Mimic the proof of Theorem 2.9 to show that \(\#(0,1)^{2}=\mathfrak{c}\). Use the fact that \(\# \mathbb{R}=\#(0,1)\) to conclude that \(\# \mathbb{R}^{2}=\mathfrak{c}\).

Data from theorem 2.9

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We have #(0, 1) = c. Proof We have to assign to every sequence (xi)iEN C (0, 1) a unique number x = (0, 1) - and vice versa. For this we write, as in the proof of Theorem 2.8, each x, as a unique infinite decimal fraction x=0.a, 1a, 2a, 3a,4... EN, and we organize the array (ak)iken into one sequence with the help of the counting scheme of Example 2.5(iv): x=0. a1,1 a1,2 a2,1 91,3 a2,2 a3,1 a1,4 a2,3 a3,2 a4,1... (the numbers refer to the corresponding diagonals in the counting scheme of Example 2.5(iv)). Since the counting scheme was bijective, this procedure is reversible, i.e. we can start with the decimal expansion of x = (0, 1) and get a unique sequence of xs. We have thus found an injection from (0, 1) to (0, 1), hence # (0, 1) < # (0, 1). On the other hand, # (0, 1) < #(0, 1) is obvi- ous, [] and an application of the Cantor-Bernstein theorem (Theorem 2.7) finishes the proof. We write P(X) for the power set {A:AC X} which is the family of all subsets of a given set X. For finite sets it is clear that the power set is of strictly larger cardinality than X. This is still true for infinite sets.