$\min$-compactness vs metacompactness

Question

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?

Answer 1

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.

Answer 2

Bing's example G, see Metrization of topological spaces, is irreducible, but not metacompact. To describe it: let $P$ be an uncountable set, say $P=\omega_1$, $Q$ its power set, and $F$ the Cantor cube $\{0,1\}^Q$. Map $P$ into $F$ by defining $f_p$ for $p\in P$ to be the function $f_p:Q\to\{0,1\}$ such that $f_p(X)=1$ iff $p\in X$. Let $F_P=\{f_p:p\in P\}$ and enlarge the product topology by making all points of $F\setminus F_P$ isolated. (It is well-known that the space $F$ is normal but not collectiowise normal.) Note that the set $F_P$ is closed and discrete.

To see that $F$ is irreducible let $\mathcal{U}$ be an open cover. For every $p\in P$ choose an open set $O_p$ such that $O_p\cap F_P=\{f_p\}$ and $O_p\subseteq U$ for some $U\in\mathcal{U}$. Let $\mathcal{V}$ be $\{O_p:p\in P\}$ together with $\{\{x\}:x\in F\setminus\bigcup_{p\in P}O_p\}$. This is an irreducible refinement of $\mathcal{U}$.

To see that $F$ is not metacompact take for every $p\in P$ the basic open set $U_p=\{f\in F:f(\{p\})=1\}$; note that $U_p\cap F_P=\{p\}$. Let $\mathcal{U}=\{F\setminus F_P\}\cup\{U_p:p\in P\}$. If $\mathcal{V}$ is an open refinement of $\mathcal{U}$ then we must have for every $p\in P$ a finite subset $Q_p$ of $Q$ and a $V_p\in\mathcal{V}$ such that $B(p,Q_p)\subseteq V_p\subseteq U_p$, where $B(p,Q_p)=\{f:f(X)=f_p(X)$ for $X\in Q_p\}$ (a basic open neighbourhood of $p$). By a $\Delta$-system argument there are points $f$ that are contained in uncountably many $B(p,Q_p)$ and hence in uncountably many $V_p$.