$B=\{f\in\{0,1\}^\mathbb N :f^{-1}(\{0\}) \,\text{and}\, f^{-1}(\{1\}) \, \text{infinite}\, \}$

Question

$$B=\{f\in\{0,1\}^\mathbb N :f^{-1}(\{0\}) \,\text{and}\, f^{-1}(\{1\}) \, \text{infinite}\, \}$$ Prove that B is not a countable set. I thought about creating a function for every $n\in\mathbb N$ so $\pi :\mathbb N\longrightarrow B$, and another function $g_n=\pi(n)$ and another one $f(n)=1-g_n(n)$ Showing that $f\neq g_n$ for all $n\in\mathbb N$. So $\pi$ won't be subjective. But I have no idea how to address the condition of the group B. TBH, it seems like a messy incomplete solution. Any ideas for how to address the question?

Answer 1

Hint: $\{f: f ^{-1}(\{0\}) \, \text { is finite}\}$ is countable and so is $\{f: f ^{-1}(\{1\}) \, \text { is finite}\}$. Remove these from the uncountable set $\{0,1\}^{\mathbb N}$ and you get an uncountable set.

Answer 2

Sketch: Consider any element $F$ of that set and consider two sequences $a_0< a_1<a_2<\cdots$ and $b_0< b_1< b_2<\cdots$ such that $F(a_i)=0$ and $F(b_j)=1$ for all $i,j\in\Bbb N$. Now, for $S\subseteq \Bbb N$, call $$\tau_S(m)=\begin{cases}b_i&\text{if }i\in S\land m=a_i\\ a_i&\text{if }i\in S\land m=b_i\\ m&\text{if }\forall i\in S, (m\ne a_i\land m\ne b_i)\end{cases}$$

Now, the goal is to prove that the map $$\Xi:\mathcal P(\Bbb N)\to B\\ \Xi(S)=F\circ\tau_S$$ is an injective function...