Given $z = (z_i)_{i \in I}$ a fixed point of $\displaystyle\prod_{I}X_i$, where each $X_i$ is non-empty and connected.
For any finite subset $K \subset I$, we define $X_K = \{ (x_i) ∈ \displaystyle\prod_{I}X_i \ | \ x_i = z_i \text{, if } i \in I \setminus K \}$
Finally, we define $A = \{X_K \ | \ K \text{ finite subset of } I\} $
How can one prove that $\bigcup A$ is dense in $\displaystyle\prod_{I}X_i$ ?
One needs to show that for any point of $\displaystyle\prod_{I}X_i$ any open set containing it will intersect $\bigcup A$.
I just don't see why it is the case.
Best Answer
If we have a (non-empty) basic open product set $B$ in $\displaystyle \prod_{i \in I} X_i$ it is of the form
$$B=\prod_{i \in I} U_i$$
where there is a finite set $F \subseteq I$ such that $U_i = X_i$ for all $i \in I\setminus F$ and all $U_i$ are (non-empty) open in $X_i$. (These are the finite intersections of sets of the form $p_i^{-1}[U_i]$ which form the standard subbase for the product topology).
Then it's clear that $X_F \cap B \neq \emptyset$ : pick $a_i \in U_i$ for $i \in F$ and define $a_i = z_i$ for all $i \notin F$. Then $(a_i)_i \in X_F \cap B$ and so $B \cap A \neq \emptyset$ as $A \supseteq X_F$ for this $F$ too.
And as all basic open sets intersect $A$, all open sets do too (they are unions of basic open sets after all). So $A$ is dense.