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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.

asd asd
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    hint: try interleaving the digits of a tuple $(x_1,\ldots,x_n)\in (0,1)^n$ – Atticus Stonestrom Nov 28 '20 at 18:02
  • I edited my question. – asd asd Nov 28 '20 at 18:19
  • The problem with this approach is that it does not work. Say $n=2$ and you have to map a pair of numbers from $(0,1)$ into $0.90909090\ldots$. What would that "something" be? –  Nov 28 '20 at 18:25
  • That's why I said it, we will stick with the decimal representation that has only infinite zeros, so as not to have that problem. (For convenience) – asd asd Nov 28 '20 at 18:32
  • @asdasd So, what pair of numbers would map into $0.9090909\ldots$? If there is no such pair (because the first element in the pair has to be $0.999999\ldots$ - and you've already decided not to use those expansions), then it means the map is not "onto". –  Nov 28 '20 at 18:37
  • Oh I understand. Any help to fix that problem? – asd asd Nov 28 '20 at 18:41

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