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Let $ z \in (0,1) ^ n $. We can express this "$ z $" as a $ n $ -tuple of real numbers $ z = (z_1, z_2, \ldots ,z_n) $, where each $ z_i $ will have its infinite decimal expansion.

We can write:

$\begin{align*}z_1&=0,z_{11}z_{12} \ldots z_{1n} \ldots\\ z_2&=0,z_{21}z_{22}\ldots z_{2n}\ldots \\ \ldots \\ z_n&=0,z_{n1}z_{n2}\ldots z_{nn}\ldots \end{align*}$

If we now take the decimals of this arrangement according to the Cantor diagonal method, we obtain $$y=0,z_{11}z_{21} z_{12}z_{13}z_{22} \ldots \in (0,1)$$ Having that function, how is it shown that this is a bijection? To have $ | (0,1) ^ n | = | (0,1) | $

Bernard
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James A.
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1 Answers1

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I suppose $(0,1)=[0,1[$.

The application is clearly a surjection.

If $y=0,z_{11}z_{21}z_{12}...=t_{11}t_{21}t_{12}...$, then you clearly have the equality of $z_1,...z_n$ with $t_1,...,t_n$.

The only problem (which is not specific to your function) comes from the fact that some numbers have two different decimal expansion : think of $0,099999...$ and $0,100...$ for instance, wich are the same numbers.

This is why we usually decide to take only one of these decimal expansion, for instance only the one that is not stationary at $9$.

Then since you have a bijection, you have that the two sets are (by definition) equipotent, which can be written $| (0,1) ^ n | = | (0,1) |$.

math
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  • Is that the only problem? In setting up a bijection from $(0,1)$ to $(0,1)^2$ what do you do with $0.101010...$? Do you send it to $(0.111...,\ 0.000...)$? – bof Jan 01 '21 at 11:59
  • Yes, but I feel you mean (0,1) meant $]0,1[$. Is that it? (I thought it was $[0,1[$) – math Jan 01 '21 at 12:07
  • Yes, that's what I meant. Personally, I'd advise just learning the Cantor-Bernstein theorem and not bothering with explicit bijections for this sort of thing. – bof Jan 01 '21 at 12:22