Poles of a meromorphic function form a closed discrete set and are at most countable

Fix $a\in P$. By [L1], $a$ is a pole, so some radius $r_a>0$ has $\mathrm{\mid }z-a\mathrm{\mid }<r_a$ contained in $\mathrm{\Omega }$ and $f$ holomorphic on $0<\mathrm{\mid }z-a\mathrm{\mid }<r_a$. If $b\in P$ and $0<\mathrm{\mid }b-a\mathrm{\mid }<r_a$, then $b$ lies in a region where $f$ is holomorphic, contradicting $b\in P$. Thus $B(a,r_a)$ contains no pole other than $a$.

Let $\mathcal{D}$ be the family of discs $B(p+iq,s)$ with $p,q,s\in \mathbb{Q}$ and $s>0$. By [L2], ${\mathbb{Q}}^3$ is at most countable, so $\mathcal{D}$ is at most countable and admits an injection $e:\mathcal{D}\to \mathbb{N}$.

Step 1.1 proves that $P$ is discrete in $\mathrm{\Omega }$. If $c\in \mathrm{\Omega }\setminus P$, then [L1] says $f$ is holomorphic on a neighbourhood of $c$, and that neighbourhood contains no point of $P$; hence $\mathrm{\Omega }\setminus P$ is open and $P$ is closed in $\mathrm{\Omega }$.

For each $a\in P$, step 1.1 gives $r_a>0$. Write $a=x+iy$. By [L3], choose rationals $p,q$ with $\mathrm{\mid }x-p\mathrm{\mid }<r_a\mathrm{/}8$ and $\mathrm{\mid }y-q\mathrm{\mid }<r_a\mathrm{/}8$, so $\mathrm{\mid }a-(p+iq)\mathrm{\mid }<r_a\mathrm{/}4$; choose a rational $s$ with $\mathrm{\mid }a-(p+iq)\mathrm{\mid }<s<r_a\mathrm{/}2$, again by [L3]. Then $a\in B(p+iq,s)\subseteq B(a,r_a)$, so the set of discs in $\mathcal{D}$ containing $a$ and contained in $B(a,r_a)$ is nonempty.

For each $a\in P$, the set $E_a:=\{e(D):D\in \mathcal{D},a\in D\subseteq B(a,r_a)\}$ is nonempty by step 2.2, so [L4] gives its least element; call it $j(a)$. If $j(a)=j(b)$, then injectivity of $e$ makes the corresponding discs equal, and that disc lies inside $B(a,r_a)$ and contains both $a$ and $b$, so step 1.1 forces $a=b$. Therefore $a\mapsto j(a)$ is injective from $P$ into $\mathbb{N}$.

The injection of step 3.1 makes $P$ at most countable, completing the proof.