essential singularity of a function Let $f:\Omega \to \Bbb C$ , and $a$ s.t $f$ has essential singularity at $a$.
Let $g : \Bbb C \to \Bbb C$ be entire. i need to prove that $g \circ f$ has essential singularity at $a$.
So , first i showed that $g(\Bbb C)$ is dense. 
Now, if we assume $a$ is a removable singularity of $g\circ f$ then it is bounded in a deleted nbhd of $a$ , so $g \circ f( B_a(r) -\{a\} ) \subset B_R(0)$
Now from  Casorati-Weierstrass Theorem $f( B_a(r) -\{a\} )$ is dense in $\Bbb C$ so $g(\Bbb C) \subset g \circ \overline{f( B_a(r) -\{a\}} )$, but this do not contradict the fact that $g \circ f( B_a(r) -\{a\} )$ is bounded.
Someone can help with that? 
Thanks.
 A: Here's an idea; correct me if I've made a mistake.
If $g \circ f$ doesn't have an essential singularity at $a$ then $lim_{z \to a}g \circ f(z) \in \mathbb{C} \cup \{\ \infty\}$.
Let's first suppose that the above limit is in $\mathbb{C}$ and call it $c$. Take some $w \in \mathbb{C}$ s.t $|w-c| > 2$ to get that by continuity of $g \circ f$ there exist $r_1, r_2 >0 $ with:
$B(a, r_1) \subset \Omega$ and $B(w, r_2) \cap g\circ f(B(a,r_1)) = \emptyset$.   
Yet the Casorati-Weierstrass theorem implies that $f(B(a, r_1) - \{a\})$ is dense in $\mathbb{C}$. And further, $g(\mathbb{C})$ is itself dense.
So let $\epsilon >0$ be with $B(w,\epsilon) \subset B(w, r_2)$. Take some $z_{\epsilon} \in \mathbb{C}$ s.t $g(z_{\epsilon}) \in B(w,\epsilon)$. Continuity of $g$ gives a $\delta >0 $ s.t $g(B(z_{\epsilon}, \delta)) \subset B(w,\epsilon)$.
Now take $z \in f(B(a, r_1))$ s.t $z \in B(z_{\epsilon}, \delta)$ and we get $g(z) \in B(w, r_2)$ - this contradicts $B(w, r_2) \cap g\circ f(B(a,r_1)) = \emptyset$.
Now if $c \in \{\ \infty\}$ just take $w = 0$.
Again, correct me if there's an error..
