How to prove that any product of separable spaces has the Souslin property A topological space $X$ has the Souslin property if every pairwise disjoint family of non-empty open subsets of $X$ is countable.
I am trying to solve the following exercise:
Prove that any product of separable spaces has the Souslin property. In particular, the space $\mathbb R^A$, has the Souslin property for any set $A$.
Any help?
Thank you!
 A: Suppose that $\{ U_\xi : \xi < \omega_1 \}$ is an uncountable family of nonempty open subsets of $\prod_{i \in I} X_i$, where each $X_i$ is separable.  Without loss of generality, we may assume that each $U_\xi$ is a basic open set: $$U_\xi = {\textstyle \prod_{i \in I}} U_{\xi,i}$$ where each $U_{\xi,i}$ is a nonempty open subset of $X_i$, and $a_\xi = \{ i \in I : U_{\xi,i} \neq X_i \}$ is finite.
By the $\Delta$-System Lemma we may assume without loss of generality that there is a finite $a \subseteq I$ such that for distinct $\xi , \eta < \omega_1$ we have $a_\xi \cap a_\eta = a$.
For $\xi < \omega_1$ consider $V_\xi = \prod_{i \in a} U_{\xi,i}$, a nonempty open subset of $\prod_{i \in a} X_i$.  Note that this is a finite product of seaprable spaces, and is therefore separable, and hence has the Souslin property.  Therefore there must be distinct $\xi , \eta < \omega_1$ such that $V_\xi \cap V_\eta \neq \varnothing$.
This means that $U_{\xi,i} \cap U_{\eta,i} \neq \varnothing$ for all $i \in a$.  But for $i \in I \setminus a$ it follows that either $U_{\xi,i} = X_i$ or $U_{\eta,i} = X_i$, and so $U_{\xi,i} \cap U_{\eta,i} \neq \varnothing$.  From this it follows that $U_\xi \cap U_\eta \neq \varnothing$.

Addendum.
The above proof essentially relies on the following two facts:

*

*If $\{ X_i \}_{i \in I}$ is a family of topological spaces, every finite product of which has the Souslin property, then $\prod_{i \in I} X_i$ also has the Souslin property.

*Finite products of separable spaces are separable (and thus have the Souslin property).

As bof mentions below, an alternative way to prove this is to consider the Knaster property:

A topology space $X$ has the Knaster property if for every uncountable family $\mathcal{U}$ of nonempty open subsets of $X$ there is an uncountable subfamily $\mathcal{U}^\prime \subseteq \mathcal{U}$ such that any two sets in $\mathcal{U}^\prime$ have nonempty intersection.

It is fairly easy to see that separable $\Rightarrow$ Knaster $\Rightarrow$ Souslin, and one can prove that any product of spaces with the Knaster property also has the Knaster property:

Given a family $\{ U_\xi : \xi < \omega_1 \}$ of nonempty basic open subsets of $\prod_{i \in I} X_i$, where $U_\xi = \prod_{i \in I} U_{\xi,i}$ and $a_\xi = \{ i \in I : U_{\xi,i} \neq X_i \}$ is finite, as above, use the $\Delta$-System Lemma to assume that the $a_\xi$ form a $\Delta$-system with root $a = \{ i_1 , \ldots , i_n \}$.
Starting with $B_0 = \omega_1$, inductively take an uncountable subset $B_{\ell} \subseteq B_{\ell-1}$ such that $U_{\xi,i_\ell} \cap U_{\eta,i_\ell} \neq \varnothing$ for all $\xi , \eta \in B_\ell$.  Then it is not too hard to show that $U_\xi \cap U_\eta \neq \varnothing$ for all $\xi , \eta \in B_n$.

