U $\subset \mathbb{C} $ is an open set and f is a meromorphic function in U. Prove that if $(z_{n})_{n\geq 1} $ is a sequence in U that converges to a certain $z^{*} \in U$ and the function f has a pole in $z_{n}$ for every n, then f has a pole in $z^{*}$.
I have been trying to prove this, knowing that this is impossible.I am trying to show the contradiction.
Here is the definition of a pole:
A function f of a complex variable z is meromorphic in the neighbourhood of a point ${\displaystyle z_{0},}$ if either f or its reciprocal function 1/f is holomorphic in some neighbourhood of ${\displaystyle z_{0}} $ (that is, if f or 1/f is differentiable in a neighbourhood of ${\displaystyle z_{0}})$ ''neighbourhood means a disc centered at $z_{0}$ of radius $\varepsilon$; $D(z_{0},\varepsilon)$ ''. If ${\displaystyle z_{0}} $ is a singularity of 1/f , then it is a pole of f.
My solution:
My argument is basically since the sequence $z_{n}$ is convergent then
$lim_{n \to \infty } |z_{n+1}-z_{n}|=0$
then if $z^{*}$ is a pole of f, then f has to be holomorphic on $D(z^{*},\varepsilon)$. But this is then impossible because no matter how small we choose $\varepsilon$ there will be always a pole in the radius we choose because of this condition $lim_{n \to \infty } |z_{n+1}-z_{n}|=0$.
So basically this is the way i was trying to prove it but i find it not rigorous enough or incomplete. I am trying to prove it in a rigorous mathematical way.