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

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?

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?