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I am trying to construct a bijection from $(0,1)$ to $(0,1)^{2}$, and I have come up with:

$$f(x)=f(0.a_{1}a_{2}a_{3}a_{4}...) = (0.a_{1}a_{3}..., 0.a_{2}a_{4}...)$$

But I think there's a problem with this function. Let $$x=0.494949... \Rightarrow f(x)=(0.444...,0.999...) =(0.444,1)$$ which is not in $(0,1)^{2}$. There are more similar problems with repeating decimals, for example: $$f(.010909)=(0,.199...)=(0,.2)=f(0.02)$$

I couldn't find how anyone fixed this problem (though I did see someone mention that it was fixable). Any help would be hugely appreciated.

Asaf Karagila
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i like math
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  • (The title of the question is different, but the answer is essentially the same.) – Asaf Karagila May 07 '21 at 21:27
  • @AsafKaragila Thank you! I must have missed it. – i like math May 07 '21 at 21:28
  • You can use the same idea with, eg, simple continued fraction representations of the reals, since this has some nicer properties than the decimal representation (uniqueness of representation mainly). – vujazzman May 07 '21 at 21:30
  • @vujazzman Thank you! One question: how do we know that the only possible different decimal representations of the same number always either end in 99999 or 00000? – i like math May 07 '21 at 21:52

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