[Math] The product topology on $\{0,1\}^X$, where $X$ is uncountable

Question

I am trying to understand the topology on $\{0,1\}^X$, where $X$ is uncountable. The topology on $\{0,1\}$ is the discrete and I am using the product topology on $\{0,1\}^X$.
My question is, who are the basic open sets? From my understanding of the definition of product topology, basic sets should either contain finite sets or cofinite sets (sets with finite complement). But from what written here nets and sequential spaces

in page 4 example 3, it seems like open basic sets must contain finite sets.
Mabe I should ask, Who are the basic sets in this topology?

Thank you! Shir

Answer 1

I'll quote the beginning of the example from the document you have linked to.

Example 3: Let $X$ be an uncountable set and let $\{0,1\}$ have the discrete topology. Consider $\mathcal P(X) = \{0,1\}^X$ with the product topology. Let $\mathcal A\subseteq \mathcal P(X)$ be the collection of all uncountable subsets of $X$. $\mathcal A$ is not open; indeed every basic open contains finite sets. However, we claim that $\mathcal A$ is sequentially open.

The important point is notice that the author identifies $\{0,1\}^X$ with $\mathcal P(X)$. Indeed, there is a very natural bijection between these two sets (and it occurs frequently in mathematics, which might be why it is mentioned in the text without any details).

For every subset $S\subseteq X$ you have the corresponding function $\chi_S \in \{0,1\}^X$ which maps elements of $S$ to $1$ and other elements to $0$ $$\chi_S(x)= \begin{cases} 1 & x\in S, \\ 0 & \text{otherwise}. \end{cases} $$ The assignment $S\mapsto \chi_S$ is a bijection between $\mathcal P(X)$ and $\{0,1\}^X$.

Now if we look at the space $\{0,1\}^X$ then the basic open sets are obtained in this way: Choose finitely many coordinates $k_1,\dots,k_n$ and choose values $v_1,\dots,v_n\in\{0,1\}$ which you want to prescribe for these coordinates. Then you get a basic neighborhood consisting of all functions with these prescribed values, i.e. $f(k_j)=v_j$.

What happens if we transfer this topology (using the bijection we mentioned above) to the set $\mathcal P(X)$? Well, for every basic neighborhood we have fixed some finite set $F$ of indices where the functions have to attain the value $1$. If this function is of the form $\chi_S$, this is the same as saying $F\subseteq S$. We also prescribed some finite sets $G$ such that for $x\in G$ we want $\chi_S(x)=0$. This is the same as requiring $S\cap G=\emptyset$.

So the basic open sets in $\mathcal P(X)$ are like this: For any finite sets $F,G\subseteq X$ you get a basic set $$\mathcal O_{F,G}=\{S\subseteq X; F\subseteq S, G\cap S=\emptyset\}.$$ You can notice, that for any choice of $F$, $G$ such that $F\cap G=\emptyset$, this basic set contains at least one finite set. (For $F\cap G\ne\emptyset$, the set $\mathcal O_{F,G}$ described here is empty.)