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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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    $\begingroup$ Does the term "anti-compact" in this sense have a history or did you just invent it? It has been used for spaces in which every compact set is finite; see Paul Bankston, The total negation of a topological property, Illinois J. Math. 23 (1979), 241-252. $\endgroup$
    – bof
    Dec 22, 2014 at 21:42
  • $\begingroup$ I was toying with the concept mentioned in the question and found an til compactness just to be a fitting name. $\endgroup$ Dec 23, 2014 at 21:18
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    $\begingroup$ The name anticompact is very confusing here as it goes against commonly accepted usage (see the Bankston article). $\endgroup$
    – PatrickR
    Nov 27, 2022 at 6:45
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    $\begingroup$ @DominicvanderZypen FYI, from the first paragraph in the answer by Eric Wofsey, your "anti-compact" is nothing else than the standard ultraconnected property. $\endgroup$
    – PatrickR
    Nov 30, 2022 at 5:49

2 Answers 2

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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, 2014 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, 2017 at 19:43
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    $\begingroup$ @WlodAA: I've added such a statement (the first sentence of the second paragraph). $\endgroup$ Jun 29, 2017 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, 2017 at 20:11
  • $\begingroup$ Nice. (I still had to learn what a specialization order means :-). $\endgroup$
    – Wlod AA
    Jun 30, 2017 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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