Density of a subset of $\{0,1\}^{\mathbb{R}}$

Question

Consider $X=\{0,1\}^{\mathbb{R}}$ where $\{0,1\}$ is a discrete space. Show that the set $A=\{\chi_D \mid D \subset \Bbb R \text{ discrete } \}$ where $\chi_D$ denotes the indicator function is a dense subset of $X$.

I need to show that $A \cap U \ne \emptyset$ for arbitary $U$ open in $X$. This will follow if I can show that $A \cap \left( \prod_{x \in F} U_x \right) \ne \emptyset$ for a basic open set $\prod_{x \in F} U_x$ with $F \subset \Bbb R$ finite.

So let $\chi_S \in X$ and let $U$ be an open set containing $\chi_S$. Then there exists a basic open set $\prod_{x \in F} U_x$ such that $$\chi_S \in \prod_{x \in F} U_x \subset U$$

Now $U_x=\{0,1\}$ for all but finitely many $x \in \Bbb R$. How can I leverage this information to show that the intersection $A \cap \left( \prod_{x \in F} U_x \right)$ is non-empty?

Answer 1

What you have to do is this:

Consider any $f \in X$ and any open neighborhood $U$ of $f$. Then find $\chi_D \in A$ such that $\chi_D \in U$.

If you want, you can write $f = \chi_S$ with a some subset $S \subset \mathbb R$, but it is not needed in the sequel.

Of course it suffices to consider a basic open $U$ which has the form $$U = \prod_{x \in \mathbb R} U_x$$ where $U_x = \{0,1\}$ for all but finitely many $x$. Note that $f \in \prod_{x \in \mathbb R} U_x$ means that $f(x) \in U_x$ for all $x$. In particular all $U_x \ne \emptyset$.

Let $F = \{ x \in \mathbb R \mid U_x = \{1\}\}$ which is a finite, and therefore discrete, subset of $\mathbb R$. Thus $\chi_F \in A$. All $U_x$ with $x \notin F$ contain $0$ (because they are non-empty and $\ne \{1\}$). Therefore $\chi_F \in U$ because for each $x \in X$ we have $\chi_F(x) \in U_x$: For $x \in F$ we have $\chi(x) = 1 \in U_x$ and for $x \notin F$ we have $\chi_F(x) = 0 \in U_x$.

Answer 2

I'd like to prove the fact using as much of the reasoning of the OP. The following statements are equivalent:

• the set $A$ is dense,
• $A\cap U\neq\emptyset$ for arbitary $U\neq\emptyset$ open in $X$,
• $A\cap B\neq\emptyset$ for arbitary $B\neq\emptyset$ from the base of the topology on $X$.

Let $B$ be a base set. We know that $$B=\prod_{x\in \Bbb R}U_x,\text{ where }\emptyset\neq U_x\subsetneq\{0,1\}\text{ for }x\in F\subset \Bbb R\text{ and }\#F<\infty.$$ Therefore $U_x=\{a_x\}$ for $x\in F$ where $a_x\in\{0,1\}$. Let $D=\{x\in F:a_x=1\}$. Then $\chi_D\in A\cap B$. This proves the third condition on the density.