Show that $\{0,1\}^\mathbb{R}$ is separable

Question

Show that $\{0,1\}^\mathbb{R}$ is separable. I was trying to think about how to show this. First I considered the subset $S=\{f \in\{0,1\}^\mathbb{R}:f(a)=0,a \in \mathbb{Q}^{c}\}$. However I was told that although this set is dense it is uncountable being of cardinality $2^{|\mathbb{N}|}$. I was given a hint to use the bijective correspondence with elements in this set, and elements in the power set of $\mathbb{R}$ and consider the collection of all finite subsets of $\mathbb{R}$. So in this case I should consider the subset $S=\{f\in\{0,1\}^\mathbb{R}:f(a)=0,\text{for all but finitely many }a\}$. This set is supposedly countable and dense.

Now I am rethinking this. I see the set of all finite subsets of $\mathbb{N}$ is countable, so maybe I was told to consider the set $S=\{f \in\{0,1\}^\mathbb{R}:f(a)=0 \ \text{for all but finitely many}\ a \in \mathbb{N}\}$? If I do this, to prove this subset is countable, would I just need to prove the equivalent statement that the set of all finite subsets of $\mathbb{N}$ is countable?

Once the set is proved to be countable, the next thing to prove is that the closure of the set is equal to $\{0,1\}^\mathbb{R}$. That the set is dense. I will try to give this an attempt.

Let $x \in \{0,1\}^{\mathbb{R}}$. Let $\prod U_{\alpha}$ be an open set containing $x$. Then $U_\alpha=\{0,1\}$ for all but finitely many $\alpha$. Now I am struggling to figure out what to do if $f(a)=1$ for only finitely many irrationals or non whole numbers. Maybe I should go back to considering the second set? The set of functions that are zero at each coordinate but finitely many? If this is the case then again Let $x \in \{0,1\}^{\mathbb{R}}$. Let $\prod U_{\alpha}$ be any open set containing $x$. The $U_\alpha=\{0,1\}$ for all but finitely many alpha. So certainly it contains a point in $S$ since it contains any function $0$ everywhere but finitely many coordinates. What would be a better way to show this?Is the set of all finite subsets of $\mathbb{R}$ actually countable?

Unfortunately I am hitting a brick wall with how to actually come up with all of the math I need to show this set is dense. How should I try to show any open set containing $x$ must contain some elements in $S$?

Answer 1

Your first set $S_1 = \{f \in \{0,1\}^{\Bbb R}\mid \forall \alpha \in \Bbb Q^c: f(\alpha)=0\}$ is uncountable (in the power-set representation it is the set of all subsets of $\Bbb Q$, really) and not dense, but closed. The argument is essentially the same as I gave in my earlier reply: if $A \notin S_1$ (in the powerset view, so $A \subseteq \Bbb R$) then $\exists \alpha \in \Bbb Q^c \cap A$ and then $[\{\alpha\};\emptyset]$ is a basic neighbourhood of $A$ that misses $S_1$. Here I use basic open sets of $P(\Bbb R)$ of the form $$[F;G]=\{A \subseteq \Bbb R\mid F \subseteq A, A \cap G = \emptyset\}$$ where $F,G$ are disjoint (WLOG) finite subsets of $\Bbb R$.

Your second attempt, $S_2 = \{A \subseteq \Bbb R\mid A \text{ finite }\}$ ( I only use the powerset view in the remainder) is a better attempt: it is at least dense, but still uncountable as already the subset $\{\{x\}: x \in \Bbb R\}$ is uncountable (size continuum). Denseness is clear: if $[F;G]$ is basic open as above, then $F \in S_2 \cap [F;G]$, so every basic open set intersects $S_2$.

To show that we indeed have a countable dense subset: let $D$ be the set of finite disjoint unions of open intervals of the form $(q,q’)$ where $q,q’ \in \Bbb Q, q < q’$. Elementary set theory tells us that $D$ is countable indeed and if $[F;G]$ is basic open, define $H = F \cup G = \{h_0,h_1,\ldots, h_n\}$ (in increasing order) and pick $q_0 < q_1 < \ldots q_n < q_{n+1}$, all rationals, such that $q_0 < h_0 < q_1 < h_1 < q_2 < \ldots < h_n < q_{n+1}$ and define $A$ to be the union of the $(q_i, q_{i+1})$ that contain all the $h_i$ that lie in $F$. It’s then clear that $A \in D \cap [F;G]$ and denseness has been shown.