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Define $\mathbb{Z}_2 := \{0,1\}$, $\mathbb{N}:=\{1, 2, \dots\}$. Consider the function $e$, that assigns to every natural number $n \in \mathbb{N}$ the indicator function: $$ e(n):\mathcal{P}\mathbb{N}\rightarrow\mathbb{Z}_2,\hspace{1cm} e(n) := \mathbb{1}_{\big\{S \subseteq \mathbb{N}\ \big|\!:\ n \in S\big\}}. $$

In other words, for every $n \in \mathbb{N}$, $e(n)$ is a function, whose domain is the collection of all subsets of $\mathbb{N}$, and such that, for every $S \subseteq \mathbb{N}$, $$ e(n)(S)=\begin{cases} 1 &, n \in S, \\ 0 &, n \notin S. \end{cases} $$

Equipping $\mathbb{Z}_2$ with the two element Boolean algebra structure, the set $\mathcal{P}\mathbb{N}\rightarrow\mathbb{Z}_2$ is naturally induced with a Boolean algebra structure, by applying the Boolean operations $\vee$, $\wedge$, and $\neg$ component-wise.

We may now form Boolean combinations of the members of $\mathcal{P}\mathbb{N}\rightarrow\mathbb{Z}_2$, where a Boolean combination is the result of evaluating a finite Boolean expression built from members of $\mathcal{P}\mathbb{N}\rightarrow\mathbb{Z}_2$ using the three Boolean operations.

Denote by $\mathcal{C}$ the set of Boolean combinations formed from members of $\mathfrak{Im}(e)$. Is $\mathcal{C}$ countable?

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This question arose from a difficulty I've had with a certain point in this answer.

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Evan Aad
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There are countably many symbols in the set: $A=\{\left(,\right),\land,\lor, \lnot ,e_1,e_2,e_3,\dots,\}$.

Thus there are only countably many strings in the set $A^*$ of strings of symbols in $A$, the set.

Any subset of those (such as the set of valid boolean expressions) is thus also countable. The set you want is the image of countable subset of the set of such strings, so it is also countable.

So the real question is, if $A$ is countable, why is $A^*$ countable?

Given a countable alphabet of symbols $a_1,a_2,\dots,a_n,\dots$ and a finite string of symbols $a_{i_1}a_{i_2}\dots a_{i_k}$ you can pick $p_1^{i_1}p_2^{i_2}\cdots p_k^{i_k}$ as a unique encoding of the string, where $p_1,p_2,\dots,$ is the set of all primes. This maps $A^*$ in a one-to-one and onto the set $\mathbb N^+$ (with the empty string going to $1$.)

Thomas Andrews
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