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Suppose the $f$ has an essential singularity at $z=a$.Prove that if $c\in \mathbb{C}$,and $\varepsilon >0$ are given,then for each $\delta >0$ there is a number $b$,$|c-b|<\varepsilon$,such the $f(z)=b$ has infinitely many solution in $B(a;\delta)$. This is an exercise from 'functions of one complex variable'.

I solved this question by using open mapping theorem and Baire Category Theorem to argue that $\bigcap_{i=1}^{\infty}f(\{z:0<|z-a|<1/n\})$ is a dense set.

Is there a solution which does not use Baire Category Theorem? Thank you.

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    I think there's a typo about $\epsilon$; the given $\epsilon>0$ is not mentioned in the rest of the theorem.2012-12-10
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    typo fixed.Thank you.2012-12-10
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    Sounds like Picard's theorem. Which states that for any except at most one of points z of C, f(z)=c has infinitely many solutions in a arbitrarily small neighborhood of it's essential singularity.2012-12-10
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    I want to avoid using Picard great thm,otherwise this problem will be trivial.2012-12-10
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    How did you make sure that there are infinitely many points such that $f(z) = b$, just using Baire's Theorem?2016-02-18

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