closure of A in the product space For each $\alpha\in I=[0,1]$, let $X_{\alpha} = \{0,1\}$ be the discrete topology. For every $\Delta\subseteq I$, define $f_\Delta=(x_\alpha)\in\prod_{\alpha\in I}X_\alpha$, where $x_\alpha=1$ if $\alpha\in\Delta$, and $x_\alpha=0$ if $\alpha\notin\Delta$. Let $A=\{f_\Delta:\Delta\text{ is a finite subset of }I\}$. Show that $f\in\operatorname{cl}A$, where $\operatorname{cl}A$ is the closure of $A$ in the product space $\prod_{\alpha\in I}X_\alpha$.
 A: Consider some arbitrary $f$, along with an open set $U$ that contains $f$ in the product topology. Recall that the product topology has a basis consisting of sets $\prod_{\alpha \in I} U_{\alpha}$, where each $U_\alpha$ is  non-empty, open in $X_\alpha$, and $U_\alpha = X_{\alpha}$ for all but finite $\alpha$. Thus, we will have:
$$f \in \prod_{\alpha \in I} U_{\alpha} \subset U$$
Let $S$ be the finite set of $\alpha$, where  for which $U_\alpha \neq X_\alpha$. Then, we let $T$ be the finite subset of all $\alpha \in S$ such that $U_\alpha$ contains $1$. Consider the sequence $f_T = (x_\alpha)$. Given some $\alpha \in T$, we note that $x_\alpha = 1$, which is in $U_\alpha$, by definition of $T$.
Given some $\alpha \notin T$, we note that $x_\alpha = 0$. In such a case, $U_\alpha$ is either not equal to $X_\alpha$, and doesn't contain $1$, so it must contain $0$, or $U_\alpha = X_\alpha$, so it also contains $0$ in this case. Thus, in either case, $x_\alpha \in U_\alpha$.
It follows that $f_T \in \prod_{\alpha \in I} U_\alpha \subset U$ and is in $A$. Thus, every open neighbourhood of $f$ intersects $A$, so $f \in \text{cl}(A)$. This completes the proof.
A: Let $f$ be an arbitrary function in $\prod_{\alpha \in I} X_\alpha$, and $O$ be an arbitrary open neighbourhood of $f$. By the definition of the product topology (via its standard base) there is a basic open set $f \in B:=\prod_{\alpha \in I} O_\alpha \subseteq O$, which is such that there is a finite subset $F$ of $I$ such that $O_\alpha = X_\alpha $ for all $\alpha \in I\setminus F$ and such that all $O_\alpha$ are open in $X_\alpha$.
Let $\Delta = f^{-1}[\{1\}] \cap F$, which is a finite subset of $I$ too, so $f_\Delta \in A$, and $f_\Delta \in B$ by the definitions: if $\alpha \notin F$, then $f_\Delta(\alpha) \in X_\alpha = O_\alpha$ trivially, and if $\alpha \in F$ we have $f_\Delta(\alpha) = f(\alpha) \in O_\alpha$ as well. So $O \cap A\neq  \emptyset$ and as $O$ was arbitrary , $f \in \operatorname{cl} A$, as required.
Note that we don't need the $X_\alpha$ to be discrete spaces.
