question about finer topology

Question

‎‎A topological space $X$ is said to be star compact if whenever $\mathscr{U}$ is an open cover of $X$, there is a compact subspace $K$ of $X$ such that $X = \mathrm{st}(K,\mathscr{U}),$ where $\mathrm{st}(K, \mathscr{U}) = \bigcup \{ U \in \mathscr{U}: U \cap K \neq \emptyset \}.$

We recursively define $\mathrm{st}^n$ for $n=0,1,2,\ldots$ by $$\begin{align} \mathrm{st}^0(K, \mathscr{U}) &= K \\ \mathrm{st}^{n+1}(K, \mathscr{U}‎) &= \bigcup ‎\{ U‎ ‎\in‎ \mathscr{‎U} : U ‎\cap\ \mathrm{st}^n(K, \mathscr{U}‎) \neq ‎\emptyset \}‎‎‎ \end{align}$$

Definition: ‎A space $X$ is said to be ‎$ ‎\omega‎$‎-starcompact if for every open cover ‎$\mathscr{U}‎$‎ of ‎‎$X$‎, there is some ‎‎$‎n ‎\in ‎\mathbb{N‎}‎‎^{+}‎$‎ and some finite subset ‎‎$B$‎ of ‎‎$X$‎ such that ‎‎$\mathrm{st}^{‎n‎}(B, \mathscr{U}) = X‎$‎.‎‎ ‎

Let‎‎ ‎$(X, ‎\tau)$ be‎ $\omega‎$‎-‎starcompact and $\tau^{*} \subset \tau$. Is $(X, \tau^{*})$ a‎ $‎\omega‎ $‎-‎starcompact ‎space?‎

Answer 1

Of course, every $\tau^\ast$-open cover is a $\tau$-open cover and we have $B$ there which still works. Many simple covering properties are preserved by going to a coarser topology, like (countable) compactness and Lindelöfness.

Answer 2

Yes, I think so!

To see this, fix $\mathscr{U} \subseteq \tau^* \subseteq \tau $, an open cover of $X$, and pick a finite set $B \subseteq X$ and whole number $n$ for which $st^n(B,\mathscr{U}) = X$.

Note $st(B,\mathscr{U})$ is the same open set (i.e. the same union of open sets from the open cover) regardless of which topology we consider. In fact, for all $k$, $st^k(B,\mathscr{U})$ is the same union of open sets from the open cover regardless of which topology we consider. As such, $st^n(B,\mathscr{U}) = X$, regardless of which topology we consider.