Is the countable product of co-countable topology Lindelöf?

Question

For $i\in\mathbb{N}$, let $(X_i,T_i)$ be the countable complement topology on $\mathbb{R}$. Let $(X,T)$ be the product topology (not box product).

Is $(X,T)$ Lindelöf? That is, does every open cover have a countable sub-cover?

I know it is Lindelöf in the case of the product of two countable complement topologies.

Answer 1

Let $Y$ denote $\Bbb R$ with the co-countable topology. Note first that $Y$ is not just Lindelöf, but hereditarily Lindelöf. In fact, for each $n\in\Bbb Z^+$ the product $Y^n$ is hereditarily Lindelöf.

Proof. The proof is by induction on $n$. Suppose that $n>1$, and $Y^{n-1}$ is hereditarily Lindelöf. Let $A\subseteq Y^n$, and let $\mathscr{U}$ be any family of open subsets of $Y^n$ covering $A$. Fix $p\in A$, and let $U_0\in\mathscr{U}$ be such that $p\in U_0$; there is a co-countable $V\subseteq Y$ such that $p\in V^n\subseteq U_0$. Let $C=Y\setminus V$; $C$ is countable and as a subspace of $Y$ has the discrete topology. Moreover, $Y^n\setminus V^n$ is the union of $n$ subspaces homeomorphic to $C\times Y^{n-1}$. It follows from the choice of $n$ that $Y^{n-1}$ is hereditarily Lindelöf, and therefore so are $C\times Y^{n-1}$ and the union of $n$ copies thereof. In particular, $Y^n\setminus U_0$ is hereditarily Lindelöf, so only countably many elements of $\mathscr{U}$ are needed in order to cover $A\setminus U_0$ and hence also $A$. Thus, $Y^n$ is hereditarily Lindelöf, as desired. $\dashv$

Now let $X=\prod_{n\in\Bbb N}Y$, and for each $n\in\omega$ let $\pi_n:X\to Y$ be the projection map. Let $\mathscr{B}$ be the usual base for the product topology: each $B\in\mathscr{B}$ is a product $\prod_{n\in\Bbb N}U_n$, where $Y\setminus U_n$ is countable for each $n\in\Bbb N$, and $\pi_n[B]=U_n=Y$ for all but finitely many $n\in\Bbb N$. For $n\in\Bbb N$ let

$$\mathscr{B}_n=\{B\in\mathscr{B}:\pi_k[B]=Y\text{ for all }k\ge n\}\;;$$

$\mathscr{B}=\bigcup_{n\in\omega}\mathscr{B}_n$.

Now let $\mathscr{U}$ be an open cover of $X$; without loss of generality we may assume that $\mathscr{U}\subseteq\mathscr{B}$. For $n\in\Bbb N$ let $\mathscr{U}_n=\mathscr{U}\cap\mathscr{B}_n$. For $U\in\mathscr{U}_n$ let

$$\widehat U=\prod_{k<n}\pi_k[U]\;;$$

$U$ has the form $\widehat U\times\prod_{k\ge n}Y$, where $\widehat U$ is an open subset of $Y^n$. Let $\mathscr{V}_n=\{\widehat U:U\in\mathscr{U}_n\}$; $\mathscr{V}_n$ is an open cover of the subset $\bigcup\mathscr{V}_n$ of $Y^n$, and $\mathscr{U}_n$ covers $\left(\bigcup\mathscr{V}_n\right)\times\prod_{k\ge n}Y$. $Y^n$ is hereditarily Lindelöf, so some countable $\mathscr{W}_n\subseteq\mathscr{V}_n$ covers $\bigcup\mathscr{V}_n$. Let

$$\mathscr{U}_n'=\left\{W\times\prod_{k\ge n}Y:W\in\mathscr{W}_n\right\}\;;$$

$\mathscr{U}_n'$ is a countable subset of $\mathscr{U}_n$, and $\bigcup\mathscr{U}_n'=\bigcup\mathscr{U}_n$. Finally, let $\mathscr{U}'=\bigcup_{n\in\Bbb N}\mathscr{U}_n'$; then $\mathscr{U}'$ is the desired countable subcover of $\mathscr{U}$.