While trying to prove "$f$ is an entire function such that $|f(z)| \to \infty, as |z|\to \infty$, then $f(\mathbb{C}) $ is closed"
I accidentally showed that
"$f$ is an entire function, then $f(\mathbb{C})$ is closed" I know this is not a correct statement, but I need to know where did I gone wrong and I need some hints to correct the proof...
the proof:-
Let $a$ be a cluster point of $f(\mathbb{C})$, then $\exists f(z_n) \in f(\mathbb{C})$ such that $f(z_n) \to a$
Now consider $$g(z): B[a,1] \to \mathbb{C}$$ $$g(z)=|f(z)-a|$$
This is a continuous real valued function on a compact set and hence attains a minimum, suppose it attains its minimum on the boundary, say the minimum is $r>0$, but since we have $f(z_n)\to a$, I can find a $f(z_k) \in B[a,1]$ such that $|f(z_k)-a|<r$ and hence the minimum is attained in the open ball $B(a,1)$, and hence by Minimum modulus principle the minimum is zero,
$$\implies \exists w , s.t f(w)=a$$
Which implies $a \in f(\mathbb{C})$ and hence $f(\mathbb{C})$ is Closed.
where did the proof gone wrong and what can I do to rectify and give a proof of the original statement.
Your seqeunce $z_n$ has a bounded subsequence (because if $|z_n| \to \infty$ the $|a|=\lim |f(z_n)|=\infty$, a contradiction). If $z_{n_k} \to z$ then $a=f(z)$ so $a \in f(\mathbb C)$.