# Cardinal equality: $\;\left|\{0,1\}^{\Bbb N}\right|=\left|\{0,1,2,3\}^{\Bbb N}\right|$

I need to prove the above equality without Cantor-Bernstein Theorem or cardinals arithmetic (i.e., a bijection must be found).

I know that for example $\;S\to 1_S=\;$ the indicator function, gives a bijection $\;P(\Bbb N)\to \{0,1\}^{\Bbb N}\;$ , so if I can find a bijection $\;P(\Bbb N)\to\{01,2,3\}^{\Bbb N}\;$ then I can compose these two and that's all. Yet this last one is making problems to me, so any help will be appreciated.

• $\{0,1\} \times \{0,1\}$ is equipotent to $\{0,1,2,3\}$. Use even and odd coordinates. – Henno Brandsma Feb 1 '16 at 21:41
• Thank you Henno, yet I still can't see your point as the left set is $\;\{0,1\}^{\Bbb N}\;$ and not the cartesian product. – DonAntonio Feb 1 '16 at 21:44

• Thank you very much Brian. Let me see if I get your point: every pair in the sequence will be translated to basis four say by a law like $$(0,0)\to 0\;,\;\;(0,1)\to 1\;,\;\;(1,0)\to 2\;,\;\;(1,1)\to3\;?$$ Thus, we'd get for instance $$(0,1,1,1,1,1,0,0,1,0,1,0,0,0,1,0,...)\to (1,3,3,0,2,2,0,2,...)\;\;?$$ – DonAntonio Feb 1 '16 at 22:03