bijection in $\mathcal P \left({S}\right)$ and $\mathcal P \left({S}\right) \times \mathcal P \left({S}\right)$

Question

given that ${S}$ is countably infinite set. is there any bijection exist between $\mathcal P \left({S}\right)$ and $\mathcal P \left({S}\right) \times \mathcal P \left({S}\right)$. Here $\mathcal P\left({S}\right)$ is powerset of $\mathcal S$.

Answer 1

Map $ P(S) $ to $ N $, and map $ P(S)×P(S) $ to $ N×N $, there exist a bijection between $ N \rightarrow N×N $. For example, $ \frac{(a+b)(a+b+1)}{2} + a \rightarrow (a,b) $.

Answer 2

Since $S$ is countably infinite i.e, has a bijection with N, the power set should have 1-to-1 mapping with R the set of real. Then we need to find a bijection from R x R $\to$ R by interleaving the pair of real to another real.

Answer 3

Here's an explicit bijection.

Identify $S$ with the natural numbers $N = \{1,2,3,\ldots\}$. Let $N = N_e \cup N_o$ denote the partition of the natural numbers into evens and odds, and define $$\phi_e(k) = 2k, \phi_o(k) = 2k - 1$$, noting that $\phi_e, \phi_o$ are bijections $N \to N_e, N \to N_o$ respectively.

We now define a bijection $\Phi : \mathcal{P}(N) \to \mathcal{P}(N) \times \mathcal{P}(N)$ as follows: given a set $A \subset N$, define $$ \Phi(A) = (\phi_o^{-1}(A \cap N_o), \phi_e^{-1}(A \cap N_e)) $$

To show this injects: suppose that $\Phi(A) = \Phi(B)$. Then $\phi_e, \phi_o$ biject, and so therefore $A \cap N_o = B \cap N_o$ and $A \cap N_e = B \cap N_e$, so that $A = B$ must hold.

To show this surjects: let $A_1, A_2$ be any subsets of the naturals; then $\Phi(A) = (A_1, A_2)$ upon defining $A = \phi_o(A_1) \cup \phi_e(A_2)$. This is actually an explicit form for the inverse of $\Phi$.

Answer 4

Hint: If $A$ and $B$ are disjoint sets then we have a bijection between $\mathcal P(A\cup B)$ and $\mathcal P(A)\times\mathcal P(B)$ given by $\langle X,Y\rangle\mapsto X\cup Y$.