# convergence of poles is an essential singularity

in Edward B. Saff, Arthur David Snider Fundamentals of complex analysis, with applications 2003, it has the following claim

Suppose we are given a function $$f$$ that is analytic and nonzero at each point of a simple closed cantour $$C$$ and is meromorphic inside $$C$$. Under these conditions it can be shown that $$f$$ has at most a finite number of poles inside $$C$$. The proof of this depends on two facts: first. the only singularities of $$f$$ are isolated singularities(poles), and, second, that every infinite sequence of points inside $$C$$ has a subsequence that converges to some point on or inside $$C$$. Hence if $$f$$ had an infinite numbers of poles inside $$C$$, some subsequence of them would converge to a point that must be a singularity, but not an isolated singularity of $$f$$.

I am struggling to understand why the bolded part is correct, why would a sequence of poles converge to an essential singularity?

If $$f$$ had infinitely many poles inside a bounded set, the sequence of poles would have a convergent subsequence which would converge to a non-isolated singularity of $$f$$.
It would not be an essential singularity which are isolated singularities ( $$\sin(1/z)$$ has an essential singularity at $$z=0$$ but $$1/\sin(1/z)$$ doesn't)
If the point (call it $$\ p\$$) were a pole of $$\ f\$$, or $$\ f\$$ were analytic at the point, then $$\ f\$$ would have either a Laurent or power series expansion in a disc $$\ D\$$ of positive radius centred on $$\ p\$$, and would therefore be analytic in the punctured disc $$\ D\setminus \{p\}\$$. This would contradict the existence of a sequence of poles of $$\ f\$$ converging to $$\ p\$$.