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I am trying to prove that if a meromorphic function is constant on an open subset $U$ then it is constant in all its domain $\Omega$. My idea is:

Let $F$ be meromorphic in $\Omega$. Then we have $F=\frac{f}{g}$ with $f$ and $g$ analytic. Since $F$ is holomorphic in $U$ we have $f-gK=0$. Now since $f-gK=0$ is holomorphic in $U$ by the identity theorem we have $f-gK=0$ in $\Omega$ and so $F=\frac{f}{g}=K.$

Is my proof correct, or I am missing something?

My main concern here is to know if every meromorphic function is of the form $F=\frac{f}{g}$ with $f$ and $g$ analytic.

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    $\begingroup$ You need to assume $\Omega$ is connected. $\endgroup$ – Robert Israel Jan 12 '20 at 4:03
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    $\begingroup$ @RobertIsrael $\Omega$ is a domain, and usually a domain is defined as a connected open subset of the complex plane. $\endgroup$ – Paul Frost Jan 12 '20 at 11:29
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Your proof is correct, but I think you can do it simpler.

A meromorphic $F$ function on a domain $\Omega$ is a function that is holomorphic on all of $\Omega$ except for a set $A$ of isolated singularities which are poles of the function. The set $B = \Omega \setminus A$ is open and connected and contains $U$, $F \mid_B$ is holomorphic, hence the identity theorem shows that $F \mid_B$ must be constant. This shows additionally that all possibly existing singularities are removable, thus $A$ must be empty.

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