\subset D_0\cup ... \cup D_n$.
Now, since $D_n$ is open there is $r>0$ such that $B(z_1,r)\subset D_n$. If $w\in B(z_1,r)$ then $f_n$ is analytic in a neighborhood of $w$ and by concatenating the path $\gamma$ and the line-segment $[z_1;w]$ joining the two points we construct a new path $\tilde{\gamma}$ for which all the above properties still hold with $w$ now being the end-point. Thus, every point in $B(z_1,r)$  is a regular point.
