How can I prove that the cardinality of $ (0,1) ^ n $ is equal to that of $ (0,1) $, giving an explicit bijective function.
I edit my question.
What I am trying to achieve is to construct a bijective function of $ \mathbb R ^ n \to \mathbb R $ by composition of bijective functions that is, $$ \mathbb R ^ n \to (0,1) ^ n \to (0,1) \to \mathbb R $$ My attempt is the following:
Let $ \psi: \mathbb R ^ n \to (0,1) ^ n $ given by $$ \psi (x_1, x_2, \ldots, x_n) = \left (\dfrac {1} {1 + e ^ {-x_1}}, \ldots, \dfrac {1} {1 + e ^ {- x_n}} \right) $$ is clearly a bijective function.
For the $ (0,1) ^ n \to (0,1) $ function I used the following. As is well known, a rational number can have two decimal representations, for convenience we will stick with the one with infinite zeros. Let $ \phi: (0,1) ^ n \to (0,1) $ as: $$ (a ^ {(1)}, \ldots, a ^ {(n)}) \mapsto b = \phi (a ^ {(1)}, \ldots, a ^ {(n)}) $$ where we use the decimal representation of $ a ^ {(m)} $ given by $$ a ^ {(m)} = \sum_ {k = 1} ^ {\infty} \dfrac {a_k ^ {(m)}} {10 ^ k} $$ So, we construct: $$ b = \sum_ {k = 1} ^ {\infty} \left (\dfrac {a_k ^ {(1)}} {10} + \cdots + \dfrac {a_k ^ {(n)}} {10 ^ n} \right) \dfrac {1} {10 ^ {(k-1 ) n}} $$ And finally, $ g: (0,1) \to \mathbb R $ given by $ g (x) = \ln \dfrac {x} {1-x} $ is clearly bijective.
I have two things left to show that I don't know how to do them.
The first is to show that the function $ \phi $ is bijective.
The second is to compose the functions so that it is explicit.
Could you help me with those 2.