Analytic function and connected region. We have the result.
Let $G$ be an open connected set in $\mathbb{C}$, and let $f : G \rightarrow \mathbb{C}$ be an analytic function. Then the following statements are equivalent.
*1. $f(z) = 0$ in $G$ for all $z$ *
2. There is a point $a \in G$ s.t. $f^n(a) = 0$ for each $n \in \mathbb{Z}$.
3. $\{z \in G : f(z) = 0\}$ has a limit point in $G$
Can we say $\mathbb{R}$ is open connected set in $\mathbb{C}$? Why or why not?
If we do not consider $G$ as a open connected region, the result will loss its validity. Please give me a counterexample of a function for which it may happen.
Thank you for your kind suggestions.

$\mathbb{R}$ is connected in $\mathbb{C}$. However, $\mathbb{R}$ is not open as a subset of  $\mathbb{C}$. In general, any set in a metric space is open whenever it is a union of discs $D_r(a)=\{x:\,d(x,a)<r\}$. 
Now, if $a\in\mathbb{R}$ there is no $r$ so that the complex disc $D_r(a)$ is a subset of $\mathbb{R}$.
That a function is analytic is a local property on the topology of $\mathbb{C}$. This means that a function $f$ defined on a set $\Omega$ is analytic provided that for any $z\in\Omega$ there exist a disc $D_r(z)\subset\Omega$ such that $f$ is analytic on $D_r(z)$. 
In particular, it is natural to assume the domain, $\Omega$, in the definition of analytic function is open.     
