Let $(X,\tau)$ be a topological space such that
 $\tau\ne\{\emptyset\ X\}.\ $
We call an open cover $\mathcal{U}$ of $(X,\tau)$ *proper* if
$\ X\notin \mathcal{U}.\ $ Moreover we say that $(X,\tau)$ is


* *anti-compact* if it does not have a finite proper cover;
* *anti-paracompact* if for every proper cover $\mathcal{U}$ there is $x\in X$ such that every neighborhood intersects infinitely many members of $\mathcal{U}$;
* *anti-metacompact* if for every proper cover $\mathcal{U}$ there is $x\in X$ such that $x$ is a member of infinitely many members of $\mathcal{U}$.


We have anti-metacompact $\Rightarrow$ anti-paracompact $\Rightarrow$ anti-compact.


Do any of the converse implications hold?



Anti-compactness
A space is anti-compact iff it has no proper covers consisting of two sets, or equivalently if the intersection of any two nonempty closed sets is nonempty. This is equivalent to the specialization order being directed downwards.


We can use this to prove any anti-compact space is anti-metacompact, so your three conditions are equivalent. Suppose $X$ is anti-compact, and let $\mathcal{U}$ be a proper cover. Choose some $U\_0\in\mathcal{U}$ and some $x\_0\in U\_0$. Let $y\_0\in X\setminus U\_0$, let $x\_1$ be a common lower bound of $x\_0$ and $y\_0$ with respect to the specialization order, and choose $U\_1\in \mathcal{U}$ containing $x\_1$. Let $y\_1\in X\setminus (U\_0\cup U\_1)$, $x\_2$ be a common lower bound of $x\_1$ and $y\_1$, and choose $U\_2\in \mathcal{U}$ containing $x\_2$. Continuing by induction, we get a decreasing sequence $x\_0>x\_1>x\_2>\dots$ and distinct sets $U\_n\in \mathcal{U}$ such that $x\_n\in U\_n$ for all $n$. It follows that $x\_0$ is in every $U\_n$ and so $X$ is anti-metacompact. In fact, since $x\_0$ was arbitrary, *every* element of $X$ is in infinitely many members of $\mathcal{U}$.


