Is the $\Sigma$-product a dense subset of $\{0,1\}^{\omega_1}$? Let $I=\omega_1$ be the first uncountable ordinal, and let $P=\{0,1\}^I$ be the uncountable product of discrete spaces of two points. Let $S$, the so-called $\Sigma$-product be its subspace consisting of all points that have at most countably many coordinates different from $0$. 
I want to proof that $S$ is dense. Could you give me any hint? Also, I wonder that is it true for any $\omega_1$-product of separable space? And why is $P$  a ccc space?
 A: Recall that the basic open sets of $P$ are the sets of the form $U = \prod_{\xi < \omega_1} U_\xi$ where $U_\xi \subseteq \{ 0 , 1 \}$ is nonempty, and $U_\xi = \{ 0 , 1 \}$ for all but finitely many $\xi < \omega_1$.  From here it is easy to see that every basic open set meets $S$.  (If $\alpha < \omega_1$ is such that $U_\xi = \{0,1\}$ for all $\xi \geq \alpha$, then construct an appropriate element $\mathbf{x} = \langle x_\xi \rangle_{\xi < \omega_1}$ of $U$ such that $x_\xi = 0$ for all $\xi \geq \alpha$.)

$P$ is ccc because it is a product of separable spaces.
Lemma.   If $\{ X_i : i \in I \}$ is any family spaces such that for each finite $I_0 \subseteq I$ the product $\prod_{i \in I_0} X_i$ is ccc, then $X = \prod_{i \in I} X_i$ is ccc.
proof.   It clearly suffices to show that any family of $\omega_1$-many basic open sets in $X$ is not pairwise disjoint.  So suppose that $\{ U_\xi : \xi < \omega_1 \}$ is a pairwise disjoint family of basic open subsets of $X$.  For each $\xi < \omega_1$ let $A_\xi$ be the set of all $i \in I$ such that the projection of $U_\xi$ on the $i$th coordinate ($\pi_i [ U_\xi ]$) is not full.  Then $\{ A_\xi : \xi < \omega_1 \}$ is a family of finite subsets of $I$, so by the $\Delta$-System Lemma there is an uncountable $B \subseteq \omega_1$ and a finite $I_0 \subseteq I$ such that $A_\xi \cap A_\eta = I^\prime$ for all distinct $\xi , \eta \in B$.  
I claim that $I_0 \neq \varnothing$.  Otherwise given distinct $\xi , \eta \in B$ since $A_\xi \cap A_\eta = \varnothing$ we have that $U_\xi \cap U_\eta \neq \varnothing$, contradicting our assumption that the $U_\xi$ are pairwise disjoint!
For $\xi \in B$ let $V_\xi = \prod_{i \in I_0} \pi_i [ U_\xi ]$.  Each $V_\xi$ is open in $\prod_{i \in I_0} X_i$, and for distinct $\xi , \eta \in B$ we have $V_\xi \cap V_\eta = \varnothing$, contradicting our assumption!       $\Box$
Corollary.   Products of separable spaces are ccc.
proof.   Every separable space is ccc, and finite products of separable spaces are separable.  The result now follows from the Lemma.       $\Box$
(Note that whether products of ccc spaces are ccc is independent of $\mathsf{ZFC}$.  In particular, a Souslin line would be a ccc space whose square is not ccc.  But $\mathsf{MA}(\aleph_1)$ implies that all products of ccc spaces are ccc.)
