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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?

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    $\begingroup$ Dominic, did you just mean "(...) there is a finite $\mathcal{V}\subset\mathcal{U}$" ? (I mean, not necessarily a subcover, otherwise the subsequent condition becomes trivial) $\endgroup$ – Pietro Majer May 29 '18 at 18:38
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    $\begingroup$ I did some googling. For regular spaces, $2$-star compactness is equivalent to $n$-star compactness for any $n\ge 2$. For Tychonoff spaces it is also equivalent to pseudocompactness. This is due to van Douwen, Reed, Roscoe and Tree in "Star covering properties" (top. app. 39 issue 1), where they prove much more than that. Maybe the examples you are looking for can be found in some of the references listed in this paper. $\endgroup$ – Mathieu Baillif May 29 '18 at 23:37
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    $\begingroup$ Examples as you ask are constructed in Section 4 of "Some generalizations of countable compactness" (Indian J. pure appl. math 17 (6)) by D.N. Sarkhel. It's too late for me to have a real look at it now, at first sight the construction does not look very complicated. $\endgroup$ – Mathieu Baillif May 29 '18 at 23:48
  • $\begingroup$ Thanks @MathieuBaillif - if you post this reference as an answer, I can accept it and then close this thread $\endgroup$ – Dominic van der Zypen May 30 '18 at 6:34
  • $\begingroup$ Right - thanks @PietroMajer, will correct this! Indeed compactness trivially implies $n$-star-compactness for any $n\in\mathbb{N}$ $\endgroup$ – Dominic van der Zypen May 30 '18 at 6:36
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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).

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