MHB Is $X$ Lindelöf? Here's this week's problem! Sorry about the delay!

[Chris L T521]

Here's this week's problem! Sorry about the delay!

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Problem: A collection $\mathcal{A}$ of subsets of $X$ has the countable intersection property if every countable intersection of elements from $\mathcal{A}$ is nonempty. Show that $X$ is a Lindelöf space if and only if for every collection $\mathcal{A}$ of subsets of $X$ having the countable intersection property, $\displaystyle \bigcap\limits_{A\in\mathcal{A}} \overline{A}$ is nonempty.

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[Chris L T521]

No one answered this week's problem. You can find the solution below.

[sp]Proof: Suppose that $X$ is Lindelöf and that there's a collection $\mathcal{A}$ of subsets of $X$ with the countable intersection property such that $\bigcap_{A\in\mathcal{A}} \overline{A} = \emptyset$. Then, the collection $\mathcal{O}=\{X\setminus\overline{A}: A\in\mathcal{A}\}$ is an open cover of $X$. If follows that there's a countable subcover $\mathcal{O}^{\prime}=\{X\setminus \overline{A_1},X\setminus\overline{A_2},\ldots\}$, that is, $X=\bigcup_{n=1}^{\infty} X\setminus\overline{A_n}$. But this implies that\[\emptyset=X\setminus \bigcup_{n=1}^\infty X\setminus\overline{A_n}=\bigcap_{n=1}^\infty \overline{A_n}\]
Since for every positive integer $n$, $A_n\subset\overline{A_n}$, we have that $\bigcap_{n=1}^{\infty} A_n=\emptyset$, which contradits the countable intersection property. Therefore $\bigcap_{A\in\mathcal{A}} \overline{A} \neq \emptyset$.Conversely, suppose that $X$ has the stated property but is not Lindelöf. Thus, there's an open cover $\mathcal{O}$ such that there's no countable subcover. Consider the collection of closed sets $A=\{ X\setminus U:U\in\mathcal{O}$. Note that this collection has the countable intersection property, for if a countable intersection of elements in $\mathcal{A}$ is empty, $\bigcap_{n=1}^{\infty} X\setminus U_n = \emptyset$, then $X=X\setminus \bigcap_{n=1}^{\infty} U_n = \bigcup_{n=1}^{\infty} U_n$ and $\mathcal{O}$ would have a countable subcover. It follows, by assumption, that $\bigcap_{A\in\mathcal{A}} \overline{A}= \bigcap_{A\in\mathcal{A}} \neq \emptyset$. However, this means that the complement of this set is not the whole space, and since $X\setminus _{A\in\mathcal{A}} A = \bigcup_{U\in\mathcal{O}} U$, this implies that $\mathcal{O}$ is not a cover, which contradicts our assumption that $\mathcal{O}$ is a cover. Therefore, $X$ is Lindelöf.$\hspace{.25in}\blacksquare$[/sp]