Is there a bijection between $\mathcal{P}(\mathbb{N})$ and $\mathcal{P}(\mathbb{N} \times \mathbb{N})$?

Question

I tried using the Cantor-Schröder-Bernstein theorem. I defined $f_1 \colon \mathcal{P}(\mathbb{N}) \to \mathcal{P}(\mathbb{N} \times \mathbb{N})$ as $\{a_1, a_2, a_3,...\} \mapsto \{(a_1,a_1), (a_2, a_2), (a_3,a_3),...\}$ and $f_2 \colon \mathcal{P}(\mathbb{N} \times \mathbb{N}) \to \mathcal{P}(\mathbb{N})$ as $\{(a_1, b_1), (a_2, b_2),...\} \mapsto \{2^{a_1-1}(2b_1-1), 2^{a_2-1}(2b_2-1),... \}$. Then, I showed that both $f_1$ and $f_2$ are injective. Therefore, by the Cantor-Schröder-Bernstein theorem, there exists a bijection $f \colon \mathcal{P}(\mathbb{N}) \to \mathcal{P}(\mathbb{N} \times \mathbb{N})$. Is this a correct approach?

Answer 1

Your $f_1$ is a function from $\Bbb N$ to $\Bbb N\times\Bbb N$, not from $\wp(\Bbb N)$ to $\wp(\Bbb N\times\Bbb N)$, and your $f_2$ is a function from $\Bbb N\times\Bbb N$ to $\Bbb N$, not from $\wp(\Bbb N\times\Bbb N)$ to $\wp(\Bbb N)$, so the Cantor-Schröder-Bernstein theorem tells you that there is a bijection $f:\Bbb N\to\Bbb N\times\Bbb N$. You still have some work to do to get a bijection from $\wp(\Bbb N)$ to $\wp(\Bbb N\times\Bbb N)$, though it’s pretty easy work: there is a very natural, easy way to use $f$ to define a bijection $F:\wp(\Bbb N)\to\wp(\Bbb N\times\Bbb N)$. Can you find it?