Show that $\{0,1\}^{\mathbb{R}}$ with the product topology is separable.

Question

Show that $\{0,1\}^{\mathbb{R}}$ with the product topology is separable. (Hint: Let $\mathcal{D}$ be the set of all finite unions of intervals $(a,b) \subset \Bbb R$ with rational endpoints. Show that the set of characteristic functions on $\mathcal{D}$ is countable and dense in $\{0,1\}^{\mathbb{R}}$).

Let $\mathcal{D}$ be the collection described above. Then $\mathcal{D} = \{U \subset \Bbb R \mid U = \bigcup_{i \in F} (a_i, b_i) \}$ where $F$ is finite and $a_i,b_i \in \Bbb Q$.

Let $A = \{ \chi_D \mid D \in \mathcal{D}\}$. Now $A$ is countable as $\mathcal{D}$ is countable. We only need to show then that $A$ is dense.

Let $O$ be a basic open set in $\{0,1\}^{\mathbb{R}}$. Then $$O = \bigcap_{x \in K} \pi_x^{-1}(V_x)$$ for $K \subset \Bbb R$ finite. Now we need to show that $A \cap O$ is non-empty.

The intersection is non-empty iff there exists $f \in A \cap O$ which is equivalent to $f \in A$ and $f \in O$. Now $f \in A$ if and only if $f = \chi_{D'}$ for $D' \in \mathcal{D}$. And $f \in O$ if and only if $f(x) \in \{0,1\}$ for all $x \in K$.

If I let $f = \chi_K$, then I think I can find intervals such that $K \in \mathcal{D}$ so $f \in A$. Also $f(x) = 1$ for all $x \in K$ so $f \in O$.

My question is that is it true that $K \in \mathcal{D}$ in order for $f$ to be in $A$? Also what is the motivation to define $\mathcal{D}$ as the set of such finite unions?

Answer 1

You interpret $\{0,1\}^{\mathbb R}$ as a set of functions $\mathbb R\to\{0,1\}$. That's fine but in the sequel of this answer I will interpret the space as powerset $\mathcal P(\mathbb R)$ so that its elements are subsets of $\mathbb R$ and its subsets are collections of subsets of $\mathbb R$.

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Let us call an ordered pair $(I,J)$ "suitable" if $I,J$ are finite subsets of $\mathbb R$ with $I\cap J=\varnothing$.

For every suitable pair $(I,J)$ define:

$$\mathcal B_{I,J}=\{A\in\mathcal P(\mathbb R)\mid I\subseteq A, J\subseteq A^c\}$$ Then $\mathcal B_{I,J}$ is an open subset of $\mathcal P(\mathbb R)$ and the collection of all sets of this form is a base for the topology.

Proving that $\mathcal D$ is dense then boils down to proving that $\mathcal B_{I,J}\cap\mathcal D\neq\varnothing$ for every suitable pair $(I,J)$.

Now let $(I,J)$ be a suitable pair. For every $i\in I$ we can find $a_i,b_i\in\mathbb Q$ such that $i\in(a_i,b_i)$ and moreover $(a_i,b_i)\cap J=\varnothing$.

Then evidently: $$\bigcup_{i\in I}(a_i,b_i)\in\mathcal B_{I,J}\cap\mathcal D$$ and we are ready.