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!


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 $U_\xi \cap U_\eta \neq \varnothing$.


The above proof essentially relies on the following two facts:

  1. 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.
  2. 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 $\xi , \eta \in B_n$.

  • $\begingroup$ A nicer way to put it: the Knaster property is preserved by products; and separable $\Rightarrow$ Knaster $\Rightarrow$ Souslin. $\endgroup$ – bof Apr 10 '14 at 9:15
  • $\begingroup$ I see, ok. Thank you for the proof! $\endgroup$ – topsi Apr 10 '14 at 9:36
  • $\begingroup$ I will also check the Knaster property. Thank you! $\endgroup$ – topsi Apr 10 '14 at 9:37
  • $\begingroup$ @bof: Yeah, that's a nice thing to mention. I think I'll add a note or two. Thanks. $\endgroup$ – user642796 Apr 10 '14 at 10:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.