(related to Proposition: Uncountable and Countable Subsets of Natural Numbers)

- The set of all functions $2^{\mathbb N}=\{f\mid \mathbb N\to \{0,1\}\}$ is equipotent to the power set $\mathcal P(\mathbb N)$, since every function mapping all natural numbers to the set $\{0,1\}$ is an indicator function (see also explanation about the connection between the power set and mapping to two elements)
- Thus, $|\mathcal P(\mathbb N)|=|2^{\mathbb N}|.$
- On the other hand, there is no surjective function of a set to its power set, and therefore neither a surjective function.
- Trivially, there is an injective function $f:\mathbb N\to\mathcal P(\mathbb N)$ (just set $f(a)=\{a\}$ for all $a\in\mathbb N.$)
- Comparing both cardinal numbers yields $|\mathbb N| < \mathcal P(\mathbb N)=2^{\mathbb N}.$
- Thus, $2^{\mathbb N}$ is uncountable.

- To be provided...∎

**Wille, D; Holz, M**: "Repetitorium der Linearen Algebra", Binomi Verlag, 1994