If a function is meromorphic on a ball, the set of poles is finite

Question

Let $X$ be a normed space, $\Omega\subseteq\mathbb C$ be open and $f:\Omega\to X$ be meromorphic in the sense that for all $z_0\in\Omega$, there is a $k_{z_0}\in\mathbb N_0$, $\left(a^{(z_0)}_k\right)_{k\ge-k_{z_0}}\subseteq X$ and a neighborhood $N_{z_0}\subseteq\Omega$ of $z_0$ with $$f(z)=\underbrace{\sum_{k=-k_{z_0}}^{-1}a_k^{(z_0)}(z-z_0)^k}_{=:\:m_{z_0}(z)}+\underbrace{\sum_{k=0}^\infty a_k^{(z_0)}(z-z_0)^k}_{=:\:h_{z_0}(z)}\tag1$$ for all $z\in N_{z_0}\setminus\{z_0\}$. Let $P$ denote the subset of $\Omega$ for which $k_{z_0}$ cannot be chosen to be $0$.

Let $B\subseteq\Omega$ be bounded. According to comments by Martin R below a previous version of this question, $F:=B\cap P$ is only finite, when $\overline B\subseteq\Omega$. Why?

I really tried to understand this, but I don't. First of all, discretness is a hereditary property of a topological space and hence there is no difference between $P$ being "discrete in $\Omega$" or "discrete in $\mathbb C$".

To make sure, we talk about the same concepts: Let $(E,\tau)$ be a topological space and $\mathcal N_\tau(x):=\{N\subseteq E:N\text{ is a }\tau\text{-neighborhood of }x\}$ for $x\in E$. Then,

• $B\subseteq E$ is called $\tau$-discrete if $\left.\tau\right|_B$ is the discrete topology on $D$;
• $x\in E$ is called $\tau$-isolated point of $B\subseteq E$ if $$\exists N\in\mathcal N_\tau(x):B\cap N=\{x\};$$
• $x\in E$ is called $\tau$-accumulation point of $B\subseteq E$ if $$\forall N\in\mathcal N_\tau(x):B\cap N\setminus\{x\}\ne\emptyset.$$

If $x\in E$ and $A\subseteq B\subseteq E$, it is straightforward to show that

1. if $x\in A$ and $x$ is an $\tau$-isolated point of $B$, then $x$ is an $\tau$-isolated point of $A$;
2. $x$ is an $\left.\tau\right|_B$-isolated point of $A$ iff $x$ is an $\tau$-isolated point of $A$;
3. $x\in E$ is an $\tau$-isolated point of $B$ iff $$\{x\}\in\left.\tau\right|_B;$$
4. $B$ is $\tau$-discrete iff every point in $B$ is an $\tau$-isolated point of $B$;
5. if $x$ is an $\tau$-accumulation point of $A$, then $x$ is an $\tau$-accumulation point of $B$;
6. $x$ is an $\left.\tau\right|_B$-accumulation point of $A$ iff $x$ is an $\tau$-accumulation point of $A$;
7. $B$ is $\tau$-discrete iff $D\cap K$ is finite for every $\tau$-compact $K\subseteq E$;
8. Each point in $B$ is either a $\tau$-accumulation point of $B$ or an $\tau$-isolated point of $B$.

From (7.), it should immediately follow that $P\cap\overline B$ is finite, since $\overline B$ is clearly compact (which, by the way, is a hereditary property as well). But if $P\cap\overline B$ is finite, then clearly $F=P\cap B$ is finite as well.

What am I missing?

EDIT: The example $\Omega=\{z\in\mathbb C:|z|<1\}$ and $P=\left\{1-\frac1n:n\in\mathbb N\right\}$, provided in the comments below, shows that something about (7.) seems actually to be wrong. But which of my assertions is wrong?

Proof of "$\Rightarrow$" in (7.): Let $K\subseteq E$ be $\tau$-compact and assume $D\cap K$ is infinite. Since $(K,\left.\tau\right|_K)$ is limit point compact, $D\cap K$ has a $\left.\tau\right|_K$-accumulation point $x$. By (5.) and (6.), $x$ is an $\tau$-accumulation point of $D$. Thus, by (8.), $x$ is not an $\tau$-isolated point of $D$; in contradiction to (4.).