Let $X$ be the set of all functions $f: A \to B$ where $|A|=\aleph_0$ and $|B|=2^{\aleph_0}$. Using some cardinal arithmetic, one can show that $|X|=2^{\aleph_0}$. However, I wanted to construct a bijection $g: P(\mathbb{N}) \to X$ explicitly.
Here is what I came up with: Letting $C^\mathbb{N}$ denote the set of sequences on $C$, I constructed a bijection $P(\mathbb{N}) \to (0,1) \to (0,1)^\mathbb{N} \to B^\mathbb{N} \to X$. For $(0,1) \to (0,1)^\mathbb{N}$, I represent each $x_n$ by its binary expansion $0.x_{n1}x_{n2}\dots$ and consider $x \in (0,1)$ given by $0.x_{11}x_{12}x_{21}x_{13}x_{22}x_{31}\dots$. I would like to simplify this, though, if possible.
I'm wondering if it is possible to construct a bijection directly, without having to go through the reals. In other words, I would like to avoid the step $(0,1) \to (0,1)^\mathbb{N}$ outlined above, or else encode it somehow in a direct bijection $P(\mathbb{N}) \to B^\mathbb{N}$. Does anyone have any suggestions?
Thanks!
Sure. Fix your favorite bijection $\langle \cdot, \cdot\rangle:\mathbb{N}^2\cong\mathbb{N}$. Now given a map $f: \mathbb{N}\rightarrow 2^{\mathbb{N}}$, let $\hat{f}:\mathbb{N}\rightarrow 2$ be given by $$\hat{f}(\langle m, n\rangle)=f(m)(n).$$