Let $X$ be a topological space with a dense subset $D\subseteq X$. Suppose that every open cover of $X$ has a finite subfamily which covers $D$. Can I conclude that $X$ itself is compact?

The answer is clearly negative in general: one can take any space $X_0$ and form a new space $X:=X_0\cup\{\infty\}$ by adjoining a new point $\infty$ such that the nonempty open sets are those of the form $\{\infty\}\cup U$ for $U$ open in $X_0$. Then the closure of $\{\infty\}$ is the whole space, so that the condition trivially holds with $D=\{\infty\}$. Now take any non-compact $X_0$.

Nevertheless, compactness of $X$ indeed follows if $X$ is completely regular. One way to show this is to argue that the condition is equivalent to every net in $D$ having a cluster point, and then using uniformizability to approximate an arbitrary net in $X$ by one in $D$. But of course, a finite subfamily which covers $D$ does not necessarily cover $X$ yet, even though a finite subcover is guaranteed to exist.

So, perhaps surprisingly, the answer is positive in particular for any metric space $X$.

Now my question is:

Are there weaker separation conditions than complete regularity on $X$ which guarantee that the finite subcover property for $X$ only needs to be checked on a dense subset $D\subseteq X$?

  • 2
    $\begingroup$ Notice that spaces such that every cover has a finite subfamily with a dense union are called H-closed. H-closed spaces are well studied, and as Ramiro de la Vega wrote, regular H-closed spaces are indeed compact. $\endgroup$ – Mathieu Baillif Jan 1 '18 at 10:56

Regularity is enough. Given an open cover $\mathcal{A}$ of $X$ we can use regularity to build another cover $\mathcal{B}$ such that for any $U \in \mathcal{B}$ there is a $V \in \mathcal{A}$ with $\overline U \subseteq V$. Now a finite subset of $\mathcal{B}$ covering $D$ naturally provides a finite subset of $\mathcal{A}$ that covers $X$.

Completely Hausdorff is not enough. Let $X=[0,1]$ with the topology generated by the usual open sets plus a new open (dense) set $D=X \setminus \{1/n:n \in \mathbb{Z}^+\}$. Since this topology is strictly finer than a compact Hausdorff topology, it is noncompact and completely Hausdorff. Moreover, any open cover of $X$ contains a finite subset that covers $D$.

  • $\begingroup$ Very nice, this is exactly what I was hoping for! $\endgroup$ – Tobias Fritz Dec 31 '17 at 17:16

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