Does replacing analytic by continuity in open mapping theorem holds the result? and is it true for closed sets?

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1) My first question depends on the open mapping theorem

Which state that a non constant analytic function maps open sets onto open sets does the result holds if we use continuous function instead of analytic???

2) Can I say that an analytic function maps closed sets onto closed sets?

3) A question here related to the poles maybe can $f(z)$ be analytic in a deleted neighborhood of z. even when $\lim (z-z.)^n$. $f(z)$ does not exist for $z \mapsto z$. For any integer n (may be this question is important even before the use of open mapping theorem

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(1) No. Non-constant continuous functions need not be open maps. For example, $f(z)=\frac12(z+\overline z)$ is continuous, but not open, as it maps the whole plane (which is open) onto the real axis (which is not).

(2) Not always. For example, consider what the function $f:\Bbb C^*\to\Bbb C$ given by $f(z)=\frac1z$ does to the closed set $\{x+iy:x>0,xy=1\}$.

(3) $f(z)=e^{1/z}$ is analytic in every deleted neighborhood of $0$, but for any $n$, $\lim_{z\to 0}z^nf(z)$ fails to exist. If a function has a point like this, it is called an essential singularity.