How to show that $X$ has property $S_1(\mathcal O,\mathcal O)$, if and only if $\Sigma_{n=1}^{\infty} X_n$ does. We say that a space $X$ satisfies $S_1(\mathcal O,\mathcal O)$, if for every sequence of open covers $\{ \mathcal U_n : n \in \mathbb N \}$, there exists a sequence of open sets $\{ U_n :  U_n \in \mathcal U_n \text{ for each } n \in \mathbb N \}$, which is an open cover of $X$.
For each $n \in \mathbb N$, define, $X_n := X \times \{ n \}$. Then, the $X_n$'s are disjoint. With $\tau$ denoting the topology on $X$, each $\tau_n := \{ U \times \{ n \} : U \in \tau \}$ is a topology for $X_n$. The set $\bigcup_{n=1}^{\infty} \tau_n$ generates a topology on $\bigcup_{n=1}^{\infty} X_n$. Let $\Sigma_{n=1}^{\infty} X_n$ denote this space. Every open cover of $\Sigma_{n=1}^{\infty} X_n$, has a refinement which consists of elements of $\bigcup_{n=1}^{\infty} \tau_n$ only. If $X$ has property $S_1(\mathcal O,\mathcal O)$, so does each $(X_n,\tau_n)$. 
I am trying to prove the following property:

$X$ has property $S_1(\mathcal O,\mathcal O)$, if and only if $\Sigma_{n=1}^{\infty} X_n$ does.

Any help?
Thank you!
 A: Let $Y=\sum_{n\in\Bbb N}X_n$, and suppose that $Y$ has property $S_1(\mathcal{O},\mathcal{O})$. Let $\{\mathscr{U}_n:n\in\Bbb N\}$ be a countably infinite family of open covers of $X$. For $n\in\Bbb N$ let $\mathscr{V}_n=\{U\times\Bbb N:U\in\mathscr{U}_n\}$; clearly this is an open cover of $Y$. Thus, for each $n\in\Bbb N$ there is a $U_n\in\mathscr{U}_n$ such that $\{U_n\times\Bbb N:n\in\Bbb N\}$ covers $Y$, and it’s clear that $\{U_n:n\in\Bbb N\}$ covers $X$. Thus, $X$ has property $S_1(\mathcal{O},\mathcal{O})$.
Now suppose that $X$ has property $S_1(\mathcal{O},\mathcal{O})$, and let $\mathfrak{U}$ be a countably infinite family of open covers of $Y$. Since $|\Bbb N\times\Bbb N|=|\Bbb N|$, we may index $\mathfrak{U}$ as $\mathfrak{U}=\{\mathscr{U}_{n,k}:k,n\in\Bbb N\}$. Without loss of generality we may assume that $\bigcup\mathfrak{U}\subseteq\bigcup_{n\in\Bbb N}\tau_n$. Fix $n\in\Bbb N$, and for each $k\in\Bbb N$ let
$$\mathscr{V}_{n,k}=\{V\in\tau:V\times\{n\}\in\mathscr{U}_{n,k}\}\;;$$
$\{\mathscr{V}_{n,k}:k\in\Bbb N\}$ is a countably infinite family of open covers of $X$, so for each $k\in\Bbb N$ there is a $V_{n,k}\in\mathscr{V}_{n,k}$ such that $\{V_{n,k}:k\in\Bbb N\}$ covers $X$. We can do this for each $n\in\Bbb N$.
Now for each $k,n\in\Bbb N$ let $U_{n,k}=V_{n,k}\times\{n\}\in\mathscr{U}_{n,k}$; by construction $\{U_{n,k}:k\in\Bbb N\}$ covers $X_n$, so $\{U_{n,k}:k,n\in\Bbb N\}$ covers $Y$, which therefore has property $S_1(\mathcal{O},\mathcal{O})$.
