It is known that every compact space is star compact ( some of authors called it star finite).
I cannot find an example for a star compact subset $\mathbb{R}$ which is not compact. Is it true that star compact subset of $\mathbb{R}$ is compact?
every star compact subset of $\mathbb{R}$ is compact
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$\begingroup$
general-topology
compactness
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0http://math.stackexchange.com/questions/2131665/a-star-compact-subset-of-mathbbr-is-closed Are you friends? – 2017-02-06
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0I'm not sure most authors equate *star compact* and *star finite*; see the discussion under [this previous Question](http://math.stackexchange.com/questions/95543/is-there-a-topological-space-which-is-star-compact-but-not-star-countable) and consider adding an explicit definition. – 2017-02-06
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0A space $X$ is star compact if for any open cover $\mathcal{V}$ of $X$, there is a finite $F\subseteq X$ such that $X= st(A,\mathcal{V}) = \cup\{ V \in \mathcal{V}: V \cap F \neq \emptyset$ – 2017-02-06
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0So you have no problem with my putting this in your Question, to make it self-contained? I would say this defines *star finite*. Note the definition given in the earlier Question for *star compact*. Of course $\mathbb{R}$ is a $T_4$ space, so there the notions are equivalent. – 2017-02-06