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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?

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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 $\mathcal{U}$.

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  • $\begingroup$ "induction" $\: \mapsto \:$ "dependent choice" $\;\;\;\;$ $\endgroup$ – user5810 Dec 22 '14 at 11:51
  • $\begingroup$ Eric, before I start truly reading, could you state the result(s) first? Say, a THEOREM? $\endgroup$ – Wlod AA Jun 29 '17 at 19:43
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    $\begingroup$ @WlodAA: I've added such a statement (the first sentence of the second paragraph). $\endgroup$ – Eric Wofsey Jun 29 '17 at 20:00
  • $\begingroup$ Thank you. (Possibly, I have clearly more than average difficulty to follow a text; however, I think that about half of the mathematicians are like this, so that most of the time they quickly give up on reading). $\endgroup$ – Wlod AA Jun 29 '17 at 20:11
  • $\begingroup$ Nice. (I still had to learn what a specialization order means :-). $\endgroup$ – Wlod AA Jun 30 '17 at 2:26
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Let me provide another proof of @EricWofsey's theorem prompted by Dominic's Question.

Theorem (Eric Wofsay)   Let $\ X\ $ be an anti-compact space, and let $\mathcal V\ $ be a proper cover. Then every $\ x\in X\ $ belongs to infinitely many members of $\ \mathcal V$.

Proof   Let $\ X\ $ be an anti-compact space, and let $\mathcal V\ $ be a proper cover, and let $\ x\in X\ $ be such that family $\ \mathcal K:=\{ G\in\mathcal V: x\in G\}\ $ is finite (a proof by contradiction).

Next, let $\ \mathcal M:=\mathcal V\setminus\mathcal K,\ $ and also $\ K:=\bigcup\mathcal K\ $ and $\ M:=\bigcup\mathcal M.\ $ Then $\ K\ne X\ne M.\ $ We have a $2$-element open proper covering $\ \{K\ M\},\ $ which is a contradiction. End of Proof

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