Let $X$ be a set. For $B\subseteq X$ and ${\cal H}\subseteq {\cal P}(X)$ we set $$\text{ST}^1(B,{\cal H}) = \{H\in {\cal H}: H\cap B\neq \emptyset\},$$ and $\text{st}^1(B,{\cal H}) = \bigcup \text{ST}^1(B,{\cal H})$. For any integer $n>1$ we inductively set $$\text{ST}^{n+1}(B,{\cal H}) = \{H\in {\cal H}: H\cap\text{st}^n(B,{\cal H}) \neq \emptyset\},$$ and $\text{st}^{n+1}(B,{\cal H}) = \bigcup \text{ST}^{n+1}(B,{\cal H})$.


Finally, a topological space $(X,\tau)$ is said to be *$n$-star-compact* if for every open cover ${\cal U}$ of $X$ there is a finite subset ${\cal V}\subseteq {\cal U}$ such that $$\text{st}^n(\bigcup{\cal V},{\cal U}) = X.$$ Obviously, any $n$-star-compact space is $(n+1)$-star-compact.


Given any $n\geq 1$, what is an example of a space that is $(n+1)$-star-compact, but not $n$-star-compact?



$n$-star-compactness
As I wrote in my comments, D.N. Sarkhel constructs such examples in Section 4 of "Some generalizations of countable compactness" (Indian J. pure appl. math 17 (6), 1986). It works as follows. Fix some $n\ge 2$. Take a partition of $[0,1]$ into pairwise disjoint dense subsets $A\_i$, $i = 1,\dots, 2n$. If $x\in[0,1]$ there is a unique $n(x)$ such that $x\in A\_{n(x)}$. Then set $E\_i=A\_i$ when $i$ is odd and $E\_i = A\_{i-1}\cup A\_i\cup A\_{i+1}$ if $i$ is even, where $A\_{2n+1} = A\_{2n}$. Then $G$ is open if for each $x\in G$ there is an interval $I(x)$ such that $x\in I(x) \cap E\_{n(x)} \subset G$. Sarkhel proves on page 782 that this space is Hausdorff and $n$-star compact but not $n-1$-star compact.


It is not possible to obtain regular such spaces since $2$-star compactness is equivalent to $n$-star compactness for any $n\ge 2$ in this class, and to pseudocompactness in the class of Tychonoff spaces. This is shown by van Douwen, Reed, Roscoe and Tree in "Star covering properties" (top. app. 39 issue 1, 1991).


