If $X\neq \emptyset$ is a set, we say ${\cal C}\subseteq {\cal P}(X)\setminus \{\emptyset\}$ is a *cover* if $\bigcup{\cal C} = X$, and ${\cal C}$ is said to be *minimal* if for all $D\in {\cal C}$ , the collection ${\cal C}\setminus\{D\}$ is no longer a cover. (Equivalently: for every member $D$ of a minimal cover, there is $x\in X$ such that $x$ is only covered by $D$.)


We say that a topological space $(X,\tau)$ is $\min$-*compact* if every open cover has an open refinement that is a minimal cover of $X$.


It is clear that any compact space is $\min$-compact. Are there implications between $\min$-compactness and metacompactness?



$\min$-compactness vs metacompactness
Using Zorn's lemma, one can show that every point-finite cover of a set has a minimal subcover. Therefore meta-compactness implies min-compactness. I claim that there are min-compact spaces that are not meta-compact (my example is not $T\_{1}$).


Suppose that $X$ is a topological space.


Then we say that a subset $A\subseteq X$ is a discrete spanner if


1. for each $a\in A$, there is an open set $U\subseteq X$ with $U\cap A=\{a\}$ (in other words, the subspace topology on $A$ is discrete), and
2. whenever $(U\_{a})\_{a\in A}$ is a system of open sets with $U\_{a}\cap A=\{a\}$ for each $a\in A$, we have $\bigcup\_{a\in A}U\_{a}=X$.


If $A$ is a discrete spanner, then condition 2 can be strengthened to say that every collection of open subsets of $X$ that covers $A$ must also cover $X$.


Proposition: If $X$ has a discrete spanner, then $X$ is min-compact.


Proof: Suppose that $\mathcal{U}$ is a cover of $X$. Then for each $a\in A$, suppose that $a\in U\_{a}\in\mathcal{U}$. Furthermore, suppose that $V\_{a}$ is a set that is open in $X$ where $V\_{a}\cap A=\{a\}$. Then $(U\_{a}\cap V\_{a})\_{a\in A}$ is a cover of $X$, but $(U\_{a}\cap V\_{a})\_{a\in A}$ is minimal since if $a,b\in A,b\in U\_{a}\cap V\_{a}$, then $a=b$. Q.E.D.


Proposition: Every $T\_{1}$-space with a discrete spanner is discrete.


The above proposition will follow from more general results that we shall prove below.


If $X$ is a topological space, then we give $X$ a pre-ordering $\leq$ known as the specialization ordering where $x\leq y$ if and only if $\overline{\{x\}}\subseteq\overline{\{y\}}$. Recall that the specialization ordering $\leq$ is a partial ordering if and only if $X$ is $T\_{0}$, and the specialization ordering $\leq$ is simply equality if and only if $X$ is $T\_{1}$.


Proposition: Let $X$ be a topological space. If $A\subseteq X$ is a discrete spanner for $X$, then $A$ is the set of all minimal elements with respect to the specialization ordering and for each $x\in X$, there is an $a\in A$ with $a\leq x$.


Proof: Suppose that $A$ is a discrete spanner. I first claim that for each $x\in X$, there is an $a\in A$ with $a\leq x$. Suppose to the contrary, $x\in X$ but there is no $a\in A$ with $a\leq x$. Then $(\overline{\{x\}})^{c}$ is an open set with $A\subseteq(\overline{\{x\}})^{c}$ but where $(\overline{\{x\}})^{c}\neq X$ which is a contradiction. Now, each element in $A$ is minimal. Otherwise, there would be $a,b\in A$ with $a<b$, and this would contradict the minimality of $A$. Furthermore, $A$ must contain all minimal elements since if $x\in X$ is minimal, then $x\geq a$ for some $a\in A$, so $x=a$ by minimality. Q.E.D.


Proposition: Suppose that $X$ is a topological space. Then a subset $A\subseteq X$ is a discrete spanner for $X$ if and only if


1. $A$ is discrete,
2. $A$ is the collection of all minimal elements in $X$ with respect to the specialization ordering, and
3. for each $x\in X$, there is some $a\in A$ with $a\leq x$.


Proof: We have already proven $\rightarrow$. For $\leftarrow$, it suffices to show that whenever $U\_{a}\cap A=\{a\}$ for each $a\in A$, we have $\bigcup\_{a\in A}U\_{a}=X$. However, if $x\in X$, then there is some $a\in A$ with $a\leq x$. Therefore, we have $x\in U\_{a}\subseteq\bigcup\_{a\in A}U\_{a}$. We can conclude that $\bigcup\_{a\in A}U\_{a}=X$. Q.E.D.


Suppose that $X$ has a discrete spanner $A$. Then $X$ must be min-compact, but $X$ is not necessarily metacompact. In fact, $X$ is metacompact if and only if
there is a system of open sets $(U\_{a})\_{a\in A}$ with $\{a\}=U\_{a}\cap A$ for each $a\in A$ but where each $x\in X$ is contained in only finitely many sets of the form $U\_{a}$.


For example, if $X$ is a topological space with a discrete spanner $A$, and there exists some $x\in X$ where there are infinitely many $a\in A$ with $a\leq x$, then $X$ may not be metacompact, but $X$ is min-compact.


