Rothberger property and semi-open sets

Question

Here is the definition of a space $X$ to be Rothberger, if for each sequence $(\mathcal{U}_{n})_{n\in\mathbb{N}}$ of open covers of $X$, there exists a sequence $(U_{n})_{n\in\mathbb{N}}$ where $U_{n}\in\mathcal{U}_{n}$ and $X=\bigcup_{n\in\mathbb{N}} U_{n}$.

Definition: A subset $A$ of $X$ is said to be semi-open if $A\subset cl(int(A))$ where $cl$ and $int$ are closure and interior operator, respectively.

The space $X$ is said to be s-Rothberger if we can find a sequence for each semi-open covers sequence of $X$.

Let $\mathbb{R}$ be set of real numbers and $p\in\mathbb{R}$ be fixed. Consider $\mathbb{R}$ with the Fort space $$\tau=\{U\subset\mathbb{R}: p\notin U \text{ or if } p\in U \text{ then } \mathbb{R}\setminus U \text{ is finite}\}.$$ Since this space is compact and scattered, it is Rothberger. Is this space also s-Rothberger? It actually may be hard to determine all semi-open sets of this space, but any help or hint will greatly be appreciated.

Answer 1

First let's identify the semi-open sets. We first note all open sets are open. A simpler definition of the open sets are (assuming $p=0$),

$$\tau=\{U\subseteq\mathbb R:0\in U\Rightarrow\mathbb R\setminus U\text{ finite}\}.$$

If $0\not\in S$, then $S$ is open and thus semi-open.

If $0\in S$ and $S$ is open, then $S$ is infinite. If $S$ is not open, then $int(S)=S\setminus\{0\}$. Then $0\in cl(int(S))$ provided $S$ is infinite. Thus the semiopen sets are exactly

$$\tau_{semi}=\{S\subseteq\mathbb R:0\in S\Rightarrow S\text{ infinite}\}.$$

So $\{\mathbb Q\}\cup\{\{x\}:x\in\mathbb R\setminus\mathbb Q\}$ is an uncountable partition of $\mathbb R$ into semi-open sets. Thus there exists no countable subcover, violating s-Lindelof (every semi-open cover has a countable subcover) and thus s-Rothberger.