Is every star compact subset of $\mathbb{R}$ closed?

Question

If $A\subseteq \mathbb{R}$ is star compact, is true that $A$ is closed?

A space $X$ is called star compact if for any open cover $\mathcal{U}$ of $X$, there is a finite $A\subseteq X$ such that $X= \operatorname{st}(A,\mathcal{U})$, where $$\operatorname{st} ( A , \mathcal{U} ) = \bigcup \{ U \in \mathcal{U} : A \cap U \neq \emptyset \}.$$

I think it is true, and I think we can prove it with the following property:

$A \subseteq{R}$ is closed iff every convergent sequence of points in $A$ converges to a point $A$.

But I have a problem to show that the limit is in $A$.

Answer 1

This is true. And more. In a metric space being star-compact is equivalent to being compact.

• Proof. In Hausdorff spaces, star-compactness is equivalent to countable compactness. (See, for example, this question.) And in metric spaces countable compactness is equivalent to compactness.

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For a more direct proof that star-compact subsets of $\mathbb R$ are closed, consider the following.

Suppose that $B \subseteq \mathbb R$ is not closed. Pick some $b \in \overline B \setminus B$. Then there is a monotone sequence $( b_n )_n$ in $B$ which converges to $b$. Without loss of generality we may assume the sequence is strictly increasing. Consider the collection $\mathcal U$ of open subsets of $\mathbb R$ consisting of the following sets:

• $( -\infty , b_1 )$,
• $( b_n , b_{n+2} )$ for all $n \in \mathbb N$,
• $( b , + \infty )$.

As $\bigcup \mathcal U = \mathbb R \setminus \{ b \}$, it follows that $\mathcal U$ covers $B$.

I claim that there is no finite $A \subseteq \mathbb R$ such that $B \subseteq \operatorname{st} ( A , \mathcal U )$. For this note that no finite subcollection of $\mathcal U$ covers $A$, since each set in $\mathcal U$ contains at most one $b_n$. Also, each $x \in \mathbb R$ is contained in only finitely many (in fact, at most two) sets in $\mathcal U$. Thus for every finite $A \subseteq \mathbb R$ the set $\operatorname{st} ( A , \mathcal U )$ is a finite union of sets in $\mathcal U$, and therefore cannot cover $B$.