If $P(\mathbb N)$ denotes the power set of the set of natural numbers, prove that $|P(\mathbb N) \times P(\mathbb N)|=|(0,1)|.$
I have thought of trying to find a bijection from P(N) x P(N) to (0,1).
Alternatively, I could try to use the fact that |P(N)|=|(0,1)| and prove that |P(N) x P(N)|=|P(N)| by finding a bijection from P(N) x P(N) to P(N).
Finally, maybe I could use the fact that |P(N x N)|=|P(N)|=|(0,1)| and try to prove that |P(N) x P(N)|=|P(N x N)| by finding a bijection between P(N) x P(N) and P(N x N).
I am struggling to think up a suitable bijection for any of these options.
Consider base three representation of the real numbers belonging to interval $(0,1)$.
Consider the example :
$$x=.001221110210010210...$$
to this representation, we associate the partition where
the zeros' and twos' positions denote the first element of $P(\mathbb{N})$,
the ones' and two's positions denote the second element of $P(\mathbb{N})$.
Remark: the positions begin at $0$.
In the case of our example, it would be:
$$A=\{0,1,3,4,8,10,12,13,14... \} \ \text{and} \\ B=\{2,3,4,5,6,8,9,12,14,15... \}$$
Remarks:
Said otherwise, the "zeros", "ones" and "twos" characterize resp. $A \setminus B, \ \ B \setminus A, \ \ A \cap B$
There is still some work ; you will have to deal with the non unique base representation of real numbers : for example, in base $3$, number $0.12222...$ with an infinite number or twos is the same as $0.20000...$ with an infinite number of zeros.